(solution Manual) Contemporary Linear Algebra By Howard Anton, Robert C. Busby 452j32

0

a21

a2.2

£l23

0 0

0 0 0 0

asz

a33

0 0 0

a43

0 0

=

0 0

a34

0 0 0

a44

a4s

0

as4 0

ass a6s

aM

0

as6

is the sum of the row and column numbers. Thus the matrix is

: .....

+j

(b) The entry aii = (- l)i+i is - 1 if i

is odd and +1 if i

-: r (c)

+j

is even. Thus the matrix is

-: 1

- 1

1

We have a;j = -1 if i = j or i = j ± 1; otherwise a.;i = l. Thus the entries on the main diagonal, and those on the subdiagonals immediately above and below the main diagonal , arc all -l; whereas the remaining entries are all + 1. The matrL'<. is

1 - 1 1 1

33. The component,s of the matrix product

[! :

- 1 -1 - 1 -1

represent the total expenditures for 3 -

purchases during each of the first four months of the year. For example, the February expenditures + (6)($2) + (0)($3) == $17.

were (5)($1)

34. (a)

45 + 30 6030 ++23 33 7f40:> ++2540] [75 93 115] 51 53 65 T he entnes of the matrnc M + J = = 65+12 45 + 11 21 77 56 [ 10 + 10 35 + 9 23 50 44 .

.

30+ 21 12+ 9 15 + 8

represent t he

total units sold in ea.ch of the categories during the months of May and June. For example, the total number of medium raincoats sold was M3 2 + J32 = 40 + 10 = 50. (b) The entries of the matrix

M - J

15 27 35] = : [ 1 30 26

represent the difference between May and

June sales in each of the categories. Note that June sales were less than May sales in each <:Me; thus

e ntries represent decreMP.S.

61

Discussion and Discovery

(c)

r:]· Then th7 wmponents of M

Let x

l

x

::] [:] =

= 15

<10

35

represent the totalnum90

ber (all sizes) of shirts, jeans, suits, and raincoats sold in May. (d ) Let y

= 11

1 1

1]. Then the components of 45

yM=[l 1

1

30

60

75]

30 = 1102 195 205J 12 65 [ 15 40 35

1)

represent the total number of small, medium, and large items sold in May.

The product yMx = [1

(e)

1

= {1

1

=492 represt>nts the

1

15 40 35

90

total number if items (all sizes and categories) sold in May.

DiSCUSSION AND DISCOVERY Dl. If AB has 6 rows and 8 columns, then A must have size 6 x k a nd B must have six k x 8; thus A has 6 rows and B has 8 columns. D2. If A

= [o1 0 o]' then AA = [o1 0o]1[o 0o] = (o00o].

03. Let A

=G

and B

=

In the following we illustrate three different methods for computing

the product AB. Met.'wd 1. Using Definition 3. 1.6. This is the same as what is later referred to as thP. column rule.

Since

we have AB

= [:

Method 2. Using Theorem 3.1 .7 (the dot product rule}, we have

Method 3. Using the row rule. Since

rt(AB) = rt(A)B = 11 we have AB = [:

.

2]

[!

= [7

Sj and r2(AB) = t2(A)B

= [1

lj

""' [4 3]

62

Chapter 3

-i .

D4. The matnx A

Since xc, (A) + yc,(A)

property. Here is a proof

A[:] , we must have xc, (A)+ yc,(A) + zc,( A)

+ zc3 ( A)

all x, y, and z. Talting x and c3 (A) =

is the only 3 x 3 with

1, y = 0, z = 0, it follows that c, (A) =

[=

for

Similarly, c 2 (A) = [

-i]

m.

D5. There is no such matrix. Here is a proof (by contradiction):

Suppo"'

A;,. 3 x 3 for which A[:] ["f] for all x, y, •nd z. Then A[:] ·

hand , we must have A [:]

D6. (a)

S1

=

and S 2 =

(b) The matrices S (c)

=

= -A [

l

=:

=-

m. Thus there

on the other

is no such matrix.

are two square roots of the matrix A= and S

=

[±:s·

are four different square roots of B

=

Not all matrices have square roots. For example, it. is easy to check that the matrix [has no square root.

07. Yes, the zero matrix A =

ha.o:; the prop erty that AB has three equal rows (rows of zeros)

for every 3 x 3 rnatrix B .

DS. Yes, the matrix A =

has the property that AB

= 28 for every 3 x 3 matrix B.

09. (a ) False. For example, if A is 2 x 3 and B is 3 x 2, then AB and BA are both defined. (b ) 'frue. If AB and BA are bot h defined and A ism x n, then B must ben x m ; thus AB is m x m and BA is n x n. If, in addition, AB + BA is then AB and BA must have the same size, i.e. m = n. (c) True. From the column rule, c 1 (AB) = Ac;(B). Thus if B has a column of zeros, then AB will have a column of zeroll.

(d) False. For example, (e) (f)

=

True. If A is n x m , then AT ism x n, AAT is n x n, and A T A ism x m. Thus AT A :1.nd AAT both square matrices, 8Jld so tr(AT A) and tr(AAT) are both defined. False. If u and v are 1 x n row vectors, then u T v is ann x n matrix.

Exercise S&t 3.2

63

DlO. The second column of AB is also the sum of the first and third columns:

s

Dll. {a) :L)aikbkj)

= a,lblj

+ a;2b'2j + · · · + aisb.•i

.1.:=1

(b) This sum represents (AB)ij, the ijth entry of the matrix AB.

WORKING WITH PROOFS is an s x n matrix andy= !Yt

Pl. Suppose B =

Ys]· Then yB is the 1

yz

yB = fy • c1(B) y · c2(B)

Xn

row vector

Y · Cn(B)]

whose jth component is y · Cj(B). On the other hand, the jth component of the vector

Ytrl (B)+ Y2r2(B) + · · · + Ysr s(B} is Ytblj

+ Y2b2j + · · · + Ysbsj

= y • Cj(B). Thus Formula (21) is valid.

P2. Since Ax= x 1 c 1 (.t1) + x 2 c 2 (A)

+ · · · + XnCn(A),

X1C1 (A)+

X2c2(A)

the linear system Ax= b js equivalent to

+ · · · + XnC,l(A) = b

Thus the system Ax == b is consistent if and only if the vector b can be expressed as a linear combination of the <.:olumn vectors of A.

EXERCISE SET 3.2 1.

(a)

(A+ B)+ C =

A+ (B r C)=

(b)

(AB)C

=

[28

flO0

l2

[j

(c)

-31 -21

u

(a+ b)C = (-3)

aC+bG

=

[

12

-2]7 + [01

5 .6

10

3

-5 8

4j = 9

36

3

3]5 [-187

1

4

11

15

!]

-6 12 -1

19

5

[-10 = 83 87

5

-222 -·67 33

-33] = [-10

1:]

26]

278 240

22

83

-222 -67

38

87

33

-9

-21 -15

240

-12 -27

28

14 6 -21] [ 0 12] [ 0 16 + -7 -49 -28 = -3 -21

20

36

-8

26]

278

6 -9]

-2 7 5

1!] 19

=

+ [: 7 -2

-62 17 -27

1 5

-6 12 -1

-2]6 [101

[01 -27

r1

[!

7 f.)

4

4

31 [10

-2

-1

-28

-1

A(BC) =

-4

-21

-35

-63

-9

-15

-9]

-12

-27

Chapter3

64

=

(d) a(B - C)= (4) 1 - 12 - 3

aB - aC

=

4

- 48

32 - 12 -20] [ 0 -8 0 4 8 4 28 [ 16 - 28 24 12 "20

- 12

=

4 - 48 -12

36

(r -3 -51 r m [-18 -62 -132] [-72 -2

2. (a)

a(BC) = (4)

= (4)

(aB)C =

0 4

(b)

(B + C)A

2 6

7

17

11

- 27

7 5

1 3

22 38

28 44

=

- 248 68

88

- 108

152

[320 -124 -20]8 [01 -27 3] [-72 4

16

B0C) =

1 - 7

- 28

24

3

28

=

9

5

-!32]

- 248 68 - 108

44

88 152

r8 -3 -5 [ 0 -8 1162] = r-7228 -248G8 -132] 88 28 0

L4

=

1 -7

2 6

1

4 12- 20

36

44

-108

152

20 -14 3] [ 20 -30 -9] [8 -5 -2][ 6 1 7

5

8 -2

15

-2

=

4

1

- 10

37

()7

-16

0

71

[26 -256 -11] [ -6 -5 2] [- 2010 BA +CA = -4 13 + - 6 31 54 =

-4

3.

(a )

(AT)T

-26

1 4

5

( b ) (A+ B)r =

riO0 L2

AT

(c) .

(3C)T

-12

1

H _T [2 04

-6 21 15

-r

-4

10

0 -2

1

4

4

T[0

12 27

=

3

-6 21 9 12

37 0

-9] 6i 71

=A

0 5 7

-2

-5

- lfi

70

!3]

4

-21] + [-38

0 5

-1

= [10 -4

7

5 -6

H

=

26

- 30

0 1 2

10

-l 6

1:] =(3)

27

05 -62] -2

7

1 7 3

4

10

!] 3C'

Exercise Set 3.2

65

20 0]

- 31 -21

T =

-2r 36

6

2

3

H-1-8]

4. (b)

-6 - 12

-2 - 3

T

=

6

38

5

4

[

6

2

-18

6

17 -27

ll

22 38

(a)

4

9

-5

,, ( [ _

(b)

, ,(3.4)

,, ( [

(c)

tr(A)

10, tr(B)

nnd tr(A+

(d) tr(AB)

tr (

and tr(BA) = t.r

=

r-18

-62 - 33

- 8

-3

1 22

38

=

0 2

6

-33

22

38

: ]) '' 2 +1 +I = 10, and

lr ( [

t r(A')

T

1

cr Br ,... 5

-2

-6 - 12 - 2 -3

3 4 9

-33

-62

7

:1

36

=

-5

[

38

8-1 1]

-1 -8

sr- cr = (d) (BC)'i' """""

36

-m :m m_! -m

= 2 + 4+4 = 10

;

6 + 12

tr

+ 12 = 30

= 8 +I

n "' :::

tc (

+6 = 15,

25; thustr(A

28-31 + 36

26

-4 ( [ -4

- 25 -

6 -26

ul ) =

13 1

3tr(;lj

tr(A ) +tr(B )"

= 33,

26 + 6-+ I

= 33; t.hll.s tr(AB) = tr(HA) .

Chapter 3

66

6 6. (c)

Lr(A - B) = tr

= -6 + 3-2 = - 5 = 10 - 15 = tr(A)- tr(B).

0_

[ -6

8 -2

-18 (d )

= tr

t.t (BC)

([

- 27

J

!!])

12 - 23 -24 ( [ 60 - 67

and tr(CB) = tr

7. (a)

38

= -1 8 + 17 + 38 = 37,

24

12 - 24 + 49 = 37. T hus h(BC)

A matrix X satisfies the equation tr(B)A

l<(CB).

+ 3X = BC if and only if 3X = BC -

tr(B) A, i.e.

8 -3 -5] [0 -2 3] [ 2 -1 [ !]

3X = 0 1 :1 - 7

=

-18

2 6

--62 17 - 42

7 [ 11

in w},ich case we have X

=

. .

(b)

1 3

4 -15 0 9 -2

7 5

-33]

[ 30

22 38

-

-- 15 60 15

0 -30

4 1

45] = [-48

75 60

7 41

-78] -53

-47 -47 - 42

-22

-48 -47 -78] 1 -47 -53 [ 4 1 - 42 -22

A ma trix X satisfies the equation B +(A+ X)T

=C

tf a.ud only if

(A +X)T =C - B A +X

= ((A+ X)T)T = (C - 8)7

X= (C- B )T- A = C T- BT - A

Thu:; X

8. (a)

9.

X

0

1

= -:'.

7 4

= B -2

r

=

2

35] - [- 83 0 - 74] - [ 20 -1 4

v

9

4

1

- 5

2

6

-2

-92

- 167

10

-282

A-1 = rl - 52

(b)

{A-t)-1

G

(d ) 2A = [

6 10

[-!0 2 -4} 1 10

2 1

7 . - 1

(b) X = CT- BT = _!_

(a) dPt.(A) = 6-5= 1

(c) A'l' =

1

-·

, det(AT)

=1

,det(2A)

=4

10

8

-

= [3

s

(AT)- 1 = [ 2 - 1

4 4 -10

= (A - l)T

-2]6 = 2 [ -52

=

Exercise Set 3.2

10. (a)

67

(b) det(B-1)

(c) BT

=[

= _!_

+ 12

= 8

202

(B-I)-1 =

20

T

2 :] , det(BT) = 20

(B )

-a

(d) 3B = [

ll. (a)

1 [ 4 s-1_20 ·-4

det(B)=8+12=20

(AB)

-9]

6

12

-1

12

, det(3B)

= 180

( 3B)

-i

-1

=

-3]

20 (_!_ [2 20 4

-•!])

1 [4

= (IJ-l)T

20 :;

1 [ 12 -12

= 180

4 -::::B

=

9]6 = 601[-44

[10 -5]-l 1[-7 5]

=

18

-7

s-lA-• =

= 20 -18

[_:

10

;o

= -l

(b) (ABC)_ 1

12 _ (a} (BC)_ 1

.,.-

.

_

[

-

(C _B)-I

2

11] -l

12

32

6 20 -4

= _!_ [

12 10 -16

36

2

=

22

3

-5

40 -122

-11] 18

12 -111 18J

= __!__ [

40 -16

= _1 [ 240 --32

-33]

r- -:]

aG

=

[

[-5 -7] [10 -5] = [-88

- 29

=

AB

-·45] 70

-4] _!_ [ 4 3] [ 2 -1] = l [ 79 -45]

= [36 33] - I

u-lc-1 n-1

14 . X=

2

__ 1 [ 79 - 40 -122

r-12 -4]6 20_!__ [ -44 3]2

2

·

45] 79

[18 16

c-ts-1 =! (b) (BCD)_ 1

70 122

6

4

18

- 7

[._:

__!__

11

66

37]

70

=

Chapter3

68

-24]

16. (a)

67

(b)

6 -14] 61[3o o]2 = sl-3 1r 6 -14]

= [2 19. {a) (b)

20. ( a )

Given that A - 1

-=

Given that (7A)- 1

=

Given (I+ 2A)A :-::

)

5

2

1[-5-1 2]1 21(13

-

[-3 -

1] 2

=

1 13

we have I+ ZA

=

22. The matrix A = [

l .

-

=

1 13

[_!

1 __

(-l)

r-:

:;

_31] __

::;.

·-.1>], ._,

and so it follows that

c f. 0, i.e. if and only if c :/:. 0, 1.

- :) is invertible if anJ only if dct(A) = -c2

+1

0, i.e. if aud only if c =/= ±L

In genera), any matrix of the form [: 3 2 3J

e

24. One such example is A =

;] . J

c

In general, any matrix of the form 3 -2

3x3

1

2

23. One such example is A =

25. Let X= [xiJJ

) ··

[Io OJ) = 131 [-9-cl]'

is invertible if and only if det(A) = c2

21 . The matrix A =

5

3

=

1

= T =

3

1

13, we have A = (A-

have A =

[-3 -1] it foll ows t hat 5A ·-lJT =-51 [-2 5) . -)

, (,ivc'? that ( 5AT) - 1 = and so A = -1 [-:.!. 5 5

(b)

1

noting that det( A-

. Then AX= I if and only if

'] ['"

0 1 0 1 1

;]. -e - !

0

X}2

xnl

"lJ

X21

X22

X23

X31

X32

X33

1 0

j

0

69

Exercise Set 3.2

i.e. if and only if the entries of X satisfy the following system of equations; -!-

Xn

=}

X3 1

+ X32

=0

+

X:tJ

+ X:z t

=:

0

=0

+ X22

=1

+ X23

Xt3

=0 =0

xz1

=0 1

+ X33 = This system has the unique solution

xu 1

x;12 = - !·Thus A is invertible and A- =

26. Let X =

lx,iJ 3X3 . Then AX = I

=

:t12

= xzz

= x23 =

'2

-- 2

'l

[-! -!]·

X31

= X33::;;

and

x 13

= x2 1 =

if anrl only if

2 0

l

X13]

[:::

X23

=

x:n

XJJ

[}0

0 1 0 0

()] 0 l

i.e. if and only ii xu = 1, :tu = 0, x13 = 0, :t21 = 0, xn = 1, r23 = 0, 2x11 + X31 = 0, 2x12 + X32 0, and 2x 13 + X33 = l. This is a very simple syst.em of equations whicb has the solution Xll

=- 1,

X12

= 0,

X13

= 0,

X21

= 0, X22

= l,

X23 =

0,

X31

= - 2,

X32 =

0,

Thu& A is invertible and A - t = [ - 2 0 l

28.

(a)

The mat
3)

-3 9] = [- 3 2 2 -6 ha;: the property that 4 -4

12

-3 2

-4

X33

= 1

=

70

Chapter 3

HJ. ([_! ; [-:]) HJ.

(b) u· Av

ATU· V

=

(U -:][-;)) HJ

29. U A is invertible and AB

is

50

HJ

-1-12+63

= AG then, using the Associative law (Theorem 3.2.2(a)). we have

and

(A-

1 30. If A invertible AC = 0, then C = IC = A)C = A- 1 (AC) = A- 1 0 = 0. 1 is invertible and AG = 0, then A= AI= A(CC- ) = (AC)C- 1 = OC- 1 = 0.

31.

(a) If A=

::::] , then det(A} = cos2 A_ 1 =

()

+ sin 2 B =

[cosO

Similarly, if C

1. Thus A is invertible and· :

- sin fJ] cosO

sinO

for every value of ().

(b) The given system can be written

i.e. x'

so

= xcosO- ysin8 andy'= xsinO + ycos£1.

32 .

(a) If A 2 :o:: I. t hen (1- A) 2 = 12 - 2A + A 2 = 1 - 2A +A = I - A ; t h us J - A is idempotent. ( b) If .42 = / , then (2 A - 1 )(2A- I)= 4A 2 - 4.4 + ! 2 = 4A- 4A + I = I ; thus 2A -I is invertible anci (2A - 1)- 1 = 2.4- I.

33.

(a) If A anc.l B are invertible square matrices of the same size, and if A + B is also invertible, then

A(A- 1 + s- 1 )B(A + B)- 1

:::

(I+ AB- 1 )B(A

+ B)- 1 =

(B

+ A)(A + B)- 1 =I

(b) From part( a) it follows that A(A- 1 + B - 1 )B;;::. A+ B, and so A -t + B - 1 = A- 1 (A+ B)B- 1 • Thus the matrix A- 1 + B- 1 is invertible nnd (A- 1 + B- 1 ) - 1 = B(A B) - 1 .4 .

34. Since u Tv = u · v:::: v · u = vru, we have (uvT) 2 (uT v )uvT. T hus, if u Tv =f - 1, we have

an(

= (u vT)(uvT ) =

so the matrix A = I+ u vT is invertible with A - 1 = I -

u (vTu)vT

vuvT.

= (vTu)uvT =

71

Discussion and Discovery

and p(x) = x 2

35. If A = [:

(a+ b}x +(ad - be) , then

-

!]

2

p(A) =

=

- {a+ b)A + (ad - bc)I

+ (ad- be)

- (a+ b) [

2

= [a2+bc ab + ba1_[a J-da ab l·db]+(ad - bc)(l 0] ca + de cb + d2 J ac + de ad + d2 0 1

36. From parts (d) and (e) of Theorem 3.2.12, it follows that

tr(AB - BA)

= tr(AB)- tr(BA) = 0

for a.ny two n x n matrices A and B. On the other hand, tr(fn ) = 1 t 1 + ·· · + I ·= n . Thus it is not possible to have AB - B A= I.,.. . 37. The adjacency matrix A and its square (computed using the row-column rule) are as follows:

s

1 1 0 0 0 0 0 1 l 0 0 0 0 l 0

0 1 0 0 0 0 0 0 0 () 0 0 0 0

4-

0 0 0 0

0 0 0

Az

0 1 1 0 0 1 0 0 1 (} 0 0

0

=:

0 0 0 0 0 0 0

0

0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

1 3 1 0 0 2 0 0 0 0 1 0 0 0 0 0 0 0 0 0

The entry in the ijth position of t he mat rix l1'2 represents t he number of ways of traveling from i to j wit h one intermediate stop. For example, t here are three such ways of traveling from J to 6 (t hrough 2, 3 or 1) and two such ways of t raveling from 2 to 7 (t hrough 5 or 6). In genern.l t.h€ entry in tjth position of the matrix An represents the number of ways of traveling from i to j with exactly n - 1 int.errnediatc s tops .

DISCUSSION AND DISCOV ERY 01.

2

and B =

(a) Let

the other hand, (A + B)( A-

(b) (A + B)(A - B) = A 2 (c) (A + B )(A - B ) = A 2

-

-

(a) If A 2 + 2A +I= 0, ihcn i A-

=-A - 21.

A2- B

2

=

On

=

+ BA

- B2 8 2 if and only if A B AB

0

1

B =

G

D2. If A is any one of the eight matrices A =

D3.

2

Then A =

= - A2

-

;J 0

= BA. A2

= f.:!.

±1

2A = A(-A- 21) ; thus A is invertible and we have

Chapter 3

72

(b) If p(A) = anA"+ a11 _ 1 A"- 1 + · · · + a 1 A

A(-

a .. An ao

1-

+ aol = 0, where a.o =I 0, then we have

an-I

ao

Thus A is invertible with A-t= -!!.li. A"- 1 '

A" - 2 -

--

oo

.. . -

al ao

J) = I

. .. -

ao

!!.!.[ .

oo

D4. No. First note that if A3 is defined then A must be square. Thus if A 3 = AA2 =I, it follows that A is invertible with A - I = A 2 . D5.

(a) False. (AB) 2 = (AB)(AB) = A(BA)B. If A and B commute then (AB) 2 = A2 B 2 , but if B A ¥: AB then this _will not in general be true. (b) True. Expanding both expressions, we ha.ve (A - B)'J. = A 2 - AB-BA + B 2 and (B- A)Z = 8 2 - BA - AB + A2 ; thus {A- B) 2 = (B- A) 2 • (c) True. The basic fact (from Theorem 3.2.11) is that (AT)- 1 = (A- 1 )7', and fr0m this it follows that (A - n)T = ({A") - 1 )T = {(A")T)- 1 = ((AT)n)-1 = (AT) - n. (d) False. For example, if A= (e)

D6.

(a)

and B =

G

then u (AB)

= tr

= 3, whereas

tr(A)tr(B) = (2)(2) = 4. False. For example, if B = -·A then A+ B = 0 is not invertible (whether A is invertible or not). If A is invertible, then the system Ax= b has a. unique solution for every vector b in R3 , namely x = A- 1 b . Let x 1 , X:t, and x 3 be the solutions of Ax= e 1 , Ax= e 2 , and Ax= e3 r espectively, and let B be the matrix having these vectors as its columns; B = !x, x2 x3}. T hen we have AB = Ajx 1 X 2 x3] = [Ax1 Ax2 Ax3] = !e, ez e3] =I. Thus A- 1 = B = [x1 Xz x3].

(b) From part {a) , the columns of the matrix A- 1 are the solutions of Ax= e1. Ax= e 2 , and Ax= e3 . The augmented matrix of Ax = e 1 is

[i H a.nd the reduced row echelon form of this matrix is 1

00 -13]

0

1

0

D7. The matrices

G

and

0

n

have determinant equal to 1. The matrices

have determinant equal to -1. These six matrices are invertible. The other ten matrices6U have determinant equal to 0, and thus are: not invertible. DS. True. If AB

= BA, then

n-•A-l = ( AB)- 1 =

(BA) -

1

= A - 1 s- 1 •

73

Working with Proofs

WORKJNG WITH PROOFS Pl. We proceed as in the proof of part (b) given in the text. It is clear that the matrices (ab)A and a(bA) ha.ve the same size. The following shows tha t corresponding entries are equal:

P2. 'The following shows that corresponding entries on the two sides are equal:

P3. The argument that the matrices A(B- C) and AB- AC must have the same size is the sa.rne as in the proof of part (b) given in t he text. T he following shows that corresponding column vectors are equal: cjiA( B - C)J = Acj(B- C) = A(cj(B)- Cj(C)) = Ac1 ( B)- Aci (C )

= Cj (AB)- c1 (AC) =

cj (AB - AC )

P1 . T h('.se t.hree rn<\t r ices clel\rly ha.ve t he same s ize. The following shows t ha t corresponding column vectors are equal:

= (aB)cj(C) = c; [(aB)CJ a(Bc;(C)) = B(acj(C)) = B(cj (aG)) = C j [B(aC)]

cj ja(BC)] = ac1 (BC) = a(Bc1 (C)) Cj!a(BC )J

= acj ( BC) =

P5. If cA = 0 a.nd c :;f 0 then, using Theorem 3.2.l(c)), we have A

= 1A =

((

Dc) A =

(cA) =

0 = 0.

P6 . (a) Tf A is invertible, then AA- 1 =I and A - 1 A = ! ; thus A- 1 is invertible and (A- 1 ) - 1 =A (b ) . If A is invertible t hen, from Theorem 3.2.8, A 2 is invertible and (A2 ) - l = A- 1 A - 1 = (A- 1 ) 2 . It t hen follows that A 3 is invertible and (A 3 ) · · l = (A2 ) - 1 A- 1 = (A .. 1 ) 2 A- 1 = {A- 1 ) 3 , et c. In general, from the remark following T heorem 3.2.8, (An ) - 1 = A- 1 • • -A- 1 A - 1 = (A- 1 )n. P7.

(a)

AA- 1

·

1

A- A

·"

a [c

=(

b] {

d

d

1[ \ad .... be -c

1

[

-bl) = --·a

l [ad · · be (t
d ...ab]) [f'cl db'J = ad -1 be [da-0 be

ad - be - c

(b) Let A =- (: ;] where ad - be = 0. Then A is invertible if and only if !.here a.re scalars e,

f,

g, and h , such Lhnt

[: !] (;

=

i.e. if ami only if the followhg system of equa •.ions is consistent:

ae + bg ce + dg

= 1

=0 af + bh = 0 cf + dh = 1

Multiply t he fi rst equation by d , multiply t he second equat ion by b, and subtract . This Jea.ds to (da - bc)e =- d

74

Chapter3

and from this we conclude that d = 0 is a necessary condition in order for the system to be consistent. It then follows (since ad- be = 0} that be= 0 and so either b = 0 or c = 0. Let us assume that b = 0 (the case c = 0 can be handled similarly). Then the equations reduce to ae = 1 ce = 0 af = 0 cf = 1 and these equations are easily seen to be inconsistent. Why? From ae = 1 we conclude that e f. 0; and from ce = 0 we conclude that c = 0 or e = 0. From this it follows that c must be equal to 0. But this is inconsistent with c/ = 1! ln summary, we have shown that if ad- be = 0, then the system of equations has no solution and so the matrix. A is not invertible.

n

tr(AB) =

L k =: l

n

=

L

n

n

=

k=l / "" 1

n

n

L L btkakt =

L(BA)u

1=1 k = l

1=1

= tr(BA)

EXERCISE SET 3.3 This matrix is elementary; it is obtained from J2 by adding -5 times the first row to the second row. (b) This matrix is not elementary since two row operations are needed to obtain it from I 2 (follow the one in p art (a,) by interchanging the rows). (c) This matrix js elementary ; it is obtained from h by interchanging the first and third rows. (d) This matrix is not invertible, and therefore not elementary, since it has a row of zeros.

1. (a)

2. (c) is elementary. {a) , (b) , and (d) are not elementary.

3. (a) Add 3 times the firs t row to the second row. {b) Multiply the third ro•.... hy ! . (c) lnt.erchange the first and fourth rows. (d) Add times the third row to the first row. 4. (a) Add -2 times the second row to the first row. (b) Multiply the first row by - 1. {c) Interchange the first and third rows. (d) Add -12 times the second row to t.he fourth row.

5. (a)

(c)

f

(b)

= 0 0 1 0 0 1

0 0

W'

=

0 0 1 0 0 l 0 0

{d)

0 1 0

= 0 1 0 0

I

- 7 0 1

0

r

0 OJ

0 1 0 0 0 3!

r

==

0 1 0 I\

v

l

7

0 1 0

EXERCISE SET 3.3

75

=

(b) 0

(d)

7. (a)

:: 1rr1r

0

1

0

j !r

fmm A by interchanging the first and third rows; thus EA

0

1

J

0 0 1 0

B where

(b) EB =A where E is the same as in part (a).

A by adding -2 times the first row to the third row; thus EA =

(c) :; :

c where

-2 0 1

(d)

EC = A where E is the inverse of the matrix in part (c) i.e. E 1

= 2 0 1

8.

(a)

E

=

(b)C

0l 0]2

(c) E = 0

0

(d)

1

!

9. Using thr. method of Example 3, we start with the partitioned rr.atrix lA I I] and perfonn row operations aimed at reducing the left side to I; these same row operations performed simultaneously on the right ::;ide will produce t.he matrix A -J. 1 [2

5 I 1 0] 20: 0 1

Add -2 times the first row to the second row.

5 : 1 0] 10 l -2 1

1 [0 Multipiy the seconci row by

then add -5 times the new second row to the first row.

1 0! 2 [0 11_! Thus A -l

= [ _ 12 5

det(A)

= 10

1

_!]1 .

I

5

.L

10

On the other hano 1 using the formula from Theorem 3. 2. 71 ami the fact

10

we have A- l

21) = 101 [ -2

-5]

1

-l]

= [ -g12 iO? ·

Chapter 3

76

10. 11.

A

-1

(a)

1 [ 1 -4

= 14

3]2

;=

l

[

-. -7

Start with the partitioned matrix [A

I I J. 0 1 0

- 1 l1

4

I I I

J

0

0 5 - 4l0

[! Interchange rows 1 and 2.

0 4

3l0

1

-1 : 1

0 0

I

5 - 4l0

Add - 3 times row 1 to row 2. Add - 2 times row 1 to row 3.

3:o

1

4 - 10:I 1 5 -10: 0

-3 -2

0

Add - 1 t imes row 2 to row 3.

3l

0 4

0 1

-3

0: -1

1

- 10:

I

1

1

Add -4 times row 3 to row 2, then interchange rows 2 and 3.

Multiply row 3 by -

1 , 10

0

3:

l

Q

0

' I

--1

0 - 10:

5

I

1 0] __

:

then add -3 times the new row 3 to row 1.

0 1

0 :I 0 : -1

_!l 10

5

1

1

: _l

7 10

S

I

2

2

11

-TO

From this we conclude that A is i nvertiblc , and that A-

1

-

[

l 7

10

(b)

Start with the partitioned matrix [A

'2

6

I IJ. 3 -4

-4

-

4 2

I

-9

l 0 0

0 1 0

Multiply row 1 by -1. Add - 2 times the new row 1 to row 2; add 4 times the new row 1 to row 3. 1

-3

0

10

[ ()

- 10

0 1 0

77

EXERCISE SET 3.3

Add row 2 to row 3.

[

1 0

-3 10

0

0

0

1 1

At this point, since we have obtained a row of matrix .4 is not invertible.

(c) Start with the partitioned matrix

on the left side, we conclude that the

!A I Jj. 1:1 0 1 :I 0 1

0 I

o:o

1

0

Add -1 times row 1 to row 3.

fl

0 1 1

1 :

1

j

0

0 1

1

0

I I

I

-1: -

Add · 1 times row 2 t o row 3, then mult ip ly the new row 3 by

1 1 1

0 1 0

j]

0 1

1

0 1

l

2

-1.

2

Add -1 times row :3 t.o rows 2 and 1.

o: l 0: - 4

0

I

1

I

1

0

I I

2

l

-2 1

2

I

I

2

2

.

-ll

r

I

-2

f\·, "" t h ;, weconchtdet hnt ,\ ;, iot •«li ble , nod that A·· ' = [-;

I

2 I

'i

12.

(a)

A

-l

(b)

A-

1

l

=

[0

0 0]

1 - 1

- 3 4

(c)

-ll A-

1

1

--1

= [0

0

0

Ol !. 3

13. As in t he inversion algorit hm, we s tart wit h the po.rtitioncd matrix [A I I J and perfo rm row operations a.irr..;)d at redudng left. side to its reduced row echelon form R. If A is invertible, then R will be the identity m atrix and the matrix pro duced on the right side will be A - l . In the more general situation the reduced matrix will have the form [R I B} = [BA I BIJ where the matrix B on the right has the property that B.1 = R. [Note that B is the product of ejementary matrices and thus i" always an inve rtible matrix, whether A is invertible or not .]

1 2 0 0 [1 2

3[1 0 0] !0 l 0

1

4:0 0 1

Chapter 3

78

Add -1 Limes row 1 to row 3.

3:

1 0 1I 0 0 1 1 -1 2

I I

0 1

0

Add --1 t imes row 2 t o row 3.

3:

2 0

1

I I I

1 0

0 : -1

0

0 1 - 1

Add - 3 times row 2 to row l.

0:

2 0 0

The reduced row echelon form of A is R pro perty that B A

0

14.

R=

1

0

=

1 -3 1 - 1

I I

0 I 0: -1 }

20] = [) 0 0 1 , 0 0 0

B=H

1

has the

- 1

R.

0 0

u

-3

and the matrix B =

0] l

8

1

15. 1f c = 0, t.hen t.hc row is a row of zeros, so the matrix is not invertible If c f- 0, then after rnult.iplying the fir5t row by 1/ c, we have:

Add -I timE'-'> the tlrst row to t he second row and to the t hird row. ]

1

0 c- 1 [ 0 0 If c = 1, then t he second and third rows are rows of zeros, and so t he matrix is not invertible. If c :j: 1, then we can divide the second and third rows by c - 1 obtaining

l :] from this it is clear that the reduced row echelon form is the identity matrix. Thus we conclude that the matrix is invertible if and only if c :f:. 0, 1. 16 . c ¥= 0,

±v2

17. The matrix B is obtained by starting with t he identity matrix and performing the same sequence

of row operation' ; thus B =

[!Hl

=

:

:J

EXERCISE SET 3.3

18

79

B

19. 1f any one of the ki's is 0, then the matrix A has a 1.ero row and thus is not invertible. If the kt.'s are all nonzero, then multiplying the ith row of the matrix [A I IJ by 1/ki for i = 1, 2, 3, 4 and then reversing the order of the rows yields 0

0

0

k3

l

k2 0 thus A is

and A -

20.

k

=r' 0;

A-

1

['

l

-p

=

p

1

(a)

1

k

\

-1?1

I

The identity matrix can be obtained from A by first adding 5 times row 1 to row 3, and then

1

(h)

A

(c)

A = E}

(a )

is the matrix in the right-hand block above.

·- k1

multiplying row 3 by

22.

0

0

k p

F

21.

I

I

1

0

0

0

k

l

l

•

and E2 =

Thus if E 1 =

;] , th€n

E 2 E 1 where £ 1 and E 2 a.re as in part (a) . 1

E2

1

=

:anT::,

then

[rof .: ::''.ad[t'fg 0 0 1

(b)

E2E1 A = I.

A

0 0 2

A -I = E2 E 1 where £ 1 and E2 are as in par t (a).

(c)

23. The identity ma.Lrix can be obtained from A by the following sequence of row operations. T he corresponding matrices and their inverses arc as indicated .

0 (1) Interchange rows 1 and 3.

E1=

s-1_ l -

1

0

0

(2) Add -1 times row 1 to row 2.

£2

=

0 1

0 l 0

H

ol E-I_ 2 -

0 1 0

80

Chapter 3

{3) Add - 2 times row 1 to row 3.

E,

{4 ) Add row 2 to row 3.

(5) Multiply row 3 by

[

0]

0 1 0 -2 0 1 0 1 1

E,

1

Ei'

j]

1 0

Es

0 1 0

0

-i·

01. 0]0

s;•

-1

E;'

]

0

0

(6) Add row 3 to row 2.

E,

:]

1

[:

0

0 1

E ()- t -_

E,

(8) A
It follows that A- 1

1

0

Ea =

LO

Ei'

-H

0

()

(7) Add - 2 times row 3 to row 1.

j]

0

[:

0 1 0

- 1

1

l

1 0

0

= EaE1 E6EsE4E3E2E1 and A= E} 1 E2 1Ej 1 Ei'£5'Ei 1E7 1E8 1 .

25 . The two systems have t he same coefficient matrix. If we augment this m:\trix with the two columns of cons tants on the right sides, we obtain

1 2 1!-1:0] [ z: 1

3

0

1

21I

3•0 I

4 l1

the reduced row echelon form of this matrix is

0

0

!-:

!-:]

t: o:

4

From this W P. conclude that the first system has the solution x 1 second system has the solution 4, x2 = -4, xa = 4.

= - 9, x = 4, :r = 0; 2

3

and the

81

EXE RCISE SET 3.3

26. 'The coefficient matrix, augmented with the two colUJnns of constants, is

2 5: -1: 2] l

-3

0

I I

-

I

1 :

I I I

1 3 :- 1 }

and the reduced row echelon form of this matrix is

1 0 0 l - 32 ! 11] 0 1 0 8 -2 [0 0 1 l 3 : - 1 Thus the first system has the solution x 1 = -32, x 2 = 8, x 3 = 3; and the second system has the solution x 1 = 11, x2 = - 2, X3 = -1. I

I

I

27.

I

(a) The systems can be. written in matrix form as 2 3 1

:] [::] n1 ::0

The inverse of the coefficient matrix A are given by

=

and

[" :] and

28.

(

b,]

[- :

2

1

- 1

-1' ] [ - 3J l

4

-l

1

H

Hl

2

- 1

- 1

-3

m

-32 - l'] . Thus the solutions

3

[-: -:] [-!] n1 (b) A- 1 !b 1

:::

]

-3 2

2 3

0] = [-9 -:4] .

The systems can be written in matrix form us

nod

Tbe inverse of the coefficient matrix tions are given by

[

-:]

J

82

29

Chapter3

0

The augmented matrix consistent if and only if

30. The augmented matrix

if and only if b1

= 0, i.e.

b1 -

btl

1 -2 5: 4 -5 8: b2 [ -:1 3 -3 I b3

0

[

= 2b2. -2 3

+ bz + b3 = 0.

3

2!

-4

7

4 , b3

system is consistent if and only if 2b1 - b2 + b3

0

= Oo

0

T hus the system is

6:

bz- 4bl 0 I - I> 1 + bz + b3

!

-2 -1 : -1 o

0

l

bl

- 12l

o

l -2 -llbll&a can be row reduced to [1o

-2

2b:z)

bt

can be row reduced to

the system is consistent if and only if -b1 31. The augmented matrix

l


bl

t 2b1

0 I 2bt - b2

l

+ 113

.

0

Thus

Thus the .

32. The augmented matrix for this system can be row reduced to

1 -1 3 2 0 1 -11 - 5 0 0 0 [0 0

0

0

0

and from tohis it foll ows that the system Ax = b is consistent if and only if the components of b satisfy the equations -b1 + bs + b4 = 0 and + b2 + b4 = 0 . The general solution of this homogeneous system can be written as b1 ::: s + t , b2 = 2s + t, bs = s, b4 = t. Thus the original

l [i]

11

system Ax

b is consistent if and only if the ve•:to' b is of the fonn b

[ ;,(

s

+t

m·

33. The matrix A can be reduced to a row echelon from R by the following sequence of row operations:

ro A= (1) Intercha.nge rows 1 and 2.

ll

-2

[:

7 3

1

3 - 5

1

3

3

1

7

-2 - 5

1

r

(2) Add 2 times row 1 to row 3.

0

8

37

1 0 1

(3) Add -1 times row 2 to row 3.

R

l:

j] j]

7

3 l

:J 3 88]

7

0 0 0

It follows from this that R = E 3 E 2 E 1 A where E 1 , E'2, E 3
A where E =

EFG R

E! 1 • F = Ei 1 ,

[!

and G = £3 1 •

u! ;][:

1 1

0 0 0 0

WORKING WITH PROOFS

83

34. The matrix A is obtained from the identity matrix by aJding a times the first row to the third row and adding b times t he second row to the third raw. If either a = 0 or b = 0 (or both) the result is an elementary matrix. On the other hand, if a =I= 0 and b =I= 0, the result is not a.n elementary matrix since two elementary row operations are required to produce it. Thus A is elen,enta.ry if and only if ab = 0.

DISCUSSION AND DISCOVERY Dl. Suppose the matrix A can be reduced to the identity matrix by a known sequence of elementary row operations, and let E 1 , E 2 , ...• Ek be the corresponding sequence o£ elementary matrices. Then we have Ek · ··E2 EtA =I, and so A= E;- 1 E2 1 · · ·Ej; 1 l. Thus we can find A by applying the inverse row operations to I in the reverse order.

D2. There i.'J not. For example, let b = 1 and a"" c = d = 0. Then

=

D3. There is no nontrivial solution. From the last equation we ste. that x 4 = 0 and, from back substitution , it follows immediately that x 3 = x'2 = x 1 = 0 also. The coefficient matrix is invertible. D4.

= [01

0 ] 0

(a)

The matrix B

(b)

Yes, t he mat rix B

(c)

The matrix B must be of size 3 x 2: it is not square and therefore not invert.ible.

D5 . (a )

:

-2

=

has the requited property.

_:

False. Only invertible mat.rkes can be

F'alse. For example t he product a single elementary row operation.

(c)

This row operation is equivalent t.o multiplying the given mat.rix by an elementary matrix: and, since any elementary matrix is invertible, the product is still invertible.

(e)

D7.

=_

as a produ<:t of elementary matrices.

(b)

(d) 'T'r11e. If A is invertible and AR

D6.

works just ns well.

cannot be obtained from the identity by

= 0, then B = IB = (A '- 1 A)B = A- 1 (AB) = A- 1 0= 0.

True. If r1. is invt!rtible then the homogeneous system Ax = 0 has only the trivial solution; otherwise (if A is singular) there are infinitely many solutions.

All of these s tatements arc true. No. An iuvertible matrix cannot have a row of zeros. Thus, for A to be invertible we must have =I= 0. But (assuming this) if we add -d/a times row 1 to row 3, and add -e/h times row 5 to row 3; we have a matrix with a row of zeros in the third row.

a#- 0 and h

WORKlNG WITH PROOFS Pl. If AB is invertible then, by Theorem 3.3.8, both A and B a.re invertible. Thus, if either A orB is singular, then the product AB must be singular. P2. Suppose that A = BC where B is invertible, and that B is reduced to I by a sequence of elementary row operations corresponding to the elementary matrices E 1 , E2, .. . , Ek · Then I= Ek · · · EzE1B, and so B - 1 = Ek · · · EzE1. From this it follows that Ek · · · E2E1 A == E,.. · · · E2BC = C; thus the same sequence of r{)w operations will reduce A to C.

84

Chapter 3

P3. Suppose Ax= 0 iff x = 0. We wish to prove that, for any positive integer k, Akx = 0 iff x = 0. Our proof is by induction on the exponent k. Step 1. If k = 1, then A 1x =Ax= 0 iff x

= 0. Thus the statement is true fork=

1.

Step 2 (induction step): Suppose the statement is true fork= j, where j is any fixed integer 2:: 1. Then A1+1 x =A) (Ax) """' 0 iff Ax= 0, and this is true iff x = 0. This shows that if the statement is true fork= j it is also true for k = j + 1. •..

These two steps complete the proof by induction.

P4. From Theorem 3.3.7, the system Ax= 0 has only the trivial solution if and only if A is invertible. But, since B is invertible, A is invertible iff B A is invertible. Thus Ax = 0 has only the trivial solution iff (BA)x = 0 has only the trivial solution. P 5. Let e1, ez, .. . , em be the standard unit vectors in Rm. Note that ell e2, ... , em are the rows of the identity matrix I= lm. Thus, for any m x n matrix A, we have

for i = l, 2, ... , m. Suppose now that E is a.n elementary matrix that is obtained by performing a single elementary row operation on I. \Ve consider the three types of row operations separately. Row Interchange. Suppose E is obtained from I by interchanging rows i and j. Then

r1(EA) = fi(E)A = eiA = ri(A) ri(EA) = rj(E)A

=

e;A = ri(A)

and rk(EA) = rk(E)A = = rk(A) fork =J: i,j. Thus EA is the matrix that is obtained from A by interchanging rows i a.nd j. Row Scaling. Suppose E is obtained from I by multiplying row i by a nonzero scalar c. Then

and rk(EA) = r,(E)A = ekA = rk(A) for all k A by multiplying row i by the scalar c.

=I= i.

Thus EA is the matrix that is obtained from

Row Replacement. Suppose E is obtained from I by adding c times row ito row j (i :f.=j). Then

ri(EA) == rj(E)A

=

(ej

+ cei)A

eiA + ceiA

= ri(A} + cri(A)

and = rk(E)A = ekA = rk(A) for all k =J:. j. Thus EA is the matrix that is obtained from A by adding c times row i to row j. P6. Let A

= [:

!] ,and let B =

Then AB

a= d and b =c. Thus the only matrices

= BA if and only if [!

=

i.e. if and only if

c0mmute with Bare those of the form A=

Suppose now that A is a matrix of this type, a.nd let G

[:

:] = [:

=

Then AC

!] .

= C A if a.nd only if

[;b 2ba], and this is true if and only if b = 0. Thus the only 2 x 2 matrices that commute

with both B and C are those of the form A=

where -oo <

see that such a matrix will commute with all other 2 x 2 matrices as well.

a< oo.

It is easy to

EXERCISE SET 3.4

85

P7. Every 111 x n ma trix A can be tra.nsforrr:ed to reduced row echelon form B by a sequence of elementary row o pe rations. Le t E 1 , E2 , . . . , be the corresponding sequen ce of elementary Then we have B = · · · E2E 1A = CA where C = E1r: · · · E 2 E 1 is a.n inver tible m:1trix. P8.

Suppose tha t A is reduced to I= In by a sequence of elementary r ow operations, a.nd that Et. Fh, .. . , n.re the correspondjng elementary matrices. Then · · · E2E1 A =I and so · · · E2E1 = A- 1. Jt fo Uows that the same sequence of elementary ro w o perations will reduce the matrix [A I B] to t.he matrix I Ek · · · E2 E1B) = [I I A- 1 BJ.

EXERCISE SET 3.4 1.

(X J ,X2)

(X t 1 X;!, X3) ::

(c) 2.

=

= t(l, -1); Xt t, X2 -t t(2, 1, - 4); X} = 2t, X2 = t, X3 = -4t (x,,.:1:2, X3,X4) = t(l, 1,-2,3); X1 t, X2 = t, X3 - 2t,

=

=

X4

= 3t

=

(x1, x2) t(O, - 4); :c1 = 0, x2 = -4t ( (b) (.t: , ,:l:
3. (a)

4.

=

(a)

(b)

(Xt,X2 , X3)

= s(4, ·

4, 2) + t(-3,5, 7);

Xt

= 5t = 0 , X3 X3

= 5t , :t4 = -

4t

= 4s - 3t, X2 =-tiS + 5t, :t3 = 2s + 7t X1 = S + 3t , X2 = 2s + 4t, :r3 = S + 5t, x 4

= s( l , 2, 1, - 3) + t(3, 4, 5, 0);

(b)

(::tt , 2'2 · X3, X4 )

(a)

(:z: 1 ,:r2 X3) :-:: s{0,- 1,0) +t(-2,1 , 9); x 1 = 2t, X2 = -s+t, X3 = 9t lXt, x2. ::3, :r4, x :;) = $(1, 5, - 1, 4, 2) + t (2, 2. 0, 1, - 4); x 1 = s + 2t, x2 = Ss + 2l , x 3 X 4 = 1s + t, X:; = 2s - 4t

(b)

·. = -3.'f

= - s,

(a) (b)

u

= - 2v ; Lhus u

u

:f:

6.

(a)

no

7.

(a)

Two vectors are linearly dependent if and only if one is a sca la r m ultiple of the other; thus

5.

is in the s ubspace span{v}.

kv fo r a ny scalar k ; thus u is not in the subspace s pa n{v}.

{b)

yes

these vectors 1\re linea.rJy dependent. (b) These vectors are linearly independent.

8.

(a)

9.

u = 2v - '\v; thus the

linear ly independe nt.

10.

u = 4v - 2 w

11.

(a) (b)

12. (a)

(b)

linearly dependent

u, v , w are linearly dependent.

A line (a 1-dilnens io nal subspace) in R 4 that es thro ugh the origin and is parallel to the vector u = (2, - 3, 1, 4) = 4(4, -6, 2, 8}. A plane (a 2-.:!ime nsional suospace) in R" that es through the origin and is parallel to the vecto rs u = (3,- 2, 2, 5) and v = (6, -4, 4, 0).

A pla ne in tha t es through the origin and is parallel to the vectors u = (6, -2, -4,8) and v = (3, 0. 2, - 4). (h) A line in R 4 tha t es through the origin and is parallel to the vector u = (6, -2, -4, 8).

Chapter 3

86

13. The augmented matrix of the homogenous system is

H

-5 -3

2

6

-6 -1 12 5 -18

a.nd the reduced row-echelon form of this matrix is 0 11 1 -8 0 0

6 0 0

Thus a general solution of the system can be written in parametric form as Xt

= -6s- llt,

or in vector form as (Xt. X2 1 x 3 , X4) =

X2

= s,

X3 :=

8t,

X4

=

t

...:

S( -61 1,0, 0) + t( -11, 0, 8, 1)

This shows that the solution space is span{ v 1, v2} where v1

= (-6, 1, 0, 0) and v2 = (-11, 0, 8, 1).

14. The reduced row ech!:!lon form of the augmented matrix is

0

0

0-1 1 1

2

- 3

0

0

0

0

0

1 -1 [

thus a. general solution of the system is given by (x 1 , x 2 , x 3 , x 4 , x 5 ) = r( l, 1, 0 , 0, 0) + s(l , 0 , - 2, l , 0) + t( -l , 0 , 3 , 0, 1} 15. The augmented matrix of the homogeneous system is 2 4

1 -1

l

1

0

1


1 0]

1 2 0 [0 0 1

i

0

Thus a general solution of the system can be written in parametric form as X1

=

-2r-

js-

X2

= T,

X3

=

.X4

= S,

X$=

t

or in vector form as

(xt, x z, .x3, X4, xs) The vectors v 1 = (-2, 1, 0,0, 0), v2 sp3ce.

0,0,0) + s(

-l ,0,

1, 0)

= (- 3. 0, -i, 1, 0), and v 3 = (-

+

0, -l,O, 1) 0,-

0 , 1) span the solution

16. The reduced row echelon form of t!1a augmented matrix is

0 1 0 0

-4 1 0 0

1 -2 1 -3 0 0 0 0

thus a. general solution is given by (xt. x2, X3, x4 , xs) = s(4, -1, 1, 0 , O) + t( - 1, 2, 0, 3, 1).

87

EXERCISE SET 3.4

17. (a) Vz is a scalar multiple ol· v 1(v2 = -5vl); thus these two vectors are linearly dependent. (b) Any set of more than 2 vectors in R 2 is linearly dependent (Theorem 3.4.8). 18. (a) Any set containing the zero vector is linearly dependent. (b) Any set of more than 3 vectors in R 3 is linearly depenrlcnt (Theorem 3.4.8).

n-: [:]-

19. (a) Those two vectors arc linearly independent since neither is a scalar multiple of the other. (b) v 1 is a scalar multiple of vz(v 1 = -3vz)i thus these two vectors are linearly dependent .

(c) These three vectors are linearly Independent since the system

has

only the trivial solution . The coefficient matrix, which is the matrix having the given vectors as its columns, is invertible. (d) These four vectors are linearly dependent since any set of more than 3 vectors in R 3 is linearly dependent.

(b) linearly independent (d) linearly dependent

20. (a) linearly independent (c) linearly dependent

21. (a) The matrix having t hese vectors as its columns is invertible. Thus the vectors are linearly independent; they do not lie in a. plane. (b) These vectors are linearly dependent (v 1 = 2v2 - 3v3); they lie in a. plane but not on a line. (c) These vectors line on a line; v 1 = 2v2 and v3 = 3vz. 22. In parts (a) and (b} the vectors are linearly independent; they do not lie in a plane. In part {c) the vectors are linearly dependent; they lie in a plane but not on a line . ......-·--...,""

23. (a)

This vectors is it is closed under scalar and addition: k(a, 0, 0) = (ka, 0, O) and (a1. a, 0) + (az, 0, 0) = (a1 + az, 0, 0). (b) This seL of ''cctors is not a subspace; it is not closed under scalar multipllcation. (c) set of vectors is a.· subspace. lf b = a+ c, then . kb = ka + kc. If b1 ;;;·a 1 + c1, and b2 = nz + cz, then (b1 + bz) = (aJ + a2) + (c1 + cz) . (d) This set of vectors is not a subspace; it is not closed under addition or scalar multiplication.

.

24. Sets (a) and (b) are not subspaces. Sets (c) and (d) are suhspaces. 25. The set W consists of all vectors of the form x = a(l,O, 1, 0); thus W = span{v} v (l,O , 1, 0) . This corresponds to a line (i.e. a 1-dimensional s ubs pace) through the origin in R 4 .

=

26 . W = 5pan{v 1 , v 2 } ·.vhcre v 1 ::: (1, 0, 2, 0, - 1) and v 2 = (0, 1, 0, 3, 0). This corresponds to a. plane (i.e. a 2-dimensional subspace) through the origin R::.. 27. (a) 7vl - 2vz + 3v3 = 0 2 3 (b) YJ = 7Y2 ·- 7V3 , Vz =

7

zVl

3 + 2YJ

1

V3

7 = -3Vl + 32 V2

!,

28. If ,\ = then the given vectors are clearly dependent. Otherwise, consider the m atrix having these vectors as its columns. Are there any other values of..\ for which it is sing\.!lar? To determine this, it is convenient to start by multiplying all of the rows by 2.

[Y 2: JJ

88

Chapter 3

Assum!ng

>. ¥-

this matrix can be reduced to the row echelon form

0 2+2.>.

l

which is singular if and only if>.= - 1. T hus the gjven vectors are d ependent iff >.= 29.

! or>.= -1.

Suppose S = { v 1 , v 3 } is a. linearly independent set. Note first that none of these vectors can be equal to 0 (otherwise S would be linearl y dependent). and so each of the sets {v1 }, {v 2}, and {v 3} is linearly independent . Suppose then that Tis a 2-element subset of S, e.g. T = {v 1 , II T is linearly dependent then there are scalars c1 and c2 , not both zero, such that c1 v 1 + c 2v z = 0 . But, if this were true, then c1 V t + c2v 2 + Ovs = 0 would be a nontrivial linear relationship among the vectors v'h v2 , v 3, and so S would-.be linearly dependent. Thus T = {v1 , v 2 } is linearly independent. T h e same argument applies to any 2-element subset of S. Thus if S is linearly independent , then each of its nonen:tpty subsets is linearly independent. (b) If S = {v1 , v 2, v 3 } is linearly dependent, then there a.re scalars c1o c2 , and c3 , not all zero, s uch that c 1 v1 + c2v2 + CJVJ = 0 . Thus, for any vector v in R", we have c 1 v 1 + c2 v 2 + c3 v3 + Ov = 0 and this is a nontrivial linear relationsbjp among the vectors v 1 , v 2, vs, v. This shows that if S = {v 1 , v 2 , v 3 } is linearly dependent then so is T = (v 1 , v 2 , v 3 , v} for any v .

(a)

30. The arguments used in Exercise 29 can easily be adapted to this more general situation. 3 1.

(u - v ) + (v - w) + (w - u ) penden t set .

= 0 ; t hus the

vectors u - v, v - w, and w - u form a linearly de-··

32. First note that the relationship between the vectors u , v and s, g can be written as

(:)

= and so the inverse relationship is given by

[s] = [0.12 1

g

0.06] _, [u] 1

v

1

= 0.9928

[

1

- 0.12

Parts (a.), (b) , (c) of the problem can now be a nswered as follows: (a)

s

= 6 9i 28 ( u -

0.06v)

(b) (c) 33.

(a) (b)

No. 1t is not closed under either addition or scalar multiplication. P 876

= 0.38c + 0.59m + 0. 73y + 0.07k

P 2t6

= 0.83m

+ 0.34y + 0.47k

P328 = c + 0.47y (c) 34.

H P e75

+ P2J6) corresponds to the CMYK vector

(a) k1 = k2 = (c)

+ 0.30k

=

tc2 4c3

(0.19,0.71 , 0.535, 0.27).

(b) k1

= k2 =

· · · = k1

=

The components of x = + + represent the averag.e scores for the students if the Test 3 grade (the final exam?) is weighted twice as heavily as Test 1 and Test 2.

89

DISCUSSION ANP DISCOVERY

35. (a) kt = kz = k3 = k4 = (c) The comp onents of x; =

i

(b) k1 = k2 = k3 = k4 = k& = represent the average total population of Philadelphia, Bucks, and Delaware counties in each of the sampled years.

+ ! r2 +

n

36. v =

L

vk

k= l

=DISCUSSION AND DISCOVERY Dl . (a) Two nonzero vectors will span ·R 2 if and only if t hey a.re do not lie on a line. (b) T hree nonzero vectors . will span R3 if and only if they d o not lie in a T wo vectors in Rn will span a plane if and only if they are nonzero and not scalar multiples of one another. (b) T wo vectors in R" will span a line if and only if they are not both zero and one is a scalar multiple of the other. (c) span {u} = span {v} if and only if one of vectors u and v is a. scalar multiple of the other.

D2. ( a )

D 3. (a)

Yes. If three nonzero vectors are mutually orthogonal then none of t.hem lit:S in the plane spanned by the other two; thus t he three are linearly independent. (b) Suppose the vectors vl> v2 , and v 3 are nonzero and mutually orthogonal; thus v, · v; = llvdl 2 > 0 fori= 1, 2, 3 and v; · vi = 0 for i=/: j . To prove they are linearly independent we must s how that if C1V1 + C2V2 + C:JVJ = 0 then Ct = c-_a = C3 = 0. Tbis follows from the fact that if c 1v1 + c 2v2 + C3 Y3 = 0, then C;

for i

2

llv;ll = vi· (c1v1 + c2v2

+ c3v3)

= V1

•

0 = 0

= 1, 2, 3.

D4. The vectors in the fi rst figure are linearly independent since none of them lies in the plane spanlled by the other two (none of them can be expressed as a linear combination of the other two). The vectors in the second figure arc linearly dependent since v 3 = v 1 + v 2 . DS. This :set is d ose>d under scalar multiplication, but uot under u = (1,2) nnd v = (-2, - 1) correspond to points in Lhe set, but

06. (a)

(b) {c)

(d) (e)

D7.

11

For example, the vectors does not.

+ v = (-1, 1)

False. For example, two of the vectors ma.y lie on a line (so one is a scalar multiple of the other), but the third vector may not lie on this same line and therefore cannot be expressed a.s a linear combination of the other two. False. The set of all linear combin ations of two vectors can be {0} (if both are 0), a line (if one is a scalar multiple of the other), or a plane (if they are linearly independent) . False. For example, v and w might be linearly dependent (scalar multiples of each other). [But it is true that if {v, w} is a linearly independent set, and if u cannot be expressed as a linear combination of v and w, then {u, v, w} is a independent set .J True. See Example 9. Ttne:. lf c1(kv1)+ cz(kv2 ) + cz(kvz) = 0, then k(c1 v1 +c2v2 + c3v2) = 0. Thus, since k =/: 0, it follows that Ct v 1 + c2 v2 + c 3 v 2 :::::: 0 and so c1 = cz = c3 = 0.

(a) False. The set {u, ku} is always a linearly dependent set. (b) False. This statement is true for a homogeneous system (b:::::: 0), but not for a non-homogeneous system. !The solution space of a non-homogeneous linear system is a translated subspace.J

Chapter 3

(c)

True. If W is a subspace, then W is and so spa.n(W} = W .

closed under scalar multiplication and addition,

·

(d) False. Forexample,ifSt = {(I ,O),(O, l}}a.nd S2 = {(1, l) ,(O, l}}.thenspan(SI) = span(S2 )=

R2 ,

but. S1

# S2.

08. Since span(S) is a subspace (already closed under scalar multiplication and addition), we have s pan(spa.n(S}) = spa.n(S). ·

WORKING WITH PROOFS Pl. Let() be the angle between u and w, and let 4> be the angle between v and w . We will show that 9 = ¢. First recall that u · w = llu!lllwll cosO, so u · w kllwll cos9. Simllarly, v • w.=lllwllcos¢. O n the other hand we have ·

=

= l(u · u) + k(u · v) = lk 2 + k(u • v) u. w = lk 2 + k(u . v ) , i.e. l!wll cos e = lk + ( u. v) . Similar calculations show (v · u) + k l ; thus Uwll cos B = Uwll cos¢. It follows that cos 9 =cos¢ and () = . u · w = u · (lu + kv)

aud ::;o kllwll that llwll cos¢ =

X belongs to WIn w2 and k is a scalar, then kx also belongs to Wt n wl since both Wt and w2 are subspaces. Similarly, if XI and X2 belong to Wt n W2 , then x, + X2 belongs to WIn w2. Thus WI n w2 is closed under scalar multiplication and addition, .i.e. wl n w2 is a subspace.

P2. If

P3. First we show t hat ·w1 + W 2 is closed under scalar multiplication: Suppose z = x + y where xis in W1 andy is in W2. Then, for any scalar k , we have kz = k(x + y) = kx + ky, where kx is in W1 and ky is in (since w, and w2 are subspaces); thus kz IS in w} + w2. Finally we show that WI + w2 is closed under addition: Suppose Z l =X) + Yl and Z 2 = X 2 + Y2 o where X) and X2 are in W1 and Y• and Y2 are in W2 . Then z 1 + Z2 == (x1 + y, ) + (x2 + Yz) = (x1 + x2) + (Yl + Y2) . where X } + X2 is in lt\ and Yl + Y2 is in w2 (since WI and w2 subspaces); thus Zl + Z2 is in

H.-\ +

w2.

EXERCISE SET 3.5 1.

(a)

The reduced r,Jw echelon form of the augmented matrix of the h omogeneous system is l

0 0 thus a general solut ion is

x1

!

Gr

(in column vector form)

EXERCISE SET 3.5

(c)

91

From (a) and (b), a general solution of the nonhomogeneous system is gjven by

(d) The reduced row echelon form of the

.. ---

-

of the nonhomogeneous system is

-

.. ,,.

2

3 0 0

f ·.:.,

1

-3

;]

0 0

T his solution is related to the one in Qa.rt (c) by the change of variable .s1

--

0

2.

(o) The <educed

echelon ro,m is [;

(d) T he rnd uced row echelon form is

ll

T

2

0

0

0

¥

1

o·

-

-g

o

+t'

-

·-

-

-

:1:3

nv ..-"

The reduced row echelon form of the augmented matrix of the give n system is

x1

=

!-

+ 1. .

:r2

0

'··..

thus a general solut ion is given by

1

t = t

-·-----··- r--•. •otNo•·-·---.0o]; thus a general solution is [Zil = t

: [:} [1J 3. (a)

= s,

!3

0

0 0

1 0

.3!.] 1

0

-1t, x2 = s, X3 = t , X4 = 1; or

1

92

Chapter 3

(b) A general solution of the

!s

--:::....

.·

::] ::::: s [

[-!] ,r-;1 \. +

l

0

' .. X4

0

and a particular solution of the given nonhomogenous system is -·········

4. (a)

··•··

= !. x2 = 0, X3 =0, x 4 =

x1

: ..o···-··- ··-·

1.

The reduced row echelon form of the augmented matrix of the given system is

0 1 0

_1: 133] 9

-7

0

0

thus a general solution is given by

(b)

A general solution of the associated homogeneo'..ls system is

is x 1 =

and a particular solution of the nonhomogeneous 5.

i,

1

x2

=

0,

x3 =

-7,

X4

= 0.

The vector w can be expressed as a linear combination of v 1 , v 2 , and v 3 , if and only if the system

1

: [: ] : : 5

-12

c3

1

is consistent. The reduced row echelon form of the augmented matrix of this system is 0 1 0

0 -2] 0 1

3 1

From this we conclude that the system has a. unique solution, and that w = -2v1

+ 3v2 + v3.

6. The vector w can be expressed as a linear combination of v 1 , v 2 , and v 3 , if and only if the system

f3 l2

-2

8 1

4

93

EXERCISE SET 3.5

is consistent. The reduced row echelon form of the augmented matrix of this system is

/

/

•,

'

/

,.'

0 l 0

•·..

---·

3 -1 0

-

10]

-6

0

...---······• ...... .

.

.

/

and from this we conclude that the :;ystem. hilS inftnitcly many given b}: c1 .:;::;.10 - .. c 2 == -6 + t, c3 = t. Thus.w can be expressed as a linear combination of v1, v2, and V3. fii particutat; taking t = 0, we have w = 10v 1 - 6vz. 7. The vector w is in span{v1 , v 2 , v 3 , v 4 } if and only if the system

H

1 0

l

-3 2

is consistent. The row reduced echelon form of the augmented mo.trix of this system is

! . . : ..•.. ... . .··

From this we conclude that the system has infinitely many solutions;

.

.

. ......

- ··.

.. ·::.::-_-;:-,..._

w is in span{v 1 , vz, v 3 , v 4 }.

,......--··-··--··--··"· ······-

8. The vector w is in span{v 1 , v 2 , v 3 } if and only if the system

is consistent.. The reduced row echelon form of the augmented matrix of this system is

0

2 -1

0

0

From this we conclude that the system has infinitely many solutions; thus w is in span{vt, v2, VJ}. 9.

(a) The hyperplane al. consists of all vectors x = (x, y) such that a • x = 0, i.e. -2x + 3y = 0. This corresponds to the line through the origin wit.h parametric equations x = y = t. (b) The hyperplane a..!. consists of all vectors x = (x, y, z) such that a· x = 0, i.e. 4x- 5z = 0. This corresponds to the plane through the origin with parametric equations x = y = s, z =t. (c) a.i. consists of all vectors x = (xbx 2, X3, x4) such that a· x = 0, i.e. Xt + 2x2- 3x3 + 7x4 = 0. This is a hyperplane in R 4 with parametric equations Xt = -2r + 3s- 7t, x2 = r, X3 = s,

it,

X4

0. (a) x (c)

= t.

= 4t, y = t

x1 = r

+ 2s, x 2 = r, x 3 =

(b) s, s;4 = t

x

=

s, y

= -3s + 6t, z = t

/

i

94

Chapter 3

11. This system reduces to a single equation, Xt + x2 x 1 = -s- t, x 2 = s, x 3 =: t; or (in vector form)

+ x3 = 0.

Thus a general solution is given by

The solution space is two-dimensional.

13. The reduced row echelon form of the augmented matrix of the system is

Thus a general solution is x 1

=

19

0

7

T

1

7

-7

l

I_( s-

x2 3

X1

7

X2

-7

+ ts + ¥t,

=

xs =r

1

X4

0 0

X f.>

1

7 0 1 0

+s

= T,

x4

= s, x 5 = t; or

-:1

19

-7

2

X3

+t

The solution space is three-dimensional.

15.

(a)

A genecal

x

=:

l m f l] nl

\;t [:

t:_y

+

. ..

+t

'

·---·····--..

(b) The solution space corresponds to the plane which es through the point fP( 1, 0, 0) il.nd is ...

16.

(a)

A general solution is x = 1- t, y

(b) The solution to the vecto( v

17. (a)

·····-· ... ··-···· ...................... ........ -

....

......_..

· ····· ·

······- ····--- ....... , ...,.

= f;

-,._______.,. ._____ ·

-

corresponds to the line which es through the point P(l, 0) and is parallel = (-1,1).

A vector x = (x, y,z) is orthogonal to a= (1, 1, 1) and b = ( -2, 3,0) if and only if

y-

g+3y (b) The solution space is the line through the

z. =

0

j to the vectors a and b.

EXERCISE SET 3.5

(c) The reduced row echelon form of the augmented matrix of the system is

·

and so a

- ..

that the vectof: l8.

.

iJ ·"-

z = t; or (x, y,z) I:!.,.

by x

(a) A vector x = (x,y,z) is orthogonal to a = ( - 3,2,-1) and h= (0, -2, -2) if and only if

+ 2y -

- 3x

=

(c)

(x, y, z )

{a}

A \'ector x

--

z =0 - 2y- 2z = 0

. ori,g!11.. tha,:t is perpendicular to

(b) The solution space is th.e llne through

.9.

l).,ote --

_: 1, 1); note that the vector v = (-

= (x 1 ,x2,x3 ,x4) is orthogonal

to v1

1, ·:.__ i, 1) is orthogonal to both a and b. ·

= (1, 1, 2,2) and vz = {5, 4,3,4) if and only if

+ 2x4 = 0 + 4x z + 3x3 1- 4X-t = 0

x 1 + xz + 2x3 5xl

(b) The solution space is the plane (2 dimens ional s ubspace) in R4 that es through the origin a.nd is perpendicular to the vectors v1 and v 2. (c) The redUl:ed row P-chelon form of the augmented matrix of t.he system is

- 5 -·1

0

7

l

6

=

and so a. general solution of the system is given by (x ,,xz,XJ ,x4) s{5, -7,1 ,0} + t(4, - 6,0, 1) . Note that t he vectors (5,-7, 1,0) and (4 ,-6,0, 1) are orthogonal to both v 1 and v2.

:o.

(a )

A vector x = (x 1,:c2,r.1,x 4, is orthogonal to the vectors v 1 = (1 , 3, 4,4, -1 ), v 2 == (3, 2, 1, 2, - 3) , and v3 == (1, 5, 0, - 2, - 4) if and only if

+ 3xz + 4x3 + + 2x2 + X3 + 2x4 x, + 5xz - 2x4 x1

3xl

-

Xs

- 3xs - 4xs

=0 =0 =0

{b) The solution space is the plane (2 dimensional subspace) in R 5 t hat. es through the origin and is perpendicular to the vectors v1, v z, a.nd v3. (c) The red uced row echelon fo r m of the augmented matrix of the system is 0

0

5

3

7 -- IO

1

0

13 - Z5 31

_33 50 21 50

0

25

and so a. general solution of the system is given by

(x1 , x2, x 3 , x -1, xs) = s( The

(- t W.

1, 0) and ( t70 ,

1, 0) + t( {o,

I)

0, 1) are orthogonal to v •• Vz , and v 3.

Chapter 3

96

DISCUSSION AND DISCOVERY Dl. The solution set of Ax = b is a. translated subspace x 0

+ W , where W

is the solution space of

Ax= 0 . D2. If vis orthogon al to every row of A, t hen ilv = 0 , and so (since A is invertible) v = 0 .

D3. T he general solution will have at least 3 free variables. Thus, assuming A i: 0, the solution space will be of dimension 3, 4, 5, or 6 depending on how much redunda.ocy there is. D4.

(a) True. The solution set of Ax= b is of the form x 0 + W where W is the solution space of Ax = O. 0 (b) False. For example, the system :r - Y = is inconsistent, but the associated homogeneous x-y=l

system has infinitely many solutions. (c) True. Each hyperplane corresponds to a single homogeneous linear equation in four variables, and there must be a least four equations in order to have a unique solution. (d) 'Tme . Every pla ne in R3 corrr.:>ponrls to a equation of the form ax + by + cz = d. (e) False. A vector x is orthogonal to row(A) if and only if x is a solution of the homogenous system Ax = 0.

WORKING WITH PROOFS P 1. Suppose that Ax = 0 has only the and t hat Ax = b is consistent. If Xt and Xz are t wo solutions of Ax= b , then A(x 1 - x 1 ) = Ax 1 - Axz = b - b = 0 anti so Xt - x z = 0 , i.e. x 1 = xz . Thus, if Ax= 0 has only the t rivial solution, t he system Ax = b is either inconsistent or has exactly one solution.

PZ . Suppose that Ax = 0 has infinitely many solutions and that Ax= b is consistent. Let Xo be any solution of Ax = b . Then, for any solution w of Ax= 0 , we have A(xo + w) = A.xo + Aw = b + 0 = b . Thus, if Ax = 0 has infinitely many solutions, the system Ax = b is either inconsistent or ha.<> infinitely many solutions. Conversely, if Ax= b has at. most. one solution, then Ax= 0 has only t.he trivial solution. P3. If x 1 is<:\ solution of Ax = band x2 is a solution of Ax = c, then A(x, + x2) = Ax1 + Axz = b + c; i.e. x 1 + X2 is a solution of Ax= h +c. Thus if Ax= band Ax ;= care consistent systems, then Ax = b + c is also consistent. This argument c2-n easily be adapted to prove the following:

Theorem. If Ax = bi is consistent for each j Ax = b is also consistent. P4 . Since (ka) · x = k(a · x) and k that. (ka)..L a ..L.

=

= 1, 2, ... , r

and if b = b 1

f. 0, it follows that (ka) · x = 0 if and only

+ b2 + · · · + b,.,

if a· x = 0. This proves

EXERCISE SET 3.6 1.

(a)

This matrix has a row of zeros; it is not invertible.

(b) A diagonal matrix with nonzero entries on the

o200] u

then

1

=

o ol

O!O· 0

0 3

EXERCISE SET 3.6

r

0 2.

(a}

-2 0

0

3

0 3.

(a)

(c)

- 1

97

0

= rl0

0

1

-2 0

_; l [!

0

0][-:2 1] -2]0= [3

0 -1 0

[-12-3

-5

-2 4.

(a)

[ - 16 - 2 - 20

5.

(a)

(b)

"

15

- 20 10

- 2 -10

0][ 10 -2] ] - 2 = ro -6 0 2

-20 10

20 2 20 - 20

-10

10 5

{b)

0

0

0 9 0

(b)

)k

(c)

11

4

0

1]

(c)

[2'

A- k

if and only if x =!= l , - 2, 4 (the diagonal entries must be nonzero).

A is

9.

Apply the invers ion a lgorit hm (see Section 3.3) to find the inve rse of A .

2

3 :1

1 0

-2:0

0 1

1 :0

0

'

Add 2 t.i111es row 3 to row 2. Add -'- 3 times row 3 t o row l. 1

0

0 1 0

0: 1

o:o I 1 :0

-;]

Add -2 times row 2 to row 1.

0

0: 1

1

olI o

0

1:0

T hus the ;nverse of the upper triangular matrix A

0 3k

0

7.

2

12]

[ - 24 - 10 3 - 10 60 20

(c)

1

0 9 0

- 2] =

0

0 4

I

-4 1 2 5

10

0

6. (a)

[ 2 1] ' 0 [ 10 -2]

(b )

2

0

4

fl

0

- 1

2 5

(b) not invertible.

-2 1 0

i 23]

[

1 0

is A-

[' -:r -2

1

=

I

0

3]

98

10.

Chapter 3

If

0

A=

11. A =

-1

A- =

; } ,lhro

4

u

0 0 - 1

r[

0 -1 4

1

0 0 1

i]

-

1 2 - 11

0

;]

- 1

4

0-8] 0 4

12.

-4 0

13. The matrix A is symmetric if and only if a, b, and c satisfy the following equations:

a - 2b + 2c = 3

r2a The augmented matrix of this system is

+- b+···· c-·,;-· -{) ..

:}.-·:

nnd the reduced row echelon form is

0 1 0

0 0 1

- 13

Thus, in order for the m atrix A to be symmetric we must have a = 11 , b = - 9, and c = -13. 14. There are infinitely ma.ny solutions: a = l

15. 'vVe hnve A - l

J. = [- 12-1 3

16. We have A- 1

= [- 2

I

+ lOt, b = 1t, c = t , d =

[3 }]

1 = ; t hus A 5 I 2

l - 2 1

-7

3 -7

4

< t < oo.

_ ts. symmetnc.

r-45

1

lll thus A -

1 -13 = - - 13 - 5

14

0, whe re - oo

1 ; 1 -3

11

1

is s ymmetric.

=

17. AB =

0

0

- 4

0

0

3

0

0

- 12

18. From Theorem 3.6.3, the factors commute if and only if the product is symmetric. Thus the factors ;n (a) do not commute, whe,eas the ractocs ;n (h) do commute.

19. If A =

]

0

20. If

A=

0

-1

0

0 0] [(! 0)4 0 , then A3

2

0

l

0

(-1 )- 5

0 ..........

2

0

=

0

( 1)-2

0

0

I

_

-1

..

_l .

EXERCISE SET 3.6

[!

21. The lower tria.ngular matrix A =

is invertible because of the nonzero diagonal entries.

1 3 1

Thus, from Theorem 3.6.5, AAT ond AT A are also invertible; furthermore, since these matrices are symmetric, their inverses are also symmetric by Theorem 3.6.4. .Following are the matric:es AAT, AT A, and their inverses.

ATA =

AAT :::::

22.

r [l

6

10

1

(AT A)-

=

[ I -3 8] -3 8

3

3

10

- 27

(AAT)- 1 = [ 74

10

- 27 10

-3

6

-27 74

-;]

Following are the matrice..<; A AT and AT A, and their inverses.

AT.4

=

['i

5 5 2

{AT,4.)-1 =

r

3 AAT =

35

-13 5

5

-2

= l-13

1 (AA ')-•

10 5

23. The fixed points of t he matrix A ==

-3 [ - 3I 10 17 5

30

are ihe solut ions of the homogeneous system (1 - A )x

=

0. The augmented matrix of this system is

- 1 [- J

o]

-1 -1

0

and from this it is easy to see tha t the system has the

fixed poi nts of A are vectors of t he form x

24.

All vectors of thcform x

25.

(a)

If

=

H] =

= t [_

where - oo

= -t.

T hus the

< t < oo.

t [-:] , whero - oo < t < oo.

=

A2 =

and the inverse of I - A =

=

solution x 1 = t, x 2

is (I- A)-

(b) The nilpotency index of A =

Thus A is nilpotent with nilpotency mdex 2, 1

= I+ A= I - A =:'

8 l 0

is - 8 -1

100

Chapter 3

26. (a)

A=

(b) A

1-A=[lO]

has nilpotency indt>.x 2.

= [0 o o2 31,]

(I - A) - t

-1 1

.

= I + A = [11

0 ] 1

=:]

has nilpotency index 3.

0 0 0

+

+

If A is invertible and skew symmetric, then (A- 1 )T is skew symmetric. (b) If A and B are skew symmetric, then

27. (a)

=

= (AT)- 1 = (-A)- 1 = -A- 1 ; thus A- 1

(A+ B)T =AT+ BT =-A- B =-(A+ B)

(A- B)T =AT- BT = -A+ B

= -(A- B)

(kA)r =kAT= k( -A)'"'"' -kA

thus the matrices A+ B, A- B, and kA are also skew-syn\met.ric.

28. (a)

If A is any square matrix, then (A+ AT)T =AT+ ATT = A'r +A= A+ AT, so A+ AT is symmetric. Similarly, (A- AT)T =-(A- AT), so A- AT is skew-symmetric.

(b)

A=

(c)

A=

HA +.A1')

AT)

+

=

29. If H =In- 2uuT then we have

30.

T

AA

['']

T

r2

=

[r, rf

['' !

= r2r1 T

rmr1

nr =

I'I- 2(uuT)T =In- 2uuT

= H; thus li is symmetric.

r;,:]

rlrr r2rr

r1rm

Tl

r,.,.rT2

fmfm

31. Using Formula (9), we have tr(Ar A}=

rz

'' · rm f2 • r rn

1

·rz

Tm • I'm

J

:r 1 • r2

r1 · r1

r2 ·

[ ''; r, rm · rl

I'm

lladl 2 + lla2!1 2 + ··· + tlanll 2 = 1 + 1 + · ·· + 1 = n.

DISCUSSION AND DISCOVERY 3an

5an

7al3]

Dl. (a) A= 3a21 Sa22 1a23 [

3a::n

Saa2

1a33

=

[au a:n

a31

a12 a22 a32

a13l [3 0 0] a23J

1

a33

0 5 0 0 0 7

= BD

(b) There are other such factorizations as well. For example, A ""

[!::: :::

3a31 an

DISCUSSION AND DISCOVERY

101

A is symmetric since a 31 = j 2 + i 2 = i 2 + j 2 for each i, J . (b) A is not symmetric .since a;;= j - i = - (i- j) = - a;; for each i, j. In fact, A is skewsymmroetric. 2j + 2i = 2i + 2j = a,i for each i, j. (c) A is symmetric since (d) A is not symmetric s ince a 1 , = zp + 2t3 f:. 2i2 + 2j3 ==-a.; if i = 1, j = 2.

D2. {a)

a,, =

In general, the matrix A= IJ(i,j)} is symmetric if and on1y if f(j , i}

= f(i,j)

for each i , j.

03. No. In fact, if A and B a.re commuting skew-symmetric matrices, t hen

(AB)T = BT AT= (-D)(-A)

= BA =

AB

and so the product AB is symmetric rather than s kew-symmetric. 04. Using Formula {9), we have tr(AT A)= lln1JI 2 + 1la21J 2 + · · · + follows that llatll 2 = 1Jazll 2 = llanll2 = 0 amlso A= 0.

= ···

2

0 5. (a) lf A =

(b)

D6. If A

Thus, if ArA

2

thus A ==- A if and only if

then A = 0

IJa,.jJ 2 .

0 d3

0

= 0,

then it

= d; for 1 = 1, Z, and

0

3, n.nd this i:> t.rue iff di - 0 or 1 for ·i = 1, 2, ancl 3. There are a. tOtal of eight such matrices (3 x 3 matrices whose diagonal entries are either 0 or 1) There are a t otal of 2n such matrices (n x n. rnatrice.s whose diagonal entries are either 0 or 1).

= [d'0 dl0],

only if d; matrix

=-

then A 2 + 5A

+ 6!2 =

0 if and only if d?

+ 5£'-. + G = 0

for i

= 1 and 2; i.e. if and

2 or - 3 for i = 1 and 2. There are a t.ota.l of fonr s uch matrices (auy 2 x 2 diagonal diagonal entries are either - 2 or - 3) .

D7. If A is both symmetric and skew symmetric, then AT= A and AT = - A ; thus A = -A and so A =0. 0 8.

In a symmetric matrix, enlries that are symmetr ically positioned across the main diagonal are equal to each other. Thus a symmetric matrix is completely determined by the entries that lie on or above the main rliagonal, and entries t hat. appear below the main diagonal a.re duplicates of entries that appear above the main diagonal. Au n x n matrix has n entriP.s on the main diagonal , n - l entries on the diagonal just above the maiu diagonal, etc . Thus there are a total of

n

+ (n- 1) + · · · + 2 + 1 = n(n 2+ 1)

entries lhat lie on or above the main diagonal. For a ;;ymmet.ric matrix, t his is the maximum number of distinct entries the matrix can have. In a skew-symmetric matrix, the diagonal entries are 0 and entries t hat are symmetrically positioned ac:ross the main diagonal arc the negatives of each other. The maximum nwnber of distinct entries can be attained by selecting distinct positive entries for the p,.,sit.ions above the 1.0ain 11 diagonal The entries in the n(n2positions below the main diagonal will then automatically be distinct from each other and from the entries on or above the main diagonal. Thus the maximum number of distinct entries in a skew-symmetric matrix is

n(n- 1} n(n- I ) --'--:---.....:. + 2 2 09. If D == I=

let

then AD

= ldt a1

e2.

d2a2J where a 1 and a2 are lhe rolumns of A. Thus AD =

are the standard unit vectors in R 2) if and only if dlaJ = e1 But this is true if and only if d1 i- 0, d2 :/; 0, a 1 = -J;e 1 , and a 2 = :, Thus

e2] (where e1 and

and dza2 =

+ 1 = n(n- 1) + 1

Chapter 3

102

A =

0]

_!. [

1

d2

where d 1 , d 2

=1-

'

0. Although described here for the case n

= 2 , it should. be clear

that the same argument can be applied to a square matrix of any size . Thus, if .1D

f

tliagonal entries d 1 , dz , ... ,d.n of D must be nonzero, and A=

..!. dt

l o :

0

D lO.

(a)

0

.l. .. .

o

d2

:

...

;

0

... ..1. d,..

= 1, then the

.

False. If A is not square then A is not invertible; it doesnt matter whether AAT (wh.ich is always square) is invertible or not. (But if A is square and AAT is invertible, then A is invertible by Theorem 3.6.5.]

(b) False. For example if A =

( c)

0

G

and B

!] ,then A + B = G

=

is

True. If A is both symmetric and triangular, then A must be a diagonal m atrix. Thus

A=

:, 0

0

and so p(A)

=

:

Un

0

0

]

is also a. diagonal matrix (both

p(d., )

symmetric and triangular). (d) True. For example, in the 3 x 3 case, we have

Dll.

(e)

True. If Ax= 0 has only the trivial solution, then A is invertible. But if A is invertible then so is AT {Theorem 3.2.11); thus Ar x =a has only the trivial sol ution.

(a)

False. For

=

(b) True. If A is invertible then Ak is invertible, and thus Ak =1- 0, for every k = 1, 2, 3, .. . . This shows that an invertible matrix cannot be nilpotent; equivalently, a nilpotent matrix cannot be· invertible. (c) True (assuming A i= 0). If A 3 = A, then .46 = A, A 9 =A , A 12 =A, . . . . Thus is it not possible to have A,.. :::: 0 for any positive integer k , since this would imply that = 0 for all j ?. k. (d) 1'rue. See Theorem 3.2.11. (e) False. For example, I is invert)ble but I -1 = 0 is not invertible.

WORKING WITH PROOFS Pl. If A and 1J are symmetric, then A 1 '

= A and BT = B.

It follows that

(AT )T = AT

+ BT = A+ B

(A+ B )T

AT

(A - B)T

= AT -

BT = A- B

(kA )T =kAT th11s t.hc matrices

= kA

Al' . A+ B , A - B , and kA are also symmetric.

EXERCISE SET 3.7

103

P 2. Our p roof is by induction on the exponent k . 1

Step 1. We have D = D Step 2

=

dJ: [0

0

....,o:·] = [dI:

] ; thus the statement is true for k

0

rl nl

st ep). Suppose the statement is t rue for k = j, where j is an integer

and so the statement is also true for k = j

= I. 1. 'rhen

+ l.

These two steps complete the proof by induction.

:, :l

D=

0

is inver tible with

0

0

1=

;2:::

P 3. If d•, d2, ... ,d11 are nonzero, then 0

d.,

n-' =

Q

}, 0

I

0 0

01

0 .. . };

0

On the other hand if auy one of 0

• · · dn

0

:i;;- .

0

the diagonal entries is zero, then D has a row of zeros and thus is not. invertible.

P 4. We will show t hat if A is symmetric (i.e. if AT == A), then ( An)T = A 11 for each positive integer . n. Our proof is by induction on the exponent n . Step L Since A is symmetric, we have (A 1) T n = 1.

A = A 1 ; thus t he statement is true for

Step 2 (induction step). Suppose the statement is true for n = j, where j is (Ai+ l f

= (AAi)T = (Ai )T A1' =

An

integer

1. Then

AJ A= AH 1

and so the statement is also true for n = j + I. These two steps complete the proof by ind uction. P5. If A is invertible, then Theorem 3.2.11 implies AT is !::1vertible; thus t he products AAT and AT A are invertible as well. On the other hand, if either A AT or A T A is invertible, then Theorem 3.3.8 implies that A is invertihle. It follows that A , AAT, and AT A are all invertible or all singular.

EXERCISE SET 3.7 1. We will solve Ax = b by first soiving Ly = b for y, and t hen solving Ux = y for x. The system Ly = b is 3yl =0 -2yl + Y2 = 1 from which, by forward substitution, we obtain Y l = 0, Y2

= 1.

Chapter 3

104

The system U x

=y

1s x1- 2x2

=0

X2

= 1

from which, by back substitution, we obtain x 1 the solution of Ax = b .

2. The solution of Ly l X 2-- '1·

=b

is y 1 = 3, Y2

=

= 2, X2 =

1. It is easy to check that this is in fact

The solution of Ux = y (a.nd of Ax= b ) is x 1 =

3. We will solve Ax = b by first solving Ly = b for y , and then solving Ux The system Ly = b is

=

3yx 2yl

+ 4y2

- 4yl -

=y

for x .

-3

= -22

+ 2y3 =

Y2

3

from which, by forward substitution, obta.in Yl = -1, Y2 = -5, YJ = - 3. The sys tem U x = y is X J - 2X2 ··· X3 -= -1 x2 + 2x3 = -5 X 3 = -3

= -2,

from which, by back s ubstitution, we obtain :x: 1 this is in fact the solution of Ax = b .

4 . T he solution of Ly = b is Yt X - ill X 37 X _ 11 l 14 ' 2 - i4) 3 - i4.

= 1, Y2 = 5, Y3 = 11.

x2

=

1, x 3

=

- 3. It is easy to check t ha.t

The solution of Ux

= y {and of A x = b ) is

5. T he matrix A can be reduced to row echelon form by the following operations:

The mult.ipliers associated wit h these operations A = LU = [

1, and

2 0]3 [10 4]1

-1

is an LU-factorization of A.

To solve the system Ax= b where b

=

we first solve Ly

=

b for y , and then solve Ux = y

for x : The system Ly ;:: b is 2yl -yl from which we obtain y 1 =:=. - 1, Y2 ::::: - 1. The system Ux = y is X1

= - 2

+ 3y2 = + 4X2 = X2

from which we obtain x 1 Ax. =:::. b.

= 3,

2

1 = - 1

x 2 = - 1. It is easy to check that thi! is in fact the solution of

105

EXERCISE SET 3.7

6.

The solution of Ly

Ax = b) is 7.

is A= LU

An LU-decomposition of the matrix A= [ - :

x 1=

4,

= b, x2

;here b =

= (-:

= - 1.

is Yl = 2,

The solution of Ux

=y

(and of

= . -1.

The matrix A can be re
-2] [ 1 -1] ['

-2 -2

A=

·-1

2 -7

2

5 -1

0

-2

2

-1

5

2

-l] ['

l

-1

4

1

0 -2 0

-7

4

-I] [' -1 -1]

-1

0 0

-7

-7

- 1

- 1 -7 5

1

0

0 0

u

1 - 1 = 0 1

4, 0 (for the second row), 1,

The multipliers associated wilh these operations arc thus

-4 , aud

0

-2

A = LU = [ - 1

is an LU-factorization of A.

To solve the •ystem Ax

=b

w ha•e

b

4

,

=[

we fi.-s.,olve Ly

=b

ro, y, and then wive U x

=y

for x:

The system Ly = b is

= -4

2yl

= - 2

- 2y2

-y! from which we obtain Y1 The system Ux = y is

= -2, Y2

+

+ 5y3 = {)

4yz

= 1, YJ = 0.

X1 -

X2 -

X3

= -2 I = 0

xz- x 3 = X3

from which we obtain x 1 Ax=b.

= -1,

= 1, X 3 =

.1.' 2

8. An LU-decomposition of the matrix A =

0. Jt is easy to check that this is the solution of

[-! 0

Th• solution or Ly

Ax = b) is

Xt

= b,

wh•,. b

m,;., y, =

= -1, xz = 1, x3 = 0.

is A = LU = 4 26

0, Y>

[-! 0

= I, Yl = 0.

0

4 - 2

1

0

The solution or Ux

0

=y

l

(and or

Chapter 3

106

9. The matrix A can be reduced to row echelon form by the following sequence of operations:

A=

0 0

0 3 -1 0

3 -2 2 1 0

-1

2 1

-1

0 -1 1 0 -1 2 0

0 - 1 3 0

- 1

0

1 -2

-1 0

0 l

;1

2

0 0

1

1

-1

2

0

1

0

-1

1

0 1 1

0 0

.•"!.:.·

0 -1 1 0 0 1 0 0

01 2J1

[10

4

0

-1 0

0 l

0

0 0

1

0

The multipliers associated with these operations are -1 , -2, 0 (for the third row), 0, -1 , and thus

A-LU-

- 1

0 0 2

0

1

0 3

[-1

0

-1

1

0 1 0

0 0

1, 0,

is an LU-decomposition of A.

To solve thesystem Ax= b where h for x : The system Ly

=b

= [-;],we first solve Ly -

b lory, and then solve Ux - y

is -yl 2yl

== 5

=-

+ 3yz Y2

+ 2y3 Y3

from which we obtain Y1 = - 5, Y2 = 3, Y3 The sy.:>tcm Ux = y is

== 3

+ 4y4 =

= 3, Y4 ==

x1

-

X3

7

1.

=- 5

X3

xz

1

+ 2x4 = + X4 =

3 3

X4

1

from which we obtain x 1 = -3, x 2 = 1, x 3 = 2, x 4 solution of Ax = b.

= 1.

=

It is easy to check that this is in fact the

107

EXERCISE SET 3.7

['-'

10. An LU-decomposition of !1 =

ro!ution o[ Ly

= b, wh"e b =

(and of Ax= b) is x 1

0 0]

A= LU = ['

12 is 2 0 -1 -4 - 5

[!lis = y,

= -1 , x 2 =

X3

0

0

4

0

0 2 0 -1 -1

=-

4, Y>

= !, X4

=

l, y,

Y<

j ,J;.

-2

0

1

0

3 1

0

0

0]0 . The

The soluUon of Ux

y

1 10 .

....,...,_....._--..._,

11. Le!, elt e2, the standard unit vectors in R3 . ,_?, ?.)__ __ c?J'!IDP..?fL2L. 1 the matriX-".A- 1S-ofitaiiiea-·E'y solvTng Ax·= ei for x = x;. Using the given LUwyj oy llrsf'solviiig ej" _then· sqlving U'x=Y-;19.!.-

x_=;= xi. (1)

The sys tem

0'_::: = 1

2yl + 4yz =0 -4yl - Y2 + 2yJ = 0 Y'2 = -

from w hich we obtain Yt =

TI. Then

Y3 = X1 -

2Xz •r

•

T

X3

the system Ux = Yt is

!

=

2"' -"3 -- -6l :r.3 --

from which we obtain x 1

=

- L xz = -1, x3 = TI· Thus X 1 =

(2) ComputatioP- of x z: T he sys tem Ly

= e2 :1yl 2yl

!11

= 0,

)/2 ::.:.

[=!]

is = 0

+ 4y2

= 1

Y2

+ 2y3 = 0

-4y1 -

from which we oblaiu

7 il

L Y3 = ! . Then the system U x = Yz is xa - 2x2 -

X3

=

0

=! X3 = 4

xz + 2:r3

fmm which we

j,., = 0, .,

(3} Computation of x 3 : The syst em Ly

m

l· Thus x,

= e3 is 3yl

2yl + 4y2 -4Yt - Y?.

0 == 0

=

+ 2y3 = 1

Chapter 3

108

from whieh we obtain y 1

""'

0, y 2

= 0, y 3

Then t he system Ux = YJ is

=

x1 - 2x2 x2

f•·om whkh

w' obtain x, - l , x,

X3

= 0

+ 2x3 = 0 X3 =

!. Thus x, [

- 1, x,

Finally, as a result of these computations, we conclude that A- 1 = [x1

0

-!] -1 .

l 12. Let e 1 , e 2 , e 3 of! the s tandard unit vectors in R 3 • The jth column x i of the matrix A - .L obtained by solving the syst em Ax = ej for x = Xj. Using the given LU-decomposition, we do this by .first solving Ly = e1 for y = y 1 , and then solving Ux = Yj for x = x;.

The sdu tion of L. y - c, io y,

• nd the solution a[ U x

[

The sohotion of Ly

c,

The solution of Ly

e 3 is Yl -

is y 3

[:].

a nd the.olution of Ux = Y2 is

m,

•nd the solution of V x

r::J·

y , is x,

x,

y 3 is x,

m. [

7

_..2._] 14

* -t. . 'l

7

13. The

nta.t l

l

r.i

ix il can 1.-c reduced to row echelon form by Lhe followin g sequence of oper ations: A:.::

H 1

i

1

0 2

-t

r

0 2

2

-11lj 0 -t

1

[-; ! -ll

2

where thfl a.o:;sociated multipliers are

0

I

2 1 0

-:]

2 I

1

2 1

--+

2

-:]

-t

=U

2, -2, 1 (for the leading entry in t he second row) , and

- J. This leads t.o the factorization A = LU where L

=

0

If instead we prefer a lower 2 1 l

t.ri<mgular factor t hat has 1s on the main diagonal, this can be achieved by shifting the diagonal entries of L to a diagonal mat rix D and writing the factorization as

EXERCISE S ET 3.7

109

=

14 . A =

-2

-4

= L'DU

l1

2 2 1

0 0 63

0

0

1

15.

(a) This is a permutation matrix; it is obtained by interchanging the rows of J.z. (b) T his is not a. permutation matrix; the second row is not a. row of h. (c) T his js a permutation matrix; it is obtained by reordering the rows of 14 (3rd, 2nd , 4th, pt ).

16.

(b) is a permutation matrix. (a) and (c) are not permutation matrices.

17. The system

b i<equivalent top->

p -> b where p->

P

[! :].

Using the given

decomposition, the system p - l A;x = p - l b can be written as 1

0

3

1 -5

LUx= [

We solve this by first solving Ly = p-lb for y, and then solving Ux = y for x. The system Ly = p - I b is Y1 = 1 Y2 = 2 3yl - 5y2 + =5 from which we c•bt..ain y 1

=

1,

'!/2

= 2, y3 x1

from which we obtaiu Xt = 18. The system Ax

= 12.

Finally, t he system Ux = y is

+ 2x2 + X2 +

1 4X3 = 2 17x3 = 12 2x3 =

= g, This is the solution of A x = b.

x2 ::::: -

h iseq uivolent to p- 'Ax

P -' b whe>·e p- '

P

i (\ 0]

[

o

1 . 1 0

Using the given

decomposition, the syst.em p-l Ax = p-lb can be written M

The solution of Ly = p - l b is y 1 = 3, y 2 X1 = X2 = 0, X3 = 0.

= 0, Y3 = 0. The solution of Ux =

y (and of

Ax= b ) is

19. If we interchange rows 2 and 3 of A, then the resulting matrix can be reduced t o row echelon form without any further row !nterchanges. This is equivalent to first multiplying A on the left by the corresponding permutation matrix P: 3

-1

-1

0

2

-1

-1

2

[3

:1 J

3

110

The reduction of P A to row echelon form proceeds as follows:

PA

=

3 0 [3

-1 2 -1

This corresponds to t he LU-decomposition

PA =

3 0

0] [3 0 0] [1

-1 2

1 = 1

[ 3 -1

0 1

0 2 3 0

0 0

of t.he matrix PA, or to the following P LU-decomposition of A

A=

Note that, since pThe system Ax

=b

1

=

!

1 0 0] [3 0 0] [1 [0 1 0 3 0 1 0 0

0

1

0

2 0

0

= p - 1 LU

1

0 as A = P LU.

decomposition ca.n also be

is equivalent to P Ax= P b =

and , using the LU-decornposjtion

obtained above, this can be written as

[3 o olll -·4 !ol [x [-2] = 1

PA x == DUx ==

Finally, the solutjon of Ly Ax = b ) is x 1 = x2 = 20.

0

2

0

0

1

3 0 }

0

0

= Pb is Yt = X3

y 2 = 2, y 3

= 3.

}

x 2]

4

X3

1

Pb

= 3, and the solution of Ux =

y (and of

If we interchange rows 1 and 2 of A, then the resulti ng matrix can be reduced to row echelon form withoul any furt her row interchanges. This is equivalent to first multiplying A on the left by the corresponding permutation matrix P:

The LU-decomposit ion of t he matrix PA is

P A = LU = 2

0

-3

0

0

-;] 1

a.nd the corresponding FLU-decomposit ion of A is

Since P

1

= P , this

<.:
111


The system Ax

!]

= b = [_

is equivalent to PAx

= Pb =

using the LU-decomposition

obtained above, this can be WTitten as

PAx = LUx = Q -3

The solution of Ly = Pb is y 1 = 5, is X1 = - 16, X2 = 5, X3 = 4. 21.

-;] [ ::] = [

1

2

=

Y2

0 y3

0

1

-2

X3

= Pb

= 4; and the solution of Ux =

y (and of

Ax= b )

(a) We have n = l 0 5 for the given systern and so, from Table 3. 7.1, the number of gigaflops required for the forward and backward phases a.re approximately Grorwa.r

=

X

10-9 =

Gback wt>rd =

n2

X

X

10-

9

w- 9 =

= (105 )'2

x

s

106

10- 9

= 10

X

;:::::

670,000

Thus if a computer can execute l gigaflop per second it will require approximately 66.67 x 104 s for the forward phase, and approximately 10 s for the backward phase. (b) If n = 104 then, from Example 4, the total number of gigaflops required for the forward and backwo.rd phasE>.s is approximately x 10 3 + 10- 1 ;::::: 667 . Thus, in order to complete task in IE'.ss than 0.5 s, a computer must be a.ble to execute at least = 1334 gigaflops per second. 22.

(a)

If A has such au LU -decomposition, it ca.n be written us

This yields the syslem of equations

x

= a,

y

= b,

w:z;

= c,

wy + z

=d

and , since a =I= 0, this system has the unique solution

x =a, (b )

y= b,

711

c

= -, a

z=

(ad- be) a

fl-orn the above, we ha\'e

[: DISCUSSION AND DISCOVERY 01. The rows of Pare obtaine
PA=

3 - 11

0 2

- 3 12 7 1

CHAPTER 4

Determinants EXERCISE SET 4.1

:1 =

1.

2.

1:

= (4)(2) _ (1){8) =

7

3. ,-S

-7 -2

4.

5.

6.

s- 8 = o

(J2)(/3) _ (4)(./6) =

I

-3v'6

a- 3 -3 a -5 2 = (a- 3)(a - 2) - (5)( -3) I -2 5

= (a2 -Sa+ 6) + 15 =

a2

3

R

4

-2

1

4

3

5 6

- 7 = ( - 20 - 7 + 72) - (20 + 8·1 + 6) = 45- 110 = -65

1 0

2 -5 2

-1

8.

3

7

3

0

2 1

- 1

9

c 2

-4

4

c- 1

11. (a) {b) (c) (d) {e) (f)

-

5a + 21

7 6 1 -2 = (- 8 - 42 + 240)- (18 + 140 + 32) = 190- 190 = 0

l

10 .

= 12 + 10 = 22

1 = (-5){-2) - (7)(-7) = 10 + 49 = 59

IV:

1.

9.

(3}(4)- (5)( - 2)

1

2

= (O- 5 + 42) -

{0 + 6 + 35) = 37 - 41 = - 4

0 5 = (12 + 0 + 0) - (0 + 135 + 0)

= - 123

-4 3

c2 = {2c -l6c2 + 6(c - l)) - (12+ c3 (c- 1) - l6)

= - c• +c3 -16c2 +Be- 2

2

{4, 1, 3, 5, 2} is a.n odd permutation interchanges). T he signed product is {5, 3, 4, 2, 1} is an odd permutation (3 interchanges). The signed product. is -atsa.z3a34a.42ast· {4, 2, 5, 3, 1} is an odd pArmuta.tion (3 interchanges). The signed product is -al4ll22llJS043ll51· {5, 4, 3, 2, 1} is a.n even permutation (2 interchanges) . The signed product is +ats<124a33a.czas1· { 1, 2, 3, 4 , 5} is an even permutation (O interchanges). The signed product is +au {1 , 4, 2,3, 5} is an even permutation (2 interchanges) . T he signed product is 11 8

Exercise Set 4.1

119

1 2. (a), (b) , (c), and (d) are even; (e) and {f) are odd. 13. det(A) = (,\- 2)(>. or >.= - 3. 14.

+ 4) + 5 = >.2 + 2>. - 3 =(..\ -1)(>. + 3}. Thus det(A) = 0 if and only if/\= 1

>. = 1, ,\ = 3, or ,.\ = - 2.

15. det(A) = (>. - 1)(>. + l }. Thus dct(A} = 0 if and only if A "" 1 or .A= - 1.

16. A = 2 or ,\ 17. We have

= 5.

1; 1-=_1xl = x( l - x) + 3 = -x2 + x + 3, and 1 0 2 x ] 3

-3 -6 X-

= ((x(x -

5) + 0- 18) - ( -3x- 18 + 0)

= x 2 - 2x

5

Thus t he given equation is if and only if -:t 2 roots of this quadro.tic equation are x = 3 {33.

*

+ :-r + 3 =

.x 2

-

2x, i.e. if 2x 2

-

3x - 3 = 0. The

18. y=3

19. (a)

(c)

1

0

0

0

0

0

- 1 0

1

2

7

0 0 0

I

-4

()

2 0

20. (a)

0 0 2 0 0 0 2

40

Mn

=1 ;

M21

=

M3t =

= (1)(1)(2) (3)

7

0

0

2 10

0 - 1 -- 23

=6

1-2 7

(b)

8

0 0

u

1

1

1

0 0 0

0

2 3

0

0

2

= (1 )(2)(3)(4) =

24

4

= ( - 3)(2)( -1)(3) = 18

3

-11 = 29. c11 = 29 4

,-2 !I= 1

0 '1

0 0 =0 3 0 3 8

a

= 23 =

100 200

21.

-3 1

0

l

{c)

(b)

1 2

0

2

= (1)(-1)(1) = - 1

0 0 1 2

G21 = 11

_ 31 = - 19, C31 = - 19 1

Ml2

=

M22

=

M32

-114 = 2l,Ct2 = - 21

!I=

13,c22

=

13

_ 31=== -19, C32 == 19 1

M13

= 21,

M23

= -5, C23 = s

M33 == 19, c33

= 27

= 19

120

22.

Chapter 4

M21

6

M12

=12,012 = -12

M1s

= 3, C1a =

3

2, C21

=- 2

M22

= 4, Cn = 4

M2a

= l , Oza =

-1

) O,Ca1

=0

Ma2 =

= 6, cn =

Mll

=

21 =

M31

6

0 0 23. (a)

M 13 = 4 1 14 4

(c)

(ct)

24.

= (0 + 0 + 12) -

(12 + 0 + 0) = 0

(f)

26. (a) (b) (c) (d) (e) (f)

0 13

4

-1 6 1· 14 = (8 - 56+24)-(24+56-8)=-96 1 2

4

l

Mn= 4

0

4

3

M 23 = 4

M21=

=0

2

0 23 = ·96

6

14 =(0+ 56 +72)-(0+8+168) = -48

0 22 = - 48

2

- 1

1

1

o

1

3

6

14 = {O+I4+18)-(0+2 - 42) = 72

c21=

-n

2

(a) M3z = - 30, C32 = 30 (c) .M41= - l,C41= l

25. (a) (b) (c) (d) (e)

Ms3 = 0, Cas = 0

3

4 1 (b)

O,C3 2 =0

{b) M44 = 13, 044 = 13 (d) M24 = 0,024 = 0

det(A) = (l) Cu + (- 2)C12 + (3)013 = (1)(29) + (- 2)(-21) + (3)(27) = 152 det(A) = ( I)C ll + (6)C21 + (-3)C31 = (1){29) + (6)(11) + (-3)(-19) = 152 det(A) == (6)C 21 + (7)C22 +(-1 )023 = (6)(11) + (7)(13) + (-1 )(5) = 152 det.(A) = (-2)C12 + (7)C22 + (1)032 = (-2)(- 21) + (7)(13) + {1){19) == 152 det(A) = ( -3)CJI + (l)Ca2 + ( 4)Caa = ( -3)( - 19) + ( 1)(19) + (4)( 19) = 152 det(A) = (3)013 + (-l)Czs + (4)033 = (3)(27} + (- 1)(5) + (4)(19) = 152 det(A) = (l)Cll + (l)Ct2 + (2)013 = (1)(6) + (1)(- 12) + (2)(3) = 0 det(A) = (l)011 + {6) C21 + (3)Ca 1 = (1)(6) + (3)(-2) + (0)(0) = 0 det(A) = (3)021 + (3)022 + (6)023 = (3){-2) + (3)(4) + (6)(-1} = 0 det(A) = (l)Ou + (3)022 + (I)Cs2 = (1)(-12) + {3)(4) + (1)(0) = 0 det(A) = (O)C31 + (l)C32 + (4)C33 = {0)(0) + (1)(0) + (4)(0) = 0 = (2)013 + ( G)02J + {4)033 = (2)(3} + (6)(- 1) + (4)(0) = 0

27. Using column 2: det{A} = 28. Using row 2: det(A)

=

;j = (5){-15+ 7) = - 40 (-4)

= (1)( -18) + (-4)(12)

= - 66


30. Jet( A) ::::: k 3

:n.

-

8k 2

-

121

IOk + 95 3

3

3 5 3 2 -2 - (3) 2 2 4 2 10

Using column 3: det(A) = ( -3) 2

5

= (- 3)(128) -

2 - 2 0

(3)( -48)

= -240

32. det(A) = 0 33. By expanding along the t hird column, we have

sinO -cosO

0 0 = (1)

cos B

sin8 s inO- cosO sinO+ cosB 1

I

sinB cosO' . =sin 2 fJ+cos 2 B = 1 -cos 8 smB

for a ll values of 0. 34. AB

=

bf)

a.nd BA

=

ce]. Thus AB = BA if and onJy if ae + bf = bd + ce , and

bd

this is equivalent to t he condition that

36. If A =

I

[ac db] , then tr(A ) = a + d, A

tr( ,1) 2l tr(A2)

1 tr(A)

I= 2 I 1

2

lb

0

-

e d ·-

cl=

f

[a +bed 2

=

b(
ab +bdl 2 b d 2 1, and tr(A ) J

co+ c c +

=a2 + 2bc + d2 .

Thus

I

2 2 a+ d 1 1 2 a.2 +Zbc+ d2 a +d =2 ((a+d) -(a +2bc+d ) =ad -bc=det(A).

DISCUSSION AND DISCOVERY D 1.

is always t\n integer , each elemf!ntary product is an integer and so of the elementa.ry products is an integer

Lhe product of integers

the

D2. The signed ele mentary p roducts will all be ±(1)(1) · · · (1) and half equal to -1 . Thus the determinant. will be zero. D3.

= ±1 , with

half of t hem equal to +1

A 3 x 3 matrix A can have as many as six zeros without having det(A) = 0. For example, let. A be a diag0nal matrix with nonzero diagonal entries.

D4. If we e xpand a long the first row, the equation

X

y

a1

bt b1

a ..2

1

= 0 becomes

1

and thjs is an equation for t he line through t he points Pt (a 1 , bt) and P2(a2, b2). D5.

If ur = (a1 ,a2, a 3) and ' vT = then each of the six elementary products of uvT is o f Lhe form (a1b;,)(a2bi?)(a3bj3 ) where {j 1 ,iz,]J} is a permutation of {1,2,3}; thus each of the elementary products is equal to (ata2as)(b1b2b3).

Chapter4

122

WORKING WITH PROOFS Xt

Pl. If t he three points lie on a vertical line then XJ = x2 = xa = a and we have

l/1

1

yz 1 :r:s l/3 1 x2

=

a l7l 1 a yz 1

= 0.

a !13 1

Thus, without loss of generality, we may assume that the points do not· lie on a vertical line. In this case the points are collinear if and only if %,q - x 1 = :1!2 - Xl The latter condition is equivalent to (Y3 - Yt)(x2 - xt) = (Y2- Y1)(:r3- x1) wh.ich can be writ ten as: (X2Y3 - X:;Yz) - (XtY3- XJYl)

+ (XJY2- X2Y1) = 0

On the other hand, expanding along the third column, we have:

·Thus the three points are collinear if and only if

Xl

Yt

x2

v2

1 1

XJ

l/3

1

= 0.

P2. We wish to pwve that for each positive integer n, t here are n! permutations of a set {it , )2, . .. , j .,.} of n distinct elements. Our proof is by induction on n . Step 1. It is clear that there is exactly 1 = 1! permutation of the set {j 1 } . Thus the statement is true for the case n = 1. Step 2 (induction srep). Suppose that the statement is true for n = k, where k is a fixed integer 1. Let S = {j1, j2, ... ,jk, )k+I }- Then a permutation of the setS is for med by first choosing one of k + 1 positions for the element and then choosing a permutation for the remaining k elements in t he remaining Jc positions. There are k + 1 possib ilities for the first choice and , by the hypothesis, k! possibilities for the second choice . Thus there are a. total of (k + l )k! = (k + 1)! of S . T his shows that if the statement is true for n = k it must also be true for n = k + l. These two steps complete the proof by induetion.

EXERCISE SET 4.2 1.

(a)

=

det(A)

=

(b) det(A)

= -8 -3 = -11 2 1

-1

2

4

5

-3

6

2 det(AT) = -1 3

2. (a) det(A)

3. (a)

1

0

1

3 0

0

0

- 8 -3= - 11

1

=

(24 - 20 - 9) - (30 - 6 - 24) = - 5 - 0

=-

5

5

2 -3 = (24 - 9 -· 20) -- (30- 6 - 24) = -5- 0 = -5 4

3

331

9

22

- 2 12 0

11 =

4

3

6

= det{AT) = 2

3

0

dct{AT) = -2 1 3

2

= (3) (!) (-2)(2) = -4

(b)

det(A) = d et(AT) = 17

123

EXERCISE SET 4.2

(b)

9 2 -3 = 0 5 3

3 - 1

1

l

3 0 0

-17

4

5 0

-2

and third columns are proportional)

1 = (3)(5)(-2)

4. (a) -2

(a)

e f g h i

a

a = (- 1) g c d

b

3a

(b)

-d 4g

(c)

(d )

6.

c

b h

= (- 1)(-1)

i e 1

3c

3b -e

4h

1:+ i

a

b

e

f

e f

g

h

i

d g

-3a

-3b

d g-4d

h- 4e

e h

I =(-1)(-1)( - 6)= - 6

=

6 - tl

=

-6 -2

b

I

a 12) d

4i

g

b

c

e f

= (- 1'2)( - 6) = 72

h

= -6

d

e

g

h

-3cl f = (-3) d i

I

(c)

:1 = - 16- 24 = -40 2 -6 -4 -2 -8 - 10

c

f = (-

i

-3a -3b

1 - 4/

i

Ia

(b) 0

(a) 6

(b) <.let(-· 2A)

d

c

h

-3c

f.

c

= (- 3) d e 4i •1g 4h

b I· h

e

b

g

a+g d

=

a

b c -e -f

(l

- ! = (3) -d 4h 4i 4g

7. (a) det(2A) = ,-2

8.

(c) 0 (proportional rows)

(b ) 0 (identical rows)

d 5.

= - 30

2 - 1 (- 2) 3 det(A) = ( - 8) 3 2

3

4

5

b e

f

9

h

l

c

= (-3)(-t)) = 18 (d) - 12

6

22 det(A)

= (-160 + 8 -

a

=4

,-1 3

2 4

1 = 4( - 4 - 6) .., - 40

288) - (-48 - 64 + 120)

= · ·440 -

1 = ( - 8)((20 - 1 + 36) - (6 + 8 - 15))

8 = - 4t18

= (- 8)(55 + 1) =

-448

(a) del(-4A) = - 224 = (-4) 2 (-14) = (-4) 2 det(A) (b) det(3A)-= -63 = (3) 2 (-7) = (3) 2 det(A)

9 . lf x

= 0, the given matrix becomes

tiona!, we have det(A) = 0. If x

A

and, since the first and third rows are r-ropor-

o

0 -5

= 2, the given matrix becomes B

and second rows are proportional, we have det(B) ::::: 0.

=

0

i0 - ;]5 and, since the first

Chapter 4

124

LO. If we replace the first row of the matrix by the sum of the first and second rows, we conclude that

det

b+c a [ 1

b+a]

b

· [a+b+ c b+c+ a c+b+a]

= det

c 1

l

a 1

b

a

1

1

=0

since the first and third rows of the latter matrix are proportional . 11. We use the properties of determinants stated in Theorem 4.2.2. Corresponding row operations are as indicated.

det(A)

=

3 0

6 0

-2

1

-9

1 -2 = (3) 0 5 -2 2 -3 0 -2 5 ·-1

1

:;:; (3) 0 0

= (3)( -1) 0

= (3)( -1)(-

10)

-2 5

A common factor of 3 was

taken from the first row.

The second and third rows were interchanged.

-1

5 0

0

-3

the first mw "''-' to the third row.

2 -3

1

12. det(A)

2 0 1

-2

= 30

=5

13. We use the properties of determinants stated in Theorerr1 4.2.2. Corresponding row operations are as indicated . 1

-3

det(A) = -2

4

5

-2

-

0 0

2

0

5

-3 -2 13

0

1

-3

0

0

1

-l

0

13

2

1

-3

0

0

1

- 2

0

0

2

= {- 2) = (-2}

= 33

-3 -2 - 2

l

=

= (- 2)

14. det(A)

0

0 l

2

2 times the first row was added to the second row.

-5 times the first row was added to the third row.

2

2

1

17

en =

-17

A fac tor of -· 2 was taken from the second row.

- 13 times the second row was added to the third row.

EXERCISE SET 4.2

1 5.

125

We use t he properties of de terminant s stated in T heorem 4.2.2. Corresponding row opera.t.ions are as indicate'i.

det(A)

=

1 5

-2 -9

3 1 3 6 2 - 6 - 2 6 8 1

-1

2

=

=

-5 times row 1 was added to row 2; row 1 was added to row 3; - 2 times row l was added to row 4.

-9 - 2 0 - 3 -1 12 0 -1

-3 1 -9 - 2 -3 - 1 108 23

1 0 0 0

2 1 0 0

1 0 0

2 -3 1 - 9 0 - 3 0 0

0

=

1

2 -3

1 0 0 0

- 12 times r ow 2 was added to row 4.

1 -2 - 1

36 times row 3 was added row 4

_. ]

- l3

= 39

=6

16.

det(A)

]7.

We usc the properties of deter minants stated in T heorem 4.2.2. Corresponding row operations are as indicated. 0 det(A )

=

1

1

1

1 l

1

1

2

2

1

'2

l

3

1

3

0

2

0

0

3 - :;l

3

1

= (- 1) 1 0 '2

3 I

2

1

3 l -3 2

l

l

1

1

I

0

I

3

2

1

3

3

0

2

0

0

l

1

1 1

2 l

1 1

- l

2

-3

1

1

0

1

0 0

(

1

- times row was added to row 3; times row 1 was added to row 4.

-3

'2

0

T he first and second rows were interchanged.

A facto< of was from the first row.

0 0

I

1

2 1

l

3

l

:l

1

:i

0

= (-1)

0

2 1

- 3 0

= (- 1)

1

2

l

3

3

2

1

2

1

l

times row 2 was added t o row 3; - 1 t.imes row 2 was added to row 4.

l

0

-3

-3

0

1

-3

2

- 3 1 0

1

2

1

1

0

0

1

0

0

- 3

l

1 I

2 2

3

A factor of from row 3.

was t aken

Chapter 4

126

= (-1)

(

= (- 1)

18. det.{A)

=

19.

(a)

a2 03

b3

a1

+ b2 + Cz = as+ b3 + C3

a1

bt

bz bJ

+ b1

0

0

1

1

0

0

0

-2

!

times row 3 was added to row 4.

1

=

=2

C)

Add -1 times column 2 to column 3.

C2

C:J

2a 1 2a2 2a3

Gt

a1 -

b1

Ct

02

02- b2

C2

03

GJ -

b3

C3

=

C2 CJ

a1

- bl

Ct

02

- b2

C2

,a3

- hJ

CJ

GJ

bl

Cl

02

b2

C2

(13

03

CJ

a1 + b1 t

+ b2t

OJ

Gt

= (1 - t 2)

(1.1

t2 ) bt Ct

+ btt

Gl-

Ot

b2 (1.3- b3

(1.2-

CI

ca

Add -1 times column 1 to column 2 .

Factor of - 1 taken from column 2.

+ b3t

a2

a1

+ b1t

(1-

+ bzt

·a3

t 2 )bl

+ b3t

bt

b2

b3

Ct

C2

CJ

a2

GJ

b:z b3 C2

J Add column 2 to column 1.

C2

Factor of 2 taken from column 1.

a3t+b3

att +b t

= (1.-

Add -1 times column 1 to column 3.

bz b2 + C2 bJ b3 + ca

.:l3

C)

= -2

bl bt + Ct

02

a1 - b1 (12 - b'J a3 - bs

az + b2

=2

(a)

1 1

al

02

a3 + b3

20.

2 1

(

+ b1 + C1

G!

=

(b)

1 1

- 2

bt b2

al

(

1 0

CJ

a2

(1 -

+ bzt ( 2)b2

a3

+

{1- t 2 }b3

Add - t times row 1 to row 2.

Factor of (1 - t 2 ) t.Rken from row 2.

Add - t iin:.es row 2 to row 1.

127

EXERCISE SET 4.2

a2

b1 + ta1 b2 + ta2

a3

b:1

al (b)

+ ta3

a3

b1 b2 b3

X

x2

a1

-

21.

a2

1 det{A) = 1 1

+rb1 + sa1 C2 + 1'b2 + sa2 C3 + rb3 + sa3

+ rbt + sa1 la1 b1 b-l c2 + + sa2 = c3 + rb3 + sa3 a a UJ c1

Ct

r2

Add -t times column 1 to column 2.

Add -s times column 1 to column 3. Add -r times column 2 to column 3.

C)

c2

c3

x2 1 X 1 x Y2 _ x2 =(y-x)(z - x) 0 1 y2 = 0 y - x z2 0 1 0 z- x z2 - x2

y ;;

1 X =( y- x)( z-x) 0 1

x2

x2 y+x z +x

I

xl

y + = (y - x)(z - x) (z- y) 0 0 z- y

+ 3b)

22.

dct(.4) :... (a - b) 3 (a

23.

If we add the first row of A to the second row, the rMult is a lfla.trix B that h as two identical row$; thus dct( A) = dct(.H) = 0.

24 .

If we add of the first four rows of A t o the last row, t.he result is a matrix B that h as a. row of thus det.(A) = det(B) = 0

(a) \Ve have dct(A) = (k- 3)(k ···· 2) - 4 = k 2 - 5k + 2. Thus, from Theorem 4.2.4, the matrix .4 is invertible if and if lc2 - 5k + 2 -f:. 0, i.e. k #- 5±

fTI .

(b )

We ha..-e dE>t.( A)

A is invertible if and only if 8 + Bk

(a)

k

tlct(2A- 1 )

= (27)(7) = 189

(a) det(-· A) (c)

&.

We have

''=

=

= (- l) 4 det{A) =

det(AB) =

0

2 6

10

"

-1

- 1

1 ) = det(A)

1

=7

¥ 1

= (8){7)

(1)(-2) = - 2

= [

2 -2

k=L.!

(b) det(A

= 24 det(A) =-32

= det(2A)

[i

=

1 z3 det(A)

T hus, from Theorem -1.2.4, the matrix

k =/' ... 1.

{h )

23 dct{A- 1 ) 1 ( d) clet((2 A)-t ) det(2A)

:8.

f 0, i.e.

i: ±2

(a) det(aA) = 33 det(A) (c)

:1 . ,. 8 + 8k.

+

=(

!

0

2

10

4 15 22

-5 2 -6 = 0 - 23 10

1 = 56 1 (b) d et.t.. 1,t - 1) = det(A} :;:;

(d) det(A 3 )

= { det(A)) 3

1

-2

= (- 2)3 = -8

t hus 22 - 23

4 - 5 3 9 2 2

= 21

3 9 2 2

) = 2(6 - 18) = -24

Chapter 4

128

On the other hand, det(A) det(A) det{B). 30.

= (1)(3)( -2) =

-6 and det(B)

=

We have AB = 0 0 2

5

= (2)(1){2) = 4. = -170.

and so det(AB)

0 l

10

0

2

det(A) = 10 and det(B) = -17. Thus det(AB)

Thus det(AB)

=

On the other hand,

= de t(A)det(B).

31. The following sequence of row operations reduce A to an upper triangular form:

-2 -9

H 2

8

-2

3 6 -6 6

-}·

3

-9 -3 -1

1 0 12

-1

0

-2 1 0

0 0

0

-2

3 -9 -3 -1

0

23

0

108

1 0 0

3 -9

1]

-2 -1

-3

0\·' -13

The corresponding LV -factorization is

A=

[!

-2 -9

-1 2

3

2

-G

8

6

-2

1

0 0

0

1

12

- 36

0

I)

'] = [ 1 1

2

and from this we conclude that det(A) = det(L) det(U) 32.

3 1 - 9

-2

0

-3

- 1

0

0

- 13

1] =LU

= (1)(39) = 39.

The following sequence of row operations reduce A to an upper triangular form : I

;r

.. l

1

3

2

0 2 1

1 1 2

-2

'2

2 1

1

'J

l

l

2

3

I

2 l

2 1

0 0

-1 1

1 2

-;]

1

-4

0 0

3

_;]

1 0

-2

2 1

6

The corres ponding L U-decomposition is 1

A=

0

2 1

3 1 1 2

0

l] "' 3

1

-2 2

0 - 1

l

1

0

3

l

0

2 1 1

2 l

0 0

0

_; ] = LU - 2 6

a.nd from this we conclude that det.(Jt) = dct(L) det(U) :::: (1)(6) = 6. 33. U we add the firs t row of A to the second row, using the identity sin 2 B + cos 2 0 :::: 1, we see that sin 2 o det( A) = cos 2 o

1

sin 2 /3 cos 2 /3 1

sin 2 1 sin 2 o 2 cos 1 = 1

1

1

sin2 f3

sin 2 1

1 1

1 1

=0

since the resulti(lg matrix has two identical rows. Thus, from Theorem 4.2.4, A. is not invertible.

34. (a) .A=3or.A=-1 35.

(b) .A=6or.A =-l

(c) >. -= 2 or >. = -2

Since det(AT) = det{A), we have det{AT A)= det(AT) det(A) = (det(A)) 2 = det(A) det(AT) = det(AAT ). (b} · Since det(AT A)= (d et (A)) 2 , it follows that det(AT A) = 0 if and only if det(A) = 0. Thus. from 'T'heorem 4.2.4, AT A invP.rt.ihiP. if ano only if A is invertible.

(a)


36. det(A- 1 BA) 37.

129

= det(A -I) det(B) dct( A) =

llxii 2 IIYII 2 - (x. y) 2 == (xi+

2

Jx

1

IYl

x

2

Y2

1

2

+ IXJ

X:J I 'YJ

YJ

+ lxz

Y2

- 2XtY2X2Yl

38.

(a) We: have det We have

T

2

:r

3

Y3

+

1

2 5 o - 1 3 2

= (x1Y2 -

:r2yd 2 + (x1y3- X3Y1) 2 + (x2Y3- X3Y2) 2

+ xi y§- 2XJY3X3Yl +

2

=

5

= det(B)

vi+ v5>- (XlYl + XzY2 + XsYa) 2

=5 -4 = 1 and det

l 2 0

(b)

dct(B } det(A}

+

= 3- 6

1 = 2(5 - 4) = 2 and

= -3;

3

0

2 - ·3

1

2X2Y3X3Y2

+

thus det(M) =(I){ -3)

= -3.

0

o=

(3)(1 )( - 4) = -12;

th\18 det(M) =

8 -4

(2)( -12) = - 24.

39.

(a)

det(M)

(b) det(M)

=

3 5 -2 6 2 3 5 2

121

=

jo

2 0 1 2 () l

1 3 = (fi + 4) 0 12 0 -4

5 12

- . 13

= (10)(1) 112 - 4

121=

- 13

-1 080

(1){1) ,.,, l

DISCUSSION AND DISCOVERY D l. The matrices sin-gular if and only if the corresponding determ inants are zero. This leads to the system of equations

from whir.h it follows t.hat s

D2. Since det(AB) det(BA).

=

and t =

= det(A) det(B) == det(B) d P.t(A) = det(BA),

D3. If .4 or B is not invertible then either det(A) = 0 or det(B ) det(A) det(B) = 0; thus AB is not invertible.

it i;, always true that det(AB)

= 0 (or both).

It follows that det{AB)

= =

130

Chapter 4

D4. For convenience call the given matri'V A n· If n = 2 or 3, then An can be reduced to the identity matrix by interchanging the first and last rows. Thus det(An) = - 1 if n = 2 or 3. lf = 4 or 5, then two row interchanges are required to reduce An to the identity (interchange the first and last rows, then interchange the second and next to last rows). Thus det (An) = +1 if n = 4 or 5. This pattern continues and can be summarized as follows:

= =- 1 det(A2k) = det(Azk+t) = +1

det(A2k)

fork = 1,3,5, .. .

fork= 2, 4,6, .. .

DS. If A is skew-symmetric, then det(A) = det(AT) = det(-A) = (-l)ndet(A) where n is the size of A. It follows that if A is a skew-symmetric matrix of odd order, then det(A) =- det(A) and so det(A) = 0. D6. Let A be an n x n matrix, and let B be the matrix that results when the rows of A are written in reverse order. Then the matrix B can be reduced to A by a series of row interchanges. If n = 2 or 3, then only one interchange is needed and so det(B) = - det(A). If n = 4 or 5, then two interchanges are required and so det(B) = +det( A). This pattern continues;

= - det(A) det(B) = + det(A) det(B)

for n

for n

= 2k or 2k + 1 where k is odd = 2k or 2k + 1 where k is even

D7. (a) False. For example, if A= I= /2, then det(J +A)= det(2/} = 4, whereas 1 + det{A) = 2. (b) Thue. F'tom Theorem 4.2.5 it follows that det(A") = (det(A))n for every n = 1, 2, 3, ... . (c) False. From Theorem 4.2.3(c), we have det(3A) =3n det(A) where n is the size of A. Thus the: stu.ternent is false except when n = 1 or riet(A) = 0 . (d) Thue. If det (A) = 0, the matrix is singular and so the system Ax = 0 has infinitely many solutions.

DB.

(a) True. If A is invertible, then det(A) =F 0. Since det(ABA) = det(A) det(B) det(A) it follows that if A is invertible and det( ABA) = 0, then det(B) = 0. (b) 'frue. If A ,.,; A - l , then since det( A - l) = det1(A) , itiollows that {det( A ))2 == 1 and so det (A) = ±1. (c) True. If the reduced row echelon form of A has a row of zeros, then A is n ot invertible. (d) 'frue. Since = det(A), it follows that det(AAT) = det(A) det(AT) = (det{A)) 2 :?: 0. (e) True. If det.(A) f 0 then A is invertible, and an invertible matrix can always be written as a prod uct of elementary matrices.

D9. If A = A 2 , then det(A) = det(A 2 ) = (det(A)) 2 and so det(A) = 0 or det(A) = 1. If A= A3, then det(A) = det(A 3 ) = (det(A)) 3 and so det(A) = 0 or det(A) = ± 1. DlO.

Each elementary product of this matrix must include a factor that comes from the 3 x 3 block of zeros on the upper right. Thus all of the elementary products are zero. It follows that det(A) = 0, no matter what values are assigned to the starred quantities.

Dll . This permutation of the columns of an n x n matrix A can be attained via a sequence of n- 1 column interchanges which successively move the first column to the right by one position (i.e. interchange columns 1 and 2, then interchange columns 2 and 3, etc.). Thus the determinant of the resulting matrix is equal to (-1)"- 1 det(A).

EXERCISE SET 4.3

131

WORKING WITH PROOFS Pl. If x

and y = [ ] then, using cofacto• expansions along the jth w lumn, we have

[]

det(G)

= (x1 + Y1)C1; + (xz + Y2)C2; +

· · · + (xn +yn)Cn;

P2. Suppose A is a square matrix, and B is the matrix that is obtained from A by adding k times the ith row to the jth row. Then, expanding along the jth row ofB , we have det(B) = (ajl

+ kao)Gjl + (aj2 + ka;2)Cj2 + · · •+ (a;n + ktJ.in )Cjn

= det(A) +kdct(C)

where C is the mat.rix obtained from .t1 by replacing the jth row by a copy of tho ith row. Since C has two identical rows, it follows that det( C) = 0, and so det( R) = det(A).

EXERCISE SET 4.3 1. The matrix of cofactors from A

)s

C

=

&

Thus adj{A)

= lr - 2

-!

2

and A-t

-! -s

det(A) = (2)(-3) + (5)(3)

+ (5)( -2) = -1.

3

= -a.dj(A) =

-

3

2 -2 -3

0 I

2

3

3

0 -1 3.

The matrix of cofactors from A is C 2 6

adj(A) =

4.

adj(A)

[

o 4

[

5.

Xt

3

2 0 0]0 , and det (A) = (2}(2) + (- 3)(0) + (5)(0) = 4. Thus [4 6 2 6 4

[2 6 4]

and A- 1 = -1 adj(.4) = -1 0

I]

1

.

1

0 0 2

0

12

-6

17 18 9 :::: - = 16 8

o

4 6 = 4 oo2

4

0 12 29

I! = 17

·1]

G 0 0 2

=

13

X2

=

j;

!I

[

1 29 12

26 13 = - =16 8

l

-·2

;]

Chapter 4

132

6. :r = .:....___ _;_

- 153 =- =3

y

- 51

7.

8.

x=

XJ

=

I t=

-4

-1

2

4

-20

-3

2

- 1 2

-20

1 -4

1

-1 2

2 -3

2

4

-20

2

-3

2

-4

1 2

1 4 2

1

-4

1

144

=- 55

1

y=

1

6

61 - 55

z=

= -

4 -1 2 -3 2

- 1

2 2 -3

4

2

6 - 1

4

-3

1

1

4

1

-2

- 1

2

-2

0

0

0

0 3

4

0

-3

1

-3

1

2 4

- 1

0

2

- 1

0

0

-3

4

0

-3

-3

]

2 - 1

0

0

-3

1

30 = - 11

- 32 - 4 1·1 -1 11 1 - .t - 2 - I -4 2 - 1 - 1

:3

2

-4 - l

1

1

-2

6

12 6 2 4

8

3

2 3

=

3 I

-4

2

1

7

9

- 2115 = - -=5 - 423

Xz

=

38

=- 11

X3

- 1 2 - 1

-32 14

2

11

1 - 1

-4 -4

3 1 2

2 - 1

-1

7

- .t

3

1

1

-4

2 7 3 1

- 32

1

- 1

-4

2

-1

ll

1

- 1

1

1

-2

=

- 1269 = 3 - 423

x4

=

1 9 1 -4 = - 3384 = 8 1 - 423 9

1

9

-4 -4

7

=

14 11

-4

- 423

- 1

1

2

4

- 1

2

4

5 -3 3 - 2 2 - l

4

8 3 Iz = 2 4 8 3

6 "'

-3

6

-2

2 3 5 3

- 1

2 =-=1 2 1

- 1

2

-3 -2

4

2

2 = -= l 2 1

- 1

2

-3 3 -2

4 2

3 5

- 1 - 1

1

2 4 2

2

-230

46

=--=-55 11

1 -3 4 2 - 1 -2 40 4 0 0 = 1 - 3 1 - 11

-2

- 32 14

- 423 4

Xz

1 9

7

-2

- I

=

=

1

-1

2

Xt

1

-4

- 1

10.

= -- 5-1 = -4

6

I

X3

= !..:-,-1

-1

4

9.

204

423 = -- = - 1 - 423

133

EXERCISE SET 4.3

=

11.

x=

4 1 3 2 -2 - 1 1 4 1

2 1

4 3 - 2 -1

3

l

2

1 2 4 2

2 2 4 4 3 0 8 5 12 3 3 6 2

4

- 2

= -= -1 2

X4

2

14

i2 6

-2 ::;: -=- 1 2

1

-2

[

0

0

15. The coefficienl matrix is 11 = solution if k #- 0 and k

i 1 2

5'

3

1

-2

2

3 1 3 -1 -1 4 - 2 1

2

(}

- 33 = -

41

+ sin2 8 =

1. Thus A is invertible

l

1

=

=

1 foro

2

tan acos2 a [

+ n1r.

:f.

cos a

0

k

0

.

sec 2 a

0

3 3 2I] , and d ct(A) = k(k- 4 ). [2k 2k k <1

0

-cos2 o ' · l.an o cos2 a

l

Thus, for these values o f

Thus the system has a unique

4. In this case the solution is given by:

3

1

k

2

2k k

(k- l)(k - 6) k(k- 4)

--- = k (k - 4 )

3 4 2k

z=

r

1

-3

0]

Using the identity 1 + tan 2 o = sec 2 a:, we have det(A!)

X

3

co38 - sin(1 sinO cos @ 0 .

adj(A) =

x =

3

-1

sin9 0], and det(A) = cos

e

-

n, t.he m a trix A is invertible and A- 1 = adj(/1.)

16.

-3

3 -2 2

5

4

1

[

14.

6

12. y=

cosO

•. :::

=

21 3 = - =-

13. The matr ix of cofactors is C =

and A- 1

8 3

2 -1 3 - 1

3 k 2k

l

formula. for the inverse is A -

1

= (2

2

k

2(k - 1)

=-k(k- 4)

k(k - 4)

(2k - 3)(k - 4) k(k- 4)

14k - 3 = ..,.-----.-,.-----:(5k- 3)(3k + 2) -

l

1 2

k(k- 4)

17. We have d et(A) = 1i- x 2 y = y(y 2

y=

3 1 4 2 2k l

y=

2k - 3

= - --

3{k2 - k + 3) (5k - 3}(3k + 2)

k

3k

z=--5k -

3

x 2 ). Thus A is invertible if a.nd only if y =I= 0 a.nd y l

. '!/ Y -

X

2

)

[·-x·y y 2

0 0

- :ry y2 _ ;r2

y2]

- xy . ;r2

=I=

± x. The

Chapter 4

134

= (s 2 -

18. We have det(A) is

t2 ) 2 . Thus A is invertible if ;md only if s =/; ± t. The formula for the inverse

1

19.

]det(A)I

=II+ 2] =

0s -t0 -t0]

s 1 0 = (s2 - t 2) -t [ 0

0

-t

s 0

0 · s

ldet(A)I = ]o- 61

20.

3

21. ]det(A)J = 1(0- 4 + 6)- (0 + 4 22. ]det(A)I

\-1

=6

+ 2)] = l-41 = 4

= ](1 + 1 + 0)- (0- 3 + 2}1 = 131 = 3

= (3, 1) as adjacent sides. Then, from Theorem 4.3.5, the area of the parallelogram is ]det(A)] = ]3- 61 = 3.

23. The parallelogram has the vectors P1 Pz = (3,2) and P1P4

24. The parallelogram has the vectors P1 Pz

----t -= (4, 0) = (2, 2) and P1P4

as adjacent sides. Let A ;;;.;

Then, front Theorem 4.3.5, the area of the parallelogram is ]det(A)]

25. area

1 . 2

=

2

0

1

3

1

1 2

1 =7 1

-1

2

27.

V

=

28.

v

=

]0- 8]

1

1 2

2

[33] 2 1

.

[22 "o].

= 8.

1 2

1 1

3 -3

1

=3

= ]-16] = 16.

thus V

]det(A)I wlsere A= [- 6 2

-

26. area

Let. A=

-2 -4

= 45

29. The vectors lie in the same plane if and only if the parallelepiped that they determine is Jegenerate in the sen.;;e that. it.s "volume" is zero. In this example, we have

v

-1 = -2

0

5 -4

-2

0

= 16

and so the \'Cctors do not lie in the same plane. 30. These vectors lie in the same plane. 31.

33.

a=± )5(0, 2, 1)

U XV=

2

2

34.

(a)

32. a= ±fsi(6, -3, 4)

j

k

3 3

-6 6

' .l --+ --+ ABxAC= -1

2

k 2

1

1

-1

(b) area

sin f)

= 36i- 24j j

IIXBII h =

=

=

l]u

vll

j!u!IJlvll

-4i +j - 3k

thus h

X

=

= )1296 + 576 = v'4§y49

area

=

=!

)1872 = 12M 49 49

x

ACII = £P

EXERCISE SET 4.3

15.

135

j k 2 - 3 6 7

(a) v x w = 0 2

= (14 + 18)i- {0 + 6)j + (0- 4)k = 3Zi k

j

{b)

U X (V X W)

=

3 2 -1 32 -6 -4

uxv = 3 0

(- 8 - 6}i - {- J2 + 32)j

+ (-18 - 64)k =

-- 14i- 20j- 82k

2 - 1 =(-6+2)i-(-9+0)j+(6 - 0)k =- 4i + 9j +6k 2 - 3 j

(uxv)xw = - 4 9 2 16.

=

k

j

(c)

6j - •lk

(a) (0, 171), ..... 2fJ4)

6

k 6 =(63-36)i-(-28 - 12)j +( - 24 - 18)k = 27i+40j-42k 7

(h) (-44,55,-22)

k 4 2 3 1 5

( (:)

( - 8, -3, -8)

j

17.

(a)

U

I

= 18i + 36j -18k = (18, 36, -18) k

j

( b)

U XV =

- 2

1 0

3 i8. (a) u x v

is orthogonal to both u and v.

5 = - 3i -3

= (0, - 6, - 3)

+ 9j - 3k = ( -3, 9, -3) is orthogonal to both u and v. (b )

U X V =

(2,-6, 12)

vzl) =

U2

.o.

u x (v

+ w)

.,.,.,

(l

= (u

t 12

v2

+ w:i

x v)

= (ku) x v

+ (u

x w)

-v xu

Chapter 4

136

43.

(a)

(b)

u >< v = 1

j -1

k .2

0

3

1

2

uxv=

- 1

= lfu x

44. (a) A

---+

45. PtP2

X

---+ plp3

li M 46.

A=

t II:PQ

= -7.i-j+ 3k

j

k

3 2

-2 (b) A=

vlf = 0

k 2 3

- 1

-5

2

0

X

llu x

vii =

Ji01

/IT4

= - 15i + 7j + IOk

=

X

J59

A= ll u x v lf = J 36 + 16 + 49 =

0 = -6i +4j +7k

j

=

A= ll u x v ii = J49 + 1 + 9 =

+ 49+ 100 =

= tJll40 =

4TI

J285

47. Recall that the dot product distributes across addition, i.e. (a+ d )· u =a· u + d · u . Thus, with u = b x c , it follows that (a+ d ) · (b x c ) =a · (b x c)+ d · (b x c) . 48.

Using properties of cross products stated in Theorem 4.3.8, we have (u + v j x (u - v ) = u x (u - v) + v x ( u- v ) =-= ( u x u)+ ( u x ( - v) ) + (v x u ) + (v x ( -v))

= (u x --

---t

49. The vcc:lo r AB x A C

+ {v x

u ) - (u x v) k .I 1 -3 =

j

= - I

8i + 4j

u ) - (v x v ) = 0 - (u x v) - (u x v)- 0 -)-

= - 2(u x v )

---t

+ 4k is perpendic ular t o AB aiH.I AC, and thus is per-

3 -1

pendic\llar to the plane determined by the points A, B, and C . 50.

(I

1131, - 1111 t1311. 111

vzl) ; thus

(a)

We h uve v x w =

(b)

lu · (v x w )l is equal to the volume of tlte parallelpiped having the vectors u , v , w as adjacent edges. A proof can be found in any standard calculus text.

51. {a)

We have adj(A)

vz W2

W3

W1

W3

W}

W2

l]

=

o , and so A- 1 = detCA)adj(A) I

0

·1

(b ) The ,.duced .ow echelon fo<mof(A

34 = -t [-14

-21

1

IT];, [:

0

0:-¥ I 0

0

I I

t :

-1

_:

A- 1 =

2 _:

0

7 0

'f

- 'i

0

-;]. 7

EXERCISE SET 4.3

(c)

137

The method used in (b) requires much less computation.

>2.

We have det(AJ.:) = (det(A))k. Thus if A''-'= 0 for some k, then det{A) = 0 and so A is not invertible.

53.

F:rom Theorem 4.3.9, we know that v x w is orthogonal t.o the plane determined by v and w. Thus a vector lies in the plane determined by v and w if and only if it is orthogonal to v x w. Therefore, since u x (v x w) is orthogona.l to v x w, it follows that u x (v x w) lies in the plane determined by

v and w. >4.

Since (u x v) x w = -w x (u x v), it follows from the previous exercise that (u x v) x w lies in the plane determined by u and v.

55.

If A is upper triangular, and if j > i, then the submatrix that remains when the ith row and jth column of A are deleted is upper triangular and has a zero on its main diagonal; thus C;; (the ijth cofactor of A) must be zero if j > i. It follows that the cofactor matrix C is lower triangular, and so adj( A) "" or is upper triangular. Thus, if A is invertible and upper triangular. then A_, = is also upper triangular.

56. If A is lower triangular and invertible, then AT is upper triangular and so (A- 1)T = (A 1'}- 1 is upper triangular; thus A - I is lower triangular. 57.

The polynomialp(x) = ax 3 if and only if

+ bx2 +ex+ d es through the points (0, 1}, (1, -1), (2, -1), and (3, 7) d= a+ b+ c+d=-1

Ra 27 a

+ '1b + 2c + d = -1

+ 9h + 3c + d =

7

Using Cra.mers Rule, the solution of this system is given by

-::::

1 -1 -1 7

0 1 8 27 0 l

8 c=

27

0 1

0 1

ll

0 1

1

2 1 9 3 12 ---1 0 0 1 - 124

1

I 2

1

4

9

3

1

0 l

4 9

b=

-·1

0 1

1

8 -l

2

1

0 0 1 1 8 4 27 9

l

1 -1 ·-1

I

0

1 1

1

7

1

12

7

127

8 -12 =-=-1 12

27

d=

Thus the interpolating polynomial is p(x) = x:l - 2x 2

-

0 1 1 9 12

x

+ 1.

-24 12

3 0 1 1 1 2 1

= - - = -2

3 0

1

1

-]

2

- 1

3

71

12 12

=-=1

Chapter 4

138

DISCUSSION AND DISCOVERY Dl.

(a)

The vector w = v x (u x v} is orthogonal to both v and u x v; thus w is orthogonal to v and lies in the plane determined by u and •t.

,. ux v I I

v

(b) Since w is orthogonal to v, we have v • w = 0. On the other hand, u · w = l!ullllwll cos9 = llullllwll sin(%- B) where() is the angle between u and w. It follows that lu · wl is equal to the area of the parallelogram having u and v as adjacent edges. D2.

No. For example, let u but vi= w.

= (1, 0, 0), v =

= (1, 1, 0).

(0; 1, 0), and w

Then u x v

=ux

w

= (0, 0, 1),

D3. (u · v) x w docs not make sense since the first factor is a scalar rather than a vector. D4. If either u or v is the zero vector, then u x v = 0. If u and v are nonzero then, from Theorem 4.3.10, we have flux vii= JluiJI!viJsinO where B is the angle between u and v. Thus if u x v = 0, with u and v not zero, then sin B = 0 and so u and v are parallel. D5. The associative law of multiplication is not valid for the cross product; that is u x (v x w) is not in general the same as (u x v) x w. · · D6. Let A

= [l -cc

-(l-

c

c)]. Then det(A) = c2 + (1 -

c) 2

= 2c2 - 2c

+ l f. 0 for all values of c.

Thus,

for every c, the system has a unique solution given by

-(l-c)l

3 x1 =

07.

(c)

1-4

'l.c2

7c- 4

c

-

2c + 1

= -2c"""2,.---2c_+_l

The solution by Gauss-Jordan elimination requires much less computation.

D8. {a) True. As was shown in the proof of Theorem 4.3.3, we have Aadj(A} = det(A)L (b) False. In addition, the determinant of the coefficient matrix must be nonzero. (c) True. In fact we have adj(A) = det(A)A- 1 and so (adj(A))- 1 = (d)

False. For example, if

(e)

True. Both sides "ne equal to

adj(A) Ut

v1 W1

=

UJ

l!'J

t13 •

W2 W3

WORKING WITH PROOFS P 1. We have u · v

= II ullllvll cos 0 and II u x vII = II ul!ll vI! sin 0; thus tan 0 = II '::.;11 .

P2. The angle between the vectors is () = o:- /3; thus u · v = U•V

Uullllvll'

Uullllvll cos( a-- {3)

or cos(a:- /3)

=

EXERCISE SET 4.4

P3.

139

(a) Using properties of cross products from Theorem 4.3.8, we have (u

+ kv)

x v = (u x v)

+ (kv x v)

= (u x v)

+ k(v x v)

= (u x v)

+ kO = u x v

(b) Using part (a) of Exercise 5U, we have u1

u · {v x w)

=

Uz

U3

V2

VJ

V3

Vt

v2

V3

Ut

1l2

1l3

W}

W2

W3

W)

W2

W3

= -v · {u x w) = -(u x w) • v

P4. If a, b, c, and d all lie in the same plane, then ax b and c x d are both perpendicular to that plane, and thus parallel to each other. It follows that (ax b) x (c x d} = 0. P5. Let Q1 = (xt,Yt,l), Q2 = (x2,y2,l), Q3 = (x3,Y3,l), and letT denote the tetrahedron in R3 · having the vectors OQ1, OQ2, OQ:.h a..c; adjacent edges. The base of this tetrahedron lies in the plane z = l and is congruent to the triangle 6P1 P2h; thus vol(T)"" !area(6P1 P 2 P3 ). On the ot.her hand, voi(T) iR equal to timP.s the volume of the p<\rallelepipe(l having OQ 1, OQ 2 , OQ3, as adjacent edges and, from part (b) of Exercise 50, the latter is equal to OQ 1 • (OQ:z x OQ3). Thus area(6.P1P2P3)

= 3voi(T)

· (OQ2 x OQ3) =

=

l

Yl

X}

X2

Y2

l

'X3

1/3

1

1"'o.

The fixed points

EXERCISE SET 4.4 1.

1has nontrivial fixed points since det(I--

1 0 0

(a) The matrix A= [ 1

are the solutions of the system ( l -.. A)x

A)""'

1°1

0 -1

= 0, which can be expressed in vector form

a..<>

x

=

where -oo < t < oo.

(b) The matrix B . [ 0

'] l 0

has nontrivial fixed points since det(J - B) =

points are the solutions of the system (I - B)x x

2.

=t

1 -l

1

=

o. The fixed

which can be expressed in vector form as

where -oo < t < co.

(a) No nontrivial fixed points.

3. We have Ax= [:

4.

= 0,

I

eigenvalue ..\

= 5.

We have Ax

=[

eigenvalue >.

==-

=:

0.

:]

m

(b)

= 5x; thus

=;] [:] m

X= [:]

thus

No nontrivial fixed points.

,,

an cigcnvedor of A corresponding to the

m,,

an eigenvector of A corresponding to the

Chapter 4

140

5.

(a)

A=[!

The characteristic equation of Thus ..X

= 3 and >.

=

- 1 are eigenvalues of A;

(b) The characteristic equation is I.X is the only

(c)

is det(.XI-

A) =1>.--=-

3 8

1

+ 1} =

1= (,X-

0.

has algebraic multiplicity 1.

=

10 ).. : 1 = 2 4

(>. - lO)(..l. + 2) + 36 =

(>. - 4)2 = 0.

Thus ,X = 4

it has algebraic multiplicity 2.

The characteristic equation is 1.>.-

2

-1

I= (..l.- 2)

0 , "'-2

2

= 0.

Thus ,X= 2 is the only

it

has algebraic multiplicity 2.

6.

(a) The characteristic equation is .X2 - 16 = 0. Thus multiplicity 1. (b) The characteristic equation is >.. 2 multiplicity 2. (c) The characteristic equation is(>.. multiplicity 2.

7.

(a)

Thus ,\ = 0 is the only eigenvalue; it has aJgebra.ic

= 0.

Thus >..

.>. - ·1

0

2

.>.- 1

o

2

0

A- 1

The characteristic l!quat io!l is 0. Thus .A = 1, >.. = 2, and >.

= 0.

=

,X = ±4 are eigenvalues; each has aJgebra.ic

= 1 is the only eigen.value; it fi8s algebraic

- 1

= ).3

-

6). 2 + 11 .A- 6 = (.A - 1)(>. - 2){>..- 3) =

3 are eigenvalues; each has .algebraic multiplicity 1. .A-4

(b) The characteristic equation is

-t

5

5

.X -1

1

3

). + 1

= A3 -

4>.2

+ 4A =..\(A -

and ). = 2 are eigenvalues; >. = 0 has algebraic multiplicity 1, and ). ).. -3

(c)

- 3

A = - 3 and ,\

-4

1

.x + 2 - 1

The characteristic equation is

-9

= 2 are eigenvalues; A= -

= ).3 -

>.. 2 - 8>.

+ 12 =

).

3 has multiplicity 1, and >..

2) 2

= 0.

Thus A= 0 ..

= 2 hFlS multiplicity 2. (,X+ 3)(A- 2) 2

= 0.

Thus

= 2 has multiplicity 2.

The characteristic equation is >. 3 + 2..\ 2 + A=>.(>.+ 1) 2 = 0 . Thus >. = 0 is an eigenvalue of multiplicity 1, and ).. = -1 is an eigenvalue of multiplicity 2. (b) The characteristic tlltuation is ).3 - 6).. 2 + 12)., - 8 = (A- 2) 3 = D; thus,\ = 2 is an eigenvalue of multiplicity 3. (c) The chara.ctcris t.ic equation is >.. 3 - 2A 2 - 15>. + 36 = (,\ + 4)(A - 3) 2 = 0; t hus A= -4 is an eigenvalue of multiplicity 1, and ,\ = 3 is an eigenvalue of multiplicity 2.

8.

{a)

9.

(a) The eigenspace corresponding to >. yields the general solution x

[:J =

=t

= t, y = 2t;

solution x

= 0, y = t;

=

This

= 2x in the xy-plane.

=: (:] =

This

thus the eigenspace consists of all vectors of the form

Geometrica.lly, this is the line x = 0 (y-a.x.is).

yields the general solution x =

(;]

to A = - 1 is found by solving the system [

(b) The eigenspace corresponding to .A

[:J

found by solving the system [

t.hus the eigenspace consists of all vectors of the form

Geometrically, this is the line y

The eigenspace yields

= 3 is

=4

=:!] [:] =

is found by solving the system [

= 3t, y = 2t; thus the eigenspace

Geometrically, this is the line y

= 3x-

.

This

consist of a ll vectors of the form

E.X ERCISE SET 4.4

141

(x]

0 (c) The eigenspace corresponding to >. = 2 is found by solving the system [ 0 0) = ( ]. This -1 0 y 0 . yields the general solution x = 0, y = t; thus the eigenspace consists of all vectors of the form

[:] = 11).

Geometrically, t his is the line x

= 0. =

(a) The eigenspace corresponding to>. = 4 consists of a ll vectors of the for m (:) the line y

[:] = t

=

this is

T he eigenspace corresponding to >. = - 4 consists of all vectors of the fo rrri this is t he line y = -

(b) The eigenspace corresponding to>.= 0 consists of all vectors of the form

+

=

this

is the entire xy-plane. (c)

The eigenspace corresponding to A. line x

11.

(a)

= 1 consists of all vectors of the form

= 0.

(:J=

[-

t

this is the

"

[ool

The eigenspace corresponding to >.

= 1 is obta ined

yields the general solution x == 0, y

= t, z = 0; thus the eigenspace consists of all vectors of the

hy solving

t [:] ; thls cones pond• to a Hne thmugh the o•igin (the y-axis) in R 3 • Slmilad y,

[

to !.

the eigenspace eigeosp""e conesponding to !.

= t [ = , and the

2 consists of all vecto" of the fo< m [:]

t [- : ]·

3 consists of all vccto" of the fo<m [:]

5

(b) The eigenspace corresponding to

This yields the general solution of the fo<m [:] = t

[!];

>. == 0 is found

x =

5t, y

= t, z

by solving the system

-·i = :H.. Thus the eigenspace

(c)

=s

+ t [:],

[=I

0

of all vectors

!:] m =

[:]

u -:][:] m

sponds to a line through t he origin .

=

This

0

z

, which cones ponds to a Hoe th•ough the odgin. The eigenspa<e found by solving

This

This yields

=

-3 -9 -3

!. = h

f:] = v

which conesponds to a plane thwugh the odgin.

T he eigenspace corresponding to>.= - 3 is fo:1nd by solving

[: ] = t [

l

this is the line through the odgin >nd the point. (5, 1, 3).

The eigenspace con...pondh;g to!.= 2 is found by •olving the system

yields [:]

:l

l f l],

[

to

whichabo con&

Chapter4

142

12.

(a) The cigenspA£e CO<J<SpOnding to A = 0 consists of vectors of the form .

concsponding to A

= -1 consist-s of vectorn oft he focm

and the

]·

m Lil =

t

T he eigenspace cnn esponding to A = - 4 consists of vectorn of t he fOTm [:] - t [ :] • and the

eigenspace conc•ponding to A = 3 oonsi•t• of vectOTS of the form

13.

= t[

[: ] = t [

(b) The eigenspa.;e conesponding to ). = 2 consists of vectors of the fe<m

(c)

m -!].

m

= t [ -:] ·

The c.ha.md.t1ristic polynomial is p(>.) = (>. + 1)(>.- 5) . The eigenvalues a.rc >. = ..:_ 1 and>.= 5. ( b ) The cha racteristic polynomial is p(A) = (.A - 3) (.X - 7)(>.- 1). 'The eigenvalues are >.= 3, ). = 7, and A= 1. (c) T he characteristic polynomial is p(A) = (>. + - l)(.XThe eigenvalues are.-\= (with multiplicity 2), ). = 1, and>.=

(a )

-s

2 ?0

14. 'l'wo examples are A ::::

.o

0

?.]v

L

and B = [ -3

0 -1

0 0

0

1

-1

u

- 1

0

15. Using t he block diagonal structure, the characterist-ic polynomial of t he given matrix is

p(),)

), - 2

-3

0

0

l

), - 6

0

0

0 0

0 0

.\+2 -1

-5

=

.::: [{>- - 2)(>. - 6)

+ 3)[(>. + 2)(>- -

T hus the eigenvalues are 16.

!). -l 2

=

--3 >. -6

If), + 2 - 1

-5

I

>.- 2

>- - 2

2)- 5J

='"

(>- 2

8), + 15)(,\2

-

5, ), = 3 (with multiplicity 2), and),

-

9) = (>-- 5)(.), - 3)2 (-X + 3)

= -3.

Using the block triangular structure, the characteristic polynomial of the given matrix is

I- 1).011). +0 2

p(>-) = I A

I= >-,

0

>- - 1

2 ().

Thus the eigenvalues of B are ). = 0 (with muitiplicity 2), ,\

+ 2)(>.- 1)

= - 2, and ), = l.

17 . The characteristic polynomial of A is

p( A)

= det( AI - A) =

>. +1

2

-1

). - 2 1

2 -1 =(.-\+ 1)(>. -1) 2 ).

EXERCISE SET 4.4

143

thus the eigenvalues are .X = -1 and ..\ = 1 (with multiplicity 2}. The eigenspace correspontJing to .>. = - 1 is obtained by sol.-ing the system

whkh yield$ [:]

l [ - :]. Similarly, the eigenspace co"esponding to !.

= I is obtained by solviug

H-: -:1m =m which has the general solution

' =t, y "

-t - ·'· ' = ' ' or (in vedorform)

= ·' [-:] + t [- 1]

The eigenvalues of A.! 5 are). = ( - 1) 25 = -1 and >. = ( 1) 25 = 1. Correspcnding eigenvectors same as above. 18.

Thoeigenvalues of A are !. = J , !. =

rll·

and

j, !. =

0, end !.

=2

the

:j. r:J. .

ng eigcnveclo" are [

0

and [

9

r"'peotively. The eigenvalues of A are A = (1)'

1, A =

(!)'

= ,:, .

:0)' =

0.

>. == {2)!} == 512. Corresponding eigenvectors arc the same as above.

l9.

The characteristic polynomial of A is p()..) = >.3 -- >. 2 - 5).. - 3 = (..>.- 3)().. + 1) 2 ; thus the cigeoare .>. 1 = 3, .A2 = -1 , >..3 = -l. We bnve det( A) = 3 and tr(A) = 1. Thus det(A) = 3:; (3)(- 1)(-1} = .A 1.A2>..3 and tr(A) = I= (3) + (- 1) + ( -1) =.A t+ .X2 + .>.3.

W.

The characteristic polynomial of A is p(>..) = >. 3 - 6.:\ 2 I· 12.>- - 8 = (.>.- 2)3; thus the eigenvalues are At = 2, -\2 = 2, A3 = 2. We have det(A) = 8 and tr(A) = (). T hus det(A) = 8 = (2)(2)(2) = ..\. 1.\2.:\3 and tr(A} = 6 = (2) + (2) + (2) = )q + .:\2 + .\3.

= 0 and

!1. The eigenvalues are .\ (

ann

= -!x andy .::::: 2x.

The eigenvalues are ..\ tors [

= 5, with associa.t.ed eigenvectors

GJ respectively. Thus the eigenspaces correspond to t he

perpendicular li nes y

:2.

.>.

v;]and [

= 2 and

>.

= -1, with associated eigenvec-

respectively. Thus the eigenspace::; correspond

to the perpendicular lines y""'

andy=

- J2x.

Chapter 4

144

23. The inv
The eigenvalues the lines y

-= 2x

2 and.\:::::: 3, with associated eigenvectors ami y

=x

g]

and

arc invariant under the given matrix.

(b)

This matrix

(c)

The only eigenvalue is A= 2 (multiplicity 2), with associated eigenvector

y

}m.':i

Thus

no rt'lal eigenvalues, so t her e are no invariant lines. Thus the line

= 0 is invariant unde r the given matrLx..

24 . The char acteristic polynomia l of A is p{A) = >..2 - (b + l )A + (b- 6a), so A has the stated eigenvalues if and o nly if p( 4) = p( - 3) = 0. This leads to the equations

6a - 4b = 12 6a + 3b = 12 from which we conclude that a = 2l b.,., 0. 25. T he characteristic polynomial of A i:; p( A) = >. 2 - (b + 3)A + (3b - 2a), so A has the s tated eigenvalW!S if l.l.lld o nly if p(2) = p(5) = 0. This leads to the equations

- 2a + b = 2 a+ b=5 from which we conclude t.hat a = 1 a nd b = 4.

26. The chl'l.r!'l.ctcristic polynomial of A is p(>. ) = (A - 3)(A2 - 2Ax + x 2 - 4). Note that the second factor in this polynomial cannot. have a double root (for any value of x) since ( -2x) 2 - 4(x 2 - 4) = 16 # 0. Thus t he o nly possible repeated eigenvalue of A is .>- = 3, and this occurs if and only if .\ = 3 is a root of t.he second factor of p(A), i.e. if and only if 9-- 6x + x 2 - 4 = 0. The roots of this quadratic eq uatio n are x = 1 and :t - 5. For t hese values of x, A = 3 is an e igenvalue of multiplicity 2. 27. lf A'l =.f. then A (x ·l- tlx) = Ax+ A 2 x :-:- Ax+ x = x +Ax; thus y = x +Ax is an e igenvector of A corresponding to A = 1. Sirni la dy, z = x ·- Ax is a.n eigenvector of A corresponding to A = -1.

28.

Accorrling to Theorern 4.4 .8, the characteristic polynomial of A can be expressed as

where >.1, >. 2 , ... , Ak arc the distinct eigenvalues of A and m 1 + m 2 + · · · + mk = n. The constant t.crrn iu this polynomial is p(O) . On the other har.J, p(O) == det( - A) = (-l)ndet(A) . 29.

(a)

lJsinf:! Formulu. (22), the d1atacteristic equation of A is .\2 + (a+ d)>..+ (ad - be)= 0. This is n

qundratric equation with discriminant

(a+ d) 2

-

4(ad - be) = a 2

+ 2ad + d2 -

Thus the eigenvalues of A are given by .A= (c)

+ 4/>c > 0 If (a- df + 4bc :-:o 0

{d)

If {n.-

{b) If (u

d)

2

4ad + 4bc = (a - d) 2

d)± J(a- d}'l

+ 4bc

+ 4bcj.

then, from (a) , the characteristic equation has two distinct real roots. then, from (a),

dV + 4bc < ()then, from

il:l one real eigenvalue {of multiplicity 2).

(a ), t here a re no real eigenvalues.

145

EXERCISE SET 4.4

30. lf (a- d)Z + 4bc > 0, we have two distinct real eigenvalues A1 and Az. The corresponding eigenvectors are obtained by solving the homogeneous system ·

bx2

a):tt - cx1

=0

+ (Ai- d)x2 = 0

Since A; is an eigenvalue this system is redundant, and (using the first equation) a general solution is given by

x 1 = t, Finally, setting t

n.

= -b. we ;,;ee that [a =b>.J

is an eigenvector corresponding to

>. =

..\i·

If the characteristic polynomial of A is p(.A) = A2 + 3>.- 4 = (>.- l){A + 4) then the eigenvalues of A are >.1 = 1 a nd A2 = - 4. (a) From Exercise P 3 below, A- 1 has eigenvalues .-\ 1 = 1 a nd ,\2 = (b)

From (a), together with Theorem 4.4.6, it follows tha t. A- 3 has eigenvalues A1 "'""(1 )3 = 1 and >-z-=- ( --

(c)

Prom P4 below, A - 41 has eigenvalues AJ

(d)

From P 5 below, 5A has eigenvalues)"

( e)

From P2( a.) below , t he eigenvalues of <'lA:r + 2I = (4A 1- 21) T arc the same !'IS those of 4A + 2J; namely A1 = 4(1 ) + 2 = 6 and ..\2 = 4(- 4) + 2 = -\4.

- o\.

=

1 --4

= 5 and ..\2 ;::: -20.

= (..\x) · x .,. >.(x · x) == .AIIxll 2 and so

l2.

If ...-lx -= Ax, where x -? 0 , then (Ax) · x

'3.

(a) The characteristic polynomial of the matrix C is

Add .>. l ime.-> the

p(A) =

= det ( >.J -

C) ""

()

0

-I

,\

0 0 0

0

0

0

- 1

0

>..

A 0

- 1

p( ,\}

= -3 and ,\2 = - 4 - 4 = - 8.

A=

(i1:1?t.

co Ct

Cz A+ Cn-

1

row to the first row, then expand by cofactors a.long the first column: 0

A2

- 1

0

Co+ CtA

>.

0 0

0

Ct

0

-1

.>.

0

cz

0

0

0

- 1

..\ +

)..'2

0

-1

..\

0 0

0

0

-1

co

+Ci A

Cz

=

C:n - 1

..\+ Cn - 1

Add A2 times t.he secvnd row to the first row, then expand by cofactors along the first column.

p(.>.)

=

0

Az

- 1

).

0 0

co+ Ct >. + cz>..2

,\2

C2

0

co + C)A+ C2A 2

= 0

0

- 1

..\ + c,._l

0

-1

A+ Cn-l

Chapter4

146

Continuing in this fashion for n - 2 s teps, we obtain

= 1,\n-l

co + Ct A + c2.\

-1

(b) The matrix C

=

[l

1 0

+ · ·· + Cn.-2>-.n- 21

>-+Cn-t

= Co+ CtA 0 0

2

+ C2 A 2 + · · · + Cn-2A

11

00 -2]3 hasp(>.) = 2 - 3>. + o

-1

1

5

-

2

+ <'-n - 1An- l + >."

.>-2 - 5>.3

+ .>.4 as its cha.ra.cteristic poly-

nomial.

DISCUSSION AND DISCOVERY Dl. (a) The characteristic polynomial p(.>.) has degree 6; thus A is a 6 x 6 matrix. (b ) Yes. From Theorem 4.4 .12, we have dct{A ) = {1)(3) 2 (4) 3 = 576 =/: 0; thus A is invertible. D2. If A is :;quare m n.trix all of whose entries arc the sl:l mc then de t(A) solutions and >. = 0 is an eigenvalue of A.

03. Using Formul a (22), the characteristic polynomial of A is p(..X) ). = 2 is the only eigenvalue of A (it has multiplicity 2).

= 0; thus Ax = 0 = ). 2

-

4.\ + 4 =

has nontrivial

(..\- 2) 2.

Thus

04. The eigenvalues of A (wit h multiplicity) arc 3, 3, and - 2, -2, - 2 . .Thus, from Theore m 4.4.12, we l1ave det(A) = (3)(3)( - 2) ( -··2)( -2) = - 72 m1d tr(1t) = 3 + 3 - 2- 2- 2 = 0. D5. The matrix A d

= 1 then

= [ac

blj satisfies the condition tr(A)

d

= det(A)

if and only if a+ d = ad- be. If

this is equation is satisfied if and only if be = - 1, e .g., A =

· · sa.t 1S1l€ . r. d 1·r an<1 o n 1y 1·r a = . equa t 1011 1s

d·tbc <J •• , e.g.,

1

A

[1

1

lf d

f.

1, then the

- 1] · 2

06 . The characteristic polynomial of A fac tors a.s p(..X) = (>.- 1)(-\ + 2) 3 ; t hus t he eigenvalues of A a re ). = 1 and).= - 2. It follo\vS from Theorem 4.4.6 t hat the eigenvalues of A 2 are ). = (J) 2 = 1 and >. = ( - 2) 2 = 4. .

D7. {a) False. For example, x = 0 sa.tisfies this condition for any A and .>.. The correct statement is t.hat if A x = ..\x for some nonzero vector x, then x is a n eigen\'ector of A. (b) True. If >. is an eigenva lue of A, then .-\2 is an eigenvalue of A 2 ; thus (.>. 2I - A1)x = 0 has uontrivial solutions. (c) False. If A = 0 is an eigenvalue of A, then the system Ax = 0 ha.s nontrivial solutions; thus A is not invertible anrl so the row vectors and column vectors of A are linearly dependent. !The statement becomes true if "independent" is replaced by "dependent" .J (d) False. For example, A=

G

has eigenva.lues ).

=1

and .\

= 2.

!But it is true tha t a

:;ymmetric matrix has real eigenvalucs.J

DB. (a) False. For example, the reduced row echelon form of A

fl OJ l.S =h 2

I =

[1 OJ . 0

1

(b) True. We havP. A(x 1 + x2) = A1x 1 + .>.2x2 and, if ..X1 f. >.2 it can be shown (sine"' x1 and must be linearly indep(:ndent) thal .)qxl

+ A2Xz f. {J(xl + X2) for any value of a.

X2

147

WOR KING WITH PROOFS

'!'rue. The characteristic polynomial of A is a. cubic polynomial, o.nd every cubic polynomial has at least one real root. (d) 1Tue. If r(A) = .>." + 1, then det(A) = (- l }np(D) = ± 1 =I 0; thus A is invertible. (c)

WORKING WITH PROOFS

b], t hen A2 = [02 +be ca + de

Pl. If A = [ac

A2

_

tr (A)A

= [be -

0

nod so p(A)....: A'l - tr(A)A

P 2.

(a)

ab + bdl and tr(A)A

cb+ d 2 j

d

ad

+ det(A)J

cb _o

ad] =

= (a+ d) [ac

(- de t(A) 0

b] d

d!]; thus

= [02ac+dc +do

ab + ad+ a-

0 ] _ det.(A) =- det(A)J

= 0.

Using previously estabJished properties, we have

111')

= det(Afr ·- AT)= dct((M- A)r) = det(M -

11)

Thus A a11d AT have t.hc *lame characteristic polynomial.

(b ) The e;gcnvalues are 2 and 3 in each case. The eigenspace of A conesponrling to ). tained by solving the system

t o >..

=

2 is obto.inf:d by solving

r:J

= 2 is ob-

t.he eigenspace of AT corresponding ': ' Thus the eigenspace of A corresponds to the

[:]

line y = 2x, whereas eigenspace of A'r corresponds to 11 = 0 . Simtlarly, fo r A 3, the eigf'nspacc of A corresponds to x = 0; whereas the eigeuspace of AT corresponds toy= P3. Suppose that Ax = .Xx where x i- 0 and A is invert.ible. Then x = A- 1 Ax = A- 1 >.x = AA- 1 x and, s1nce ,\ :f; 0 (because .·1 is invertible), it follows that A- 1 x = }x. Thus is an eigenvalue of .t\ - t and x is cormsponding

l

P4. Suppose A

P 5.

tlt;H

Ax

.\.x whcr<' x -:f. 0 . T hen (A - sJ)x = Ax - six = >.x - sx a r.orresponding eigenvector.

= (A- s}x .

Thus

sis an C'i);»nvaluc: nf A - sf and

Suppo:;e tha t, Ax = AX where x :f 0. Then (sA)x value of sA and x is a. corres ponciing eigenvector.

=

s(Ax}

=s(.>.x) = (s>..)x.

P 6. If the ma trix A = [: :] is symmP.tric, then c :=: b a.nd so (a- d) 2

+ 4b<: =

Thus s.>. is a a eigen-

(a - d) 2 ·+- 4b2 .

ln the case that A has a repeated eigenvalue, we must have (a- d) 2 + 4b2 = 0 and so a= d a.nd b = 0. T hus t he o nly sytnructric 2 x 2 matrices wi Lh repeated eigenvalues arc those of the form A """ al . [;ud1 a. ntatrix has .>. = a as its only eigenvalue, and the corresponding eigenspa<:e is R 2 . This prov<•s part (a) of Theorem 4.4..\ I. If (a- d) 2 + 4b 2 > 0, t hen A has two distinct real eigenvalues .>. 1 a nd eigenvectors x 1 and x2 , g.iven by:

.>.1 =

+d)+ J(a- d)2

+ 4b2)

..\2,

with corresponding

Chapter 4

148

The eigenspaces correspond to the lines y

(a- >. 1 ){a- >.2 )

= m 1x

andy = mzx where m;

= (H(a- d)+ J(a- d)2 + 4b2J) (t!(a = t((a -

d)

2

-

(a - d)

2

-

2

4b

l=

=

d) - J(a - d)2

j

= .1, 2. Since

+ 4b2])

2

-b

we have m 1 m 2 = - 1; thus the eigenspaces correspond to perpendicular lines. This proves part (b) of Theorem 4.4.11. Note. It is not possible to have (a - d) 2 + 4b2 < 0; thus the eigenvalues of a 2 x 2 symmetric matrix must necessarily be real P7. Suppose that Ax = .XX a.nd Bx = x . Then we have ABx = A(Bx) = A(x) = >.x and BAx = B (Ax) = B (>.x) = .XX. Thus >. is an eigenvalue of both AB and BA, and xis a corresponding eigenvector.

CHAPTER 6

Linear Transformations EXERCISE SET 6.1 1.

(a) TA: R2 (b) TA: R3 (c) TA: R3

---? ---? ---?

R 3 ; domain = RZ, codomain = R 3 R 2 ; domain = R3 , codomain= R2 R?; domain= R3 , codomain= R 3

3. The domain of Q , the codomain ofT

T(l , - 2)

= (-1,2,3) .

4 . The domain ofT is R3 , the codomain ofT is R 2 , o.nd T (0, - 1,4) = (-2, 2).

-2xl + :!.:7 + 4XJ]

-2

6. (a) T(x) =

=

[

+ 5x:z 6x1 -

[

+ 7xa

(b) T(x) =

7.

(a) We have

= b !f and only if xis a solution of the linear syst,em

12 [2 5 -3 0 -l

3

[XJX 2] = [- 1 ] 1 3

X;!

The reduced row echelon form of the augmented matrix of the above system is

0 1

6

-3

-111

o o oJ

nl

and it follows that the system ha..o; the general solut.ion x 1

Thus any vee to< of the fo<m (b) We have TA (x)

X

[-

= b if a nd only if x

2

:J +

t

= -1 -

6t, x2 = 1 + 3t, x 3

will have the pwpe
is a. solution of the linear system

5 -3 165

= XJ

1

[-

= t.

J

Chapter6

166

The reduced row echelon form of the augmented matrix of the above system is

0

0

0

1

and from the last row we see that the system is inconsistent. Thus there is no vector x in

R' 8.

(a)

for which TA(x)

We have TA(x)

=b

l:J· if and only if xis a solution of the linear system

[-1

2

0

5

[=:] = xa

4

The reduced row echelon form of the augmented matrix of the system is 0 1 0

2 -1 1 2

0

0

and it fo llows t.ho.t the system has general solution x 1 = 2- 2s + L, .r2 X3

S,

x,

that TA(x )

(h )

(.

m =r] +

any v.,.;tor ofthe form X

+s [

t [ - ;]

= 3- s- 2t,

will have t he p

=

We have TA (x ) = b if and only if x is a solution of the linear system

[

-1

2

! [::] = 0

5

X3

3

T he red uced row er.hclon form of the augmented matrix of t he system is

0

0

0

Thus the system is inconsistent; there is no vector x in R4 for which TA(x)

=

9. (a). (c), and (d) are linear t ransformations. (b) is not linear; neitht!r homogeneous nor 10. (a) and (c) are linear t ransformations. (b) and (d) are not linear; neither homogeneous nor addit ive. 11. (a)_and (c) are linear transformations. (b) is not linear; neither homogeneous Mr additive. 12.

(n) nml (c) o.re

(b) is not linear ; nPit her homogene<ms nor add:tive.

exercise Set 6.1

167

13 . T his lmnsfonn• lion can

written in

[! =: :] l::J ;

[o,m "' [ : : ]

it is a linear trans-

formation with domair@. and

The domain is R 2 o.nd the codoma in is R:i. ·.(__ __ - ·

• .,.,.--J· . . _ _ _ _ - - -

- -·-·----

- · - . - - · ···-· ...

=

15. This transformation can be written in matrix form as

- l 2 -4 -1

1U3

transformation with domt

16 .

('

nd

1.._

'

17. [TJ "" !T(eJ) T (e 2 )] =

(a) We have T( l , O)

-

.

18. [T] 4

19.

it is a linear

XJ

clomain:ls R 4 land the codomain(is R 2 . !

···--:_)

·...

[=:];

=

= [T(et)

T(e:z) T(e3)};:;;

0

[!

2 2 IJ

(-1 , 0) a nrl T(O, 1)

ing the m atrix, we have T ( [-

= (1, 1);

=

= [:]. This a grees with direct calculation ( -( -1) + 4, 4) = (5,4).

= [

using t he given formula: T{ -1 , 4) =

thf' st
[ -:]

(b) We have T (l,O,O) = (2, 0,0), T(O, 1,0) = (-1, 1,0), and 1'(0,0, 1) = (1,1,0) ; t hus the stan-

"G-: l

[

T his

[1'] =

(b)

ITl = [0

21. ( a)

= (2(2) -

( 1) I (- 3j, 1 + ( -3), 0) = (0, -2,0)

Ul

r([_m [:,

(a)

3

([J) [i -l ll [_rJ

usmg the given formula :

T (2, l , - 3)

20.

U
0 ] -5

T ( [

=

[

The stn.ndard mat.rix of the transformation is IT]

= [: -

i

(b ) If x -= (- 1, 2,4) t.hen, using the equations, we have

T(x)

= (3(-1) + 5{2)- (4),4(- 1)- (2) + (4) , 3(-1) + 2(2) -

On th e other hand, using t he matrix , we luwe

(4)}

= (3 , - 2, -3)

=

168

Chapter 8

22.

(a)

23.

(a)

1] .

{T] = [2 -3 3 5 - 1

25.

'26.

(a)

72

4

[- -

(a)

[ -1] 4 [•] [2J3+ 3 =

(d)

[14.598 946]

2

28.

(a )

(b )

(d )

[-!

!

3

-

41 ':i

'

.

=

-

I

liL __ [cos 28

L

2 = (:Jin9 coscotJ 8 9

(b)

2VJ -

=

1.964

..

[2+¥+ ] - [ 4598] [ l 4] rJ [ +JJ] [2482] 4 = ¥

[ l

-:1/- 4

3 -

-2 /3

+3

4

-1.964

4.299

0.518J

I

2

2

sin28] __ [cos 8-sin 8 2sin8cos0 2

;:

sin8cos9] oinl(J

4 -

[-0.2991

72] corresponds to He = 1

[>f-! :/,f!] [3]- [ >f'+•][4598] +

- l.9G4

- 7l

[

r:J= r-:J

corresponds to R 8 = [ . 88 -sin 8] where B = S i ll COS 8

sm28 - cos 28

p

r-:]

-2- ¥ l [-4598]

-2v'3 +

1

-

Thus Hr.=

{b)

[

3 =

rhe matnx A

31. (a)

=

[19641 4.598.

2

-4] [•] [1- ¥ l

. 29. 1, he matnx A = [-

30 .

[-l3'{3++

4 =

[•]

[

=

= [-

{b )

(c)

(a)

[-

(b)

4.950

= [=!J

(c)

27.

nl

=

(d)

[to -t-Jn= [-t>J [-070l 72

_:]

= GJ

(b)

=

72

[

(d)

=

(c)

=

l-1 -1)0 [-2]1 = [-1]2

=

(a)

([-m

(b) r o

=

(c) 24.

( b) T

S Jn

28 -

2sin9cos8]

. sm .. 8 - cos 2 8

m;r:_ 1] . = l+,....'l 1 [1 m

[cos28 .

m]

m2

sin29) COS 28

where B =

wherecosB

3;

(135°).

i (22.5° ).

= v l +m• a.nd sin8 =

'1/JTrn-

•

Exercl&e Set 6.1

32.

(a) We

169

have m

(b ) We have m 33. (a)

==

2;

thus H

= HL =

= 2; thus P = h

=

We bave m = 3; thus H = HL = the line y

[-! ;J and H ( [!]) [!] = t [z:] = !] !] ) 1[!] = ! ( = [::!]· fo [-: !] = !] = g !] [:] k !] fo :] [:] = =

[

and P ( [

=

and the reflection of x = [: ] about

= 3x is given by H ([:])

(b) We have m = 3; t hus P

= PL =

·

=

=

1 10

and P ([:]) =

=

34. If Tis defined by the formula. T(x,y} = (0,0), then T (cx, cy) = (0,0) = c(O, O) = cT(x,y) and T(x1 + X2, Yt + Y2) = (0, 0) = (0, 0) + {0, 0) = T(x1 , Y1) + T(x2 , Y2); thus Tis linear.

If T is defined hy T(x , y).::: (1, 1), then T(2x, 2y) = (1, I) f. 2(1, 1) = 2T(x, y) and T(x 1 + x 2 , Yl + Y2) ::..., (1, l) =I (1 , 1) + (1 , 1) T(x 1 , yl) + T(xz, Y2); thus T is neither homogeneous nor adJitive.

=

36. The given equations can we writt.en in matrix form us

r:J = -

·--

Thus t he image of the 2s .:.-=: 1. 37. (a) 38.

(a)

[T]

= [1

0

.. .

<''__........_ ... .... ..

line· ·:e + y =

r

[:] = [;], and so

r:J = l··--3: ....

. ...

.

1 corresponds to {-s + 4t ) + (3s- t) = 1, i.e. to Lhe line ·-- - -- -- -·-··- ·---

{b) IT] = 0

(iJ

1

(c)

[T] =

(b)

.f

!

Chapter 6

170

(d)

(c) X

39. T(x, y)

[-1 0]

= (- x, 0); thus IT} =

0 0

.

40. T(x,y)

= (y, - x ); thus IT]=

[-i


(a) (b) (c) (d)

False. True. True. True.

For example, T(x,y} = {x2 ,y2 ) satisfies T(O) =0 but is not linear. Such a. transformation is both homogeneous and additive. This is the transformation with standard matrix (T] = -[. The ·z.cro transformation T(x, ·y) o-:: (0, 0) is the only linear transformation with this

property.

(e)

= v o f. 0.

False. Such a transformation cannot be linear since T(O)

D2. The eigenvalues of A are >.= 1 and .>. = - 1 with corresponding eigenspaces given (respectively)

by

where -oo < t

and

< oo.

D3. From familiar trigonometric identities, we have A =

29 -sin 28] __ R-.·e .

cos [ sin 29

cos 28

"

Thus muJtiplication

by A corresponds to rotation about the origin through the angle 2fJ. D4. Jf

A ·/ 1 ·- ..

R o -_ [cos O .. sin sin9

81

cosO'

t h en AT _ -- [ cos 0 sin 8] ..__ [cos( -9) -sin9 cosO sin(-8)

= R- o·

Thus multipli-

cation by !1.,. corresponJs to rotation t hrough the ang le -0. D5 . Since T(O) :::: x 0 =f:. 0, this transformation is not linear. Geometrically, it corresponds to a. rotation

followed by a translation. D6. 1f b = 0, theu geneous.

f is both addi t ive and homogeneous.

If b =f:. 0, then j is neither additive nor homo-

D7 . Since Tis linear, we have T (x 0 +tv) = T(x0 ) + tT(v). Thus, if T(v) =f:. 0, the image of the line x = x o +tv is the line y = Yo+ tw where y 0 = T (xo) and w = T(v) . lf T(v) = 0 , then the image of x = xo + tv is the point Yo = T (xo).

EXERCISE SET 6.2 1.

ArA =

[_:I] 21

2 aQ 2 . AT A= [ ll !.1

_

29 !Q

(_:

iJ·

J [ £! :ollJ - rl0 0 _llJ ·thus A is orthogonal and A - 1 =AT= [aQ l ' ll

' D

29 !Q

-

29

'iG

-

1

29 29

29

.aQ

zg

•

Exercise Set 6.2

171

u

- !s

12] [ 54

25

4

5 _g

-

5

5

l2

25

100]0 ;

0

= [0

25

2S

·u

25

3

5

1

thus A is orthogonal and A - l

=AT =

0 0 l

16

25

1]. 16

25

I

2

76

5.

r-t _

(a) AT A = A

= [

1

73

thus A is orthogonal. We have det(A)

=

= Ro where B = 3; .

= 1, and

Thus multiplication by A corresponds to CO\tnterclock-

wise rotation about the origin through the a ngle 3;.

(b)

[

J [ ;

A= [

=He w bere fJ

thus A is o'thogonal. We have det(A )

= i · Thus multiplication by A corresponds to ref!ed ion

the line origin through the origin m aking an angle

6.

(a)

AT A= [

A

(b)

=

rlT A= [

1.

(a)

A=

8 . (a)

A=

[

2

Y,}l = (o1 a v;l [./3 _! _ J_

2

A=

2

<

1 0 ]; thus 1

!2

2

2

0

A is orthogonal. We have det (A)

] · Lhus A is or thogonal.

1 •

= l,

and

We have dct(A )

= -1,

a nd so

./3] = He where 0 = %· 2

.:£1 - ! 7. 2

[t ;]

(b)

A=

(b)

A=

[!

9. (a) Expansion in the x-direction with factor ::! (c) Shear in the x-direction with factor 4. 10. (a) (b)

with the p ositive x- axis.

Jl

[

-1]

_v'4J 1-J * [:a l = [0 v'Jl = Re where () = 2

-1 , and so

Compression in the y-direction with factor Dilation wjtb factor 8. (c) Shear in the x -direction with factor - 3. (d) Shear in the y-clirflct.ion with factor 3.

(c)

(c)

(d)

A=

n

(d)

A=

[I 2

{b) Contraction with factor (d) Shear in they-direction with factor - 4.

172

Chapter 6

11. The action of T on the stanciatd unit vectors is e 1 = [ -t

-;

thus [T] = [Te1

-t

-t

-t -t

='Fe, , and e, =

-t

-+

-t

=

Te1 . and

-t

e2

= Te,; thus

4

and

= Te, • e, =

m m-; 4

111 =

14. The action of T on the standard unit. vectors is e 1

=

[i] -}

= T et,

e2

t hus [T]

0 -1

15.

( a)

16.

17.

y

=

[-!

m m-; m

mmm -+

=

=Te2; thus IT) = [Te1 Te2) =

13. The actioo of T on the "anda
m

J-t [

-t [

Te2) =

12. The action ofT on the staudard unit vectors is e 1

e2 =

]-t [

( b)

(c)

(a}

(b)

(c)

(a)

(b)

(c)

iL

y

=

Exercise Set 6.2

1 8. (a)

(b)

(c)

19. (a)

9 20.

(a)

[TJ(:

·,, ...0... _.....1

·..

22.

24.

I)

1

(o) [T] _, o o [

(b)

- ·'

[:

(b)

IRI =

[

(c)

!RI

[·

(a)

!Ill = [:

-!

Hl

(c)

10OJ r-21 = r-20 [000 1 1

(c)

0 0 1

,.-,................ ,._,

3

3

(b) [T) = [

,.0

(01,1 ]

(b)

[Tj =

0 \) 1

0o 0-1]o [ 1

L

0

n! m ! nJ m

(c) [T) =

-1 0 0

0 - 1

tnt

23. (a)

-3

nJ nl

(a)

I ' " ·••''

21.

{b )

s

0 - 1

!11.

173

0

(c)

IT]=

[!

f .iJ

(b) [Rj

= [ _; I

7z

(c)

[RJ =:

[

25. The matrix A is a. rotation matrix since it is orthogonal anrl det(A) == 1. Tbe axis o f rotation ;, found by .olv;ng the system (/- A)x ofth;s sys<em ;s x =

[:J = t

= 0 , whkh ;s

thus the matdx A

co:,.;;••:,

r=J=

A geneml solutioo

to• a mt:.;on about the x-a>ds.

Choosing the positive orientation and comparing with Table 6.2.6, we see tha.t the angle of rotation is

!I =

Chapt&r6

174

26. The matrix A is a rotation matrix since it is orthogonal and found by solving the of t his system is :<.

s[:sltem

=

=

u -n m.

A)x = 0 , which is

t

= l. The axis of rotation is =

A

genenJ solution

; thus the axis of rotation is t he line pnssing through the origin

and the point (1, 1, 1). The plane ing through the origin that is perpendicular to this line has equation x + y + z = 0, and w == (-1, 1, 0) is a vector in this plane. Writing w in column vector form, we have

W=

·-1] l , [ 0

Thus the rotation angle 8 relative to the orientation u cos 0 -_ w-Aw _ - 21 , and so 1Ll 7_ - 32>r(120o) .

=w

x Aw = (1,1, 1) is detenn.ined by

27. We have tr(A) = 1, and so Formula (17) reduces to v = Ax + ATx. Taking x

v

= Ax+ ATx = {A + Ar)x

=

=

000

= e 1 , this results in

'j

From this we conclude th at t he x-axis is t.he ax.is of rotation. Finally, using Formula (16), the rotation angle is determined by cosO = = 0, and so 0 = 28.

We have tr(A)

= 0, and so Formula ( 17) reduces to v

= Ax + AT x

+ x . Ta king x

= e1. this results

in

v = Ax+ ATx 1- x = (A 1

llll

+ 1 )x =

-

Thus the axis of rotation is the line through the origin and the point (1,1,1). Using Formula {16), the rotation angle is determined by cos()= = thus 8 = 2; .

29.

(a)

We have T, ([:]) =

Similad y, M,

=

!

m =

and M 3

thus T, is a lineA< ope < and 1!, =

=

(b) If x= (x , y , z), then T 1x · (x-T1x) = (x,O,O) · (O, y,z)

= 0;

thogonal for every vector x in R 3. Simllarly for T2 and T3 .

30.

(a)

We have S(e 1 )

= S(l, 0, 0) = {1, 0, 0), S{e2 ) = S(O, 1, 0) =

(k, k, l) ; thus the standard matrix for S is [S] =

[

oI 0l

kkl .

0 0 1

th,;;; T 1 x and x-T1x are or-

(0, 1, 0), and S(e3 )

= S(O, 0, 1) =


175

(b) The shear in the xz-directio n with factor k is defined by S(x, y, z) = (x + ky,y, z + ky); and its s t andard matrix

is lSI =

lkO] . l o . Similarly, the shear in the yz-direction with factor k [0 1

is dcnned by S{x, y , z) = (x , y

o

+ :x , z + kx); anti its standard matrix is lS) =

k 0 1

31. If u = (1, 0,0) then , on substituting a= 1 and b = c = 0 into Formula ( 13) , we have

l

0 () -sinO 0 cos sin{) cos £J The other entries of Table 6.2.6 are obtained similarly.


(a) (b) (c) (d)

The Tho The T he

unit umt unit unit

square is mapped onto th<'! seg ment 0 $ x s; 2 along the x-ax.is (y = 0) . square is rnappC(J onto the 0 y 3 alon g they-axis (x = 0). square is mapped onto the rectl'ingle 0 5 x $ 2, 0 5 y S 3. square is rotated a hout. the o r igin through an angle () -·-= (30 degrees clockwise) .

i

D2. I n order for the m a t rix to be orthogonal we rnust have

and from this it follows that o. = 0, b = -

-jg, c = "73

or a. ""' 0, b =

r.

=-

These are the

only p orss ibilit.ies.

D3. The t.wo columns of t.his matrix a.re or thogonal for any values of a a nd b. T hus, for t he m a trix to be orthogonal, all t ha t is required is t ha t the column \·ectors be of length I. T hus a and b must satisfy (<1 + b} 2 + (a - b) 2 = 1, or (equiv<\lcntly) 2a 2 + 21} = L D4 . If A is an o rthogonal matrix and Ax :=. rel="nofollow">.x . t hen L values of A (if any) must be of ah;;olutc D 5.

(a)

llxll = I!Axll = 11>-xll = I.XIIIxll.

Thus the eigen-

VectQrs paralle l to the line y = x will be eigenvectors corresponding to the eigenvalue>.= 1. Vectors perpendicular to y x will be eigenvect ors corres p o nding to the eigenvalue ,\ = -1.

=

(b) E very nonzero vector is an eigenvecto r corresponding to the eigenvalue ,X = DO. The shel.'lr in the x .. dircction with factor - 2; thus 1'(x, y)

D7.

From thP. polarization identity,

we have x · y =

= (x -

2y, y) and

Hllx + yJ]2 - l!x -

llx + Yll = llx - Yll tho.n the parallelogram having x and y as adjacent edges has diagonals of equal length a.nd mus t t herefore be a rectangle.

D8. !f

yll

2

)

-n

ITl =

= i(l 6 -

°

4)

= 3.

y-x

X+Y

X

Chapter 6

176

EXERCISE SET 6.3 1. (a)

ker(T) = {(O,y)J -' oo < y < vo} (y-a.xis), ran(T) = {(x,O)l- oo < x < oo} transformation T is neither 1-1 nor onto.

= {{0, y, z)l -

(b) ker(T) = {(x, 0, 0)1- oo < x
ker(T)

={

(d) ke<(T) = {

··

}• ran(T) = R2 • The transformation Tis both

m}

ran(T) =

R". The transfonnotion T

2. (a) ker(T) = {(x, 0)[ - oo < x < oo} (x-axis), ran(T) transformation T is neither 1-1 nor onto. (b) ker(T) == {(0, y, 0)1- co< y < oo} (y-a.xis), ran(T) Tho transformation T is neither 1-1 nor onto. {c)

··

ker(T) = {

(d) ker(T} = {

ran(T)

The

oo < y, z < oo} (yz-plane). and onto.

i• both J.] Md onto.

=

{(O,y)l- oo < y < oo} (y..,a,-xis).

=

{(x,O, z)l - oo < x,z < oo} (xz-plane) . ·

The

= R2 The transformat.ion Tis both 1-1 nnd onto. 3

}· ran(T) = R . The transformation Tis both 1-1 and onto.

3. The kernel of the tro.nsformation is the solution set of the system Ax = 0 . The augmented matrix of thi.s system can be reduced to x =

where - oo

t;hus the solution set consists of all vectors of the form

< t < oo.

4. The kernP.l of the transformation js the solution set of

2:l 0]

matrix of this system can be reduced t.o vectors of the form x

=t[

_: -

-5 0

where -oo < t

=t [

-:J

where -oo

The augmented

< oo .

matrix of this system can he reduced to

of the form x

=

thus the solution set consists of all

0 ;

:o

5. The kernel of the t
!]

0

-:J [:] m. =

The augmented

thus the .solution set consists of all vectors

0 : 0

< t < oo.

6. The kernel of the t ransformation TA : R 3 -t R4 is the solution set of the linear system

[: =; -:] [::] 3

4 - 3

XJ

=

J

0

177

Exercise Set 6.3

0

-! : 3 I

The augmented matrix of-this sy,tem can be <educed to [:

consists of aJI vcct.ocs of the form x = t [

7.

_:

=l]

thus

0

I

0 :

l:l m. =

The augmented matrix

has only the trivial solution and so

o:o

} .

of 1' is eqml.! to the solution space of the system

Thr.

thus the solution set

o:

0

oo

!m]; 0 0

ker(T) = {

t:

whe<e -oo <

The kernel of T is equal to the solution space of [

of this system can be reduced to

8.

-il

b

l

= J.

. l 0 1 OJ [0 1 0 i0 ; thus kcr(T) cons1sts of all vectors of the form 1

matrix of this system can be redlJced to x

9.

=t[

(a)

where · oo < t < oo.

The vector b is in the column space of A if and only if the llnE\ar system Ax = b is consistent. The a tJgrnented matrix ()f Ax = b is

H 2 2

0

8

l

1

and the reduced row ochf:lon form of this matrix is

0 I

L

6

0

0

Frorn this we conclude tha.t the system is inconsistent; thus b is not in the column space of A. (b) ThP. augmented matrix of the system Ax = b is

0:1• -13]

H 2

2 8

l

1: 8


0

-k

1

6

0

0

I

1] 6

0

Chapter6

178

From this we conclude that the system is con'listent, with general solution

4+ 3ltl = [4]3 + t [ 31] [ 3

X =

Taking t == 0, the vector b can expressed as a linear combination of the column vectors of A as follows:

=b =

8 10.

3

+

+ Oc3(A) =

6

+

3168

is

(a) The vector b in the column spttee of A if a.nd only if the linear system Ax = The augmented matrix of Ax = b is

[31 -21 0

1

1 5 5 -3

2 i] I I

}

1 -1 : 0

an d the reduced row echelon form of this matrix is 0 1 0

i 0]

1 1 - 1 •0 I 0 0 : 1

From trus we conclude t hat the system is inconsistent; thus b is not in the column space of A. (b) The augmented matrix of the system Ax = b is

3 -2 1 4 [0 1 and

1 5: 4]

5 -3 :

(j

1 -1 : 1

reduced row echelon form of t his matrix is 0 1

1l -11:: 2l ]

0

0

0 : 0

Prow this we conclude that the system is consistent, with general solution

Taking s = t = 0, the vector b can expressed as a linear combination of the column vectors of A as follows:

m b

2c , (A)

+ lc,{A) + Oc, (A) + lk, (A)

2

[!] nl +

Exercise Set 6.3

179

11. The vector w is in the range of the linear operator T if and only if the linear system

2x- y =3 x +z = 3 y - z=O is consistent.. The augmented matrix of t.his system can be nxiuced to 0 1 0

Thus the system h as a unique solution x

00 :: 21] I

1 : 1

= (2, 1,1), a.nd we have Tx = T(2, 1, 1) =

(3,3, 0)

= w.

12. The vector w is in the range of the linear operator T if and only if the linear system X -

y

=

1

x +y+z = 2 :t

+2z =- l

.is consistent. The augmented matrix of thls system 0 1

0

T hus the system has a unique solution x = (1, 2, - 1) = w . 13. The operator can be written as [:: Since df't(A)

J= (

be reduced to

O 0:' I] J 3

I

4

1

I

I _§

3

I

G, -

-n [::J,

and we have T x

=T (

=

-

t hus the standard mat rix is A = [

= 17 :f. 0, the operator is both 1·1 and onto.

14. The operator can be written as [::] =

[::] ;

th11s t he standard matrix is A =

Since

det(A) '" 0, the operu.tor is neither 1-l nor onto.

'WI] [wz =

15. The operator can he written as

:\ 2]

o

4 .

3

G

1.113

Siw;e

A)

[-1 3 2] [Xi] 2

1

o

:3

"

6

2:2

:

t hus the stanrlard ma.trix is A

=

::Z:J

= 0, the operator is neither l - 1 nor onto.

16. The operator can be written as

'WI] [l 23 ] [XJ] [ = 12 50 83 1.112

:z:2

1.1/3

X3

;

thus the s tandard matrix is A

=

Since dct(A) = - 1 f. 0, the operator is both 1-1 and onto. 17. The operator can be written as w =TAx , where A=

The_ran_gr. w

=

TA consists

vedors of

is not in the range of TA .

Since det(A)

,- -- ·- - -. w = t (j_:here

-oo

= 0, TA

is not onto.

< t < oo. Jn particular,

180

onto. The range of T,1 consists of all vectors w

4

= 1

2

=

-2 1]

1.

= [5

3 . Since det(A) = 0, TA is not 2 (w1 , w2, w3) f0r which the linear system

18. The operator can be written as w =TAX, where A

- 1 1

4

[::] =

W3

X3

is consistent. The augmented matrix of this system can be row reduced as follows:

5 -1

3l

1:

W2

1

2:

W3

1 -2 [4

Wt]

-t

[1 -2 1:l 0 0

9 -2 9 -2 ]

Wt ] 5wl

W2 -

W3 - 4w1

[1 -2 1\

-+ 0 0

9 -2 : 0 0:

W1

W3 -

Thus the system is consistent {w .is in the range of 1A} if ar:d only if w 1 partiClllar, the vector w = (1 , 1,1 ) is not in the range of TA . 19.

(a)

]

5w1 W2 +. 'tJIH ·

W2 -

-

w 2 + w3

= 0.

In

Th,, linear t.rl\.nsformation T A : R 2 -+ R 3 is one-to-one if A-nd only if the linear system Ax = 0 has only the t rivial solution. The augmented matrix of Ax = 0 is

and the

row echelon form of this matrix is

From t his we conclude that Ax = 0 has only the trivial solution, and so TA (b) The augmented matrix of the system Ax -=-= 0 is

IS

one-to-one.

2 3 0 -4

[

a.nd t he reduced row echeJon form of t his matrix is

oJ

o

}4 :I 1 -2 : 0

Fwm this we conclude that .4x

0 has the gene'"! solution x = t [

In p"'ticula<, the

systerr, Ax = 0 has nontrivial solutions and so the transformation TA : R 3 -+ R 2 is not onet.o-one. 20.

(a)

The range of the t ransformation : R2 -: R3 consists of those vectors w in Jl3 for which the linear system Ax = w is consistent. The augmented matrix of Ax = w is 1 -1 [

2

! Wt]

0 :

W2

3 -4 :

W3

Exercise Set 6.3

181

l [} -}

and this matrix can be row reduced as follows:

1-1 : WJ] [1.-: : [ 2

0 :

'W2

3 -4 :

W3

,

0

W)

2 :

W2 - 2WI

0 -1 :

W3 - 3w1

0

,

1 -1

,

2

0 [

0 -2

0

l

2:

o: I

Wz Wt - 2Wt 2w3 + w2- 8w1

From t his we conclude that Ax = w i..; consistent if and only if - 8w1 + w 2 + 2w., = 0; thus the transformation TA is not onto. (b) The range of the transformation TA : R3 R 2 consists of those vectors win R2 for which the 1inear system Ax = w is consistent. T he augme nted matrix of Ax = w is 2 3 0 - 4

l Wt] W2

1

and this· matrix can be row reduced as follows: 2

3 :

WJ

l

2 -1 : u.•2 +w1 . r

F'rorn this we conclude that Ax = w is consistent for every vector w in R 2 ; thus TA is onto. 21.

(a)

The augmented matrix of the system Ax= b can be row reduced as follow:;:

b,l [I

-2 -1

3 6 -2 3 3

4

0 ,

['

b2 b:J

0 0

-2 - 1 1

l -1

0

0

0

8 - 8 ; 02 - 2bl 6 -6 : 03 - 3bl

6

b, - {bl

I I I I

Ql

\f ib3 -

- ib

transformation T A consists

= 0, or 6/ rel="nofollow">1 + 3bz -

4b3 = 0.

b o tne form

[-ls:+1']=' [-!]

+t

(c)

]

kbz -

I

It follows that Ax = b is consistent if and only ( b) The range of

k

I I

3

0

-l 3: b, l

-2

m

Note. This is just one possibility; jt was obtained by solving for b1 in t.erms of 02 a nd b.> and then making bz and o3 into parameters. The augmented matrix of the system Ax ::; 0 can be row reduced to

(l

0 1 0

1 L: 0]

l

1 -1 0 0 0 : 0

Thus the kernel of TA (i .e. the solution s pace of Ax = 0) consists of all vectors of the form

tl = [-1]+ [-1]

-s + t -s-

X=

[

s t

s

-1 1

0

t

1 0

1

l

Chapter 6

182

DISCUSSION AND DISCOVERY DL

= 0 if and only if x = 0; thus T( u - v) = 0 implies u - v = 0 and u = v. (b) True. lf T : Rn -+ Rn is onto, then (from Theorem 6.3.14) it is one-to-one and so the argument given in part (a) applies. (c) 'ITue. See Theorem 6.3.15. (d) 'ltue. If TA is not one-to-one, then the homogeneous linear system Ax = 0 has infinitely many nontrivial solutions. (a) True. If T is one-to-one, then Tx

(e) True. The standard matrix of a shear operator T is of the form A = [ either case, we have det(A) =:= 1 D2.

'f:. 0 and

or A =

. In

so T = TA is one-to-one.

No. The transformation is not one-to-one since T(v ) to a .

= ax v =

0 for all vectors v tb.at:a.re parallel

D3 . T he transformation TA : R"-+ Rm .is onto.

D 4. No (assu ming vo is not a scalar multiple of v ). T he line x = vo +tv does not p ass through the origin and thus is not a subspace of It. follows from Theorem 6.3.7 that this line Calmot be flQ ual t o range of a linear operator.

nn .

WORKING WITH PROOFS P l. Jf Bx = 0, then (11B)x

= A(Bx)

= AO = 0 ; thus x is in t he nullspace of

AB.

EXERCISE SET 6.4 -3 1.

= BA =

!Tn

0

!][:

-2

[:

-3

-2

[TA oToJ = A E3

=

[:

1

4

2

=BA =

6

7

fTA oTaf = AB

=

-3

6

2 -3

.J 2

8

r

38 -18

6

[!

=

=

T2(Tt (x 1, x2)) = (3xt

+ Jx2, 6x1 -

{b) [72 c T1]

0 5

0

-9

=

10

18

-3

31

20] 18 43

22] 16 . 58

[12) =

3. (a) iT!) =

(c)

0

30 -1][ 4 1 -1 -3

44 - 11 45

3 -1]1 = [4012

[ - 41 5 2 -3

25

45

:11] = [-5-8 - -153 - 8I]

0

0

2. [Ts o TAJ

0] = [105 -1- 8 2!]

1 -3 2 4

ITt oTz) 2x2)

=

T1 (T2(x1t x2))

=

_:]

= (Sx1 + 4x2, x1 -

4x2}

Exercise Set 6.4

4. {a) !T1]

183

0 1

= - 1

{b) [12 o T1]

=

[T2]

-3

0 - 1

2 0][ 0 - 1 0 -1

:=

{T1 o T3]

2 ()

[-

u 2 0] [ 4 8 0] 2

0 l

4

-2 - 1 -3

0 1

0 -1 0 -1

-1 -3

=

3 3

=

-2 -4 -1

- 1 -2

3

(c) T2{T,(x1,x2,x3)) = (2x2,:r1 +3x2,17:r1 +3x2) T1 (T2(x1, x2, x3)) = (4:rt + 8x2, -2x1 - 4x2 - x3, -xi - 2x2 + 3x3) 5.

(a)

The s t andard matrix for the rotat1on is A 1 tion is A2

=[

=[ -

and the standard matrix for the

Thus t he standard m atrix for the rotation followed by the reflection is

A2 A 1 =

[01 01] [01 -1]0 [l0 - OJ1 =

(b) The s t andard matrix for the projection followed by the contraction is

(c)

The s tandard matrix for the reflection followed by the dilation is

.

.il·z A1

6.

(a)

=

[:l0 3OJ [10 -1OJ = [30 -301

T he s t andard matrix for the r.omposition is

A,A, A,

_J]

[

( b ) The s tandard matrix for the composition i:s

(c)

The composition corresponds to a counterclockwise rotation of 180°. The standard matrix is

R z<. R 1n R_ .n 3

7.

(a)

12

!2

=

R

11

7rr

r.

12+J2+ 3

=R rr =

[-1 0] 0 - 1

The s tandard matrix for the reflection followed by the projection is

184

Chapter 6

{b) The standard matrix for the rotation followed by the dilation is

0o] [ o 01 o v'2 (c)

8.

The

-72

0

=

[ 1o V20 -1

0

matrix for the projection followed by the reflection is

(a) T he standard matrix for the reflection followed by the projection is

(b) T he st.andard matrix for the rotation followed by the contraction is

A2A1

=

l 000] ['0 {!0--!0] = .[!0 :If0 0] [00 -3 0 ! 0.! 1

2

D.

,/3 2

6

,/3 6

(c)

The standard matrix for the projection followed by t he reflection is

(a)

The standard ma.trix for the composition is

_ fl 16

16

.1. ]

.l.. _fl16

16

l

1-

8

(b) T he standard matrix for the composition is

10.

(a) The standard matrix for the composition is

Exercise Set 6.4

185

(b) The standard matrix for the composition is

11.

(a)

=

We have A=

thus multiplication by A corresponds to expansion by a

factor of 3 in the y-direction and expansion by a factor of 2 in the x-direction.

(b) We have A =

=

multiplication by A corresponds to reflection about

the line y = x followed by a shear in the x-d irection with factor 2. Note. The fac toriut.t.ion of A as a product of elementary matrices is not unique. This is just one possibility.

(c)

We have

multiplicat ion by

=

Acorresponds to

reflection about t he x-axis, followed by expansion in the y-direct.ion by a fad or of 2, expansion in the x-direct ion by a factor of 4, and reflection about the line y = x. (d)

We havcA= [: -:]

=

in the x-dircction with factor -3, followed hy expans ion in the y-dircction with factor 18, and a shear in t he y-d irection with factor 4. 12.

(a)

We have

A=[! ;] =

by a factor of

!]

Acorresponds t.o compression

in the x -direct.ion and compression by a factor of

n;

=

(b) We have

multipHcatlon by

thus multiplication by

Acorresponds to a shear in the

:.r.-direction with f(l(:t(Jt 2, followed by reflection about t he line y = (c)

We have

=

A= [

g

in t he y -<.lirection.

:r;.

thus multiplk:a Lion by

A corresponds

to

reflection about the line y = x . followed by followed by expans ion in the x-direct ion by a factor of 2, expansion in the y-direction by a factor of [) , and reflection about t.he x-axis. (d)

We have

:] =

g

multiplicat ion by

A corresponds

x-dircction with factor 4, followed by a shear in they-direction with factor 2. Reflection of R.2 about the :r -axis. Rotation of R 2 about the origin through an angle of Contraction of R 2 by a factor of Expansion of R2 in the y-direc tion wjth factor 2.

13.

(a) (b) (c ) (d )

14.

(a ) Refl ection about they-axis. (b j Rotation about the origin through an angle of (c) Dilation by a fact.or of 5. (d) Compression ill x-direct.ion with factor

a shear in the

186

Chapter 6

15. The st andan.l matrix fqr the operator

the standard matrix fo r

[!

Tis A=

r - 1 is A- 1 = [- :

Since A is invertible, Tis one-to-one and

thus

r - 1(w1, w2 ) = (-w1 + 2'W2, tv 1 -

16. The standard matrix for the opP.rator Tis A = the standard matrix for

y-l

is A-

1

Since A is invertible, Tis one-to-one and

!l

=

thus r

17. The standard matrix for the operator Tis the sto.ndn.rd matrix for r - l is A- 1

19. The standard mat r ix for the operator T is A

r- 1 is

A- t

21.

=[

1-2 2] [

=

2 1

1 1

3

-i [

1 . 0

7

I .

3

0

3

I

I

-2

Since A is invertible, T is

Ill

=1 ;thus the fo<mula fo< r -

Since A is inve
26

26

thus the formula for

5

T3

=

r- 1 is

26

lL is easy to see d irectly (from the geometric definitions) that T 1 o T 2 a J.qn fo llows fTo m

is

:J.

lr-*fi --ri-J5=7] ; I

1

'.i

22. The standa
(a)

= (W2, - wl)·

4-1] Since A is inver tible, T is one- to-one

=

and the standard matrix for T - 1 is A - I =

23.

A is invertible, T 1s one-t(rone and

Since A is not invertible, Tis not one-to-one.

3

is A -l =

+ iw2)·

41 t hus the formula for T -l is -sJ ..

-2 -2

1

1 12wl

-2 2 -3 ;

=

The st.andarrl matrix for the operator T is A

the stand ard mat rix for T -

=
Since A is not invertible, Tis not one-to-one.

-1

20. The standard matrix for 7 ' is A

(wl,w2)

thus r- 1(wl, w2)

18 . The standard matrix for the operator Tis A =

a n d thf' standard matrix for

1

-

Since

=[

w2) .

and IT2} lTd=

=0 =

T 2 o T 1 • This

=

187

Exercise Set 6.4

(b) It js easy to see directly that the composition of Tt and T2 (in either order) corresponds to rotation about the origin through the angle 81 + Bz; thus 11 o Tz = T2 o T1 . This also follows from the computation carried out in Example l.

lTd = rlO l

(c) We have

0 0 cosO t -sin9t

0

£11

sin

l

and [12]

[cos 0 sin 0 2

=

r.:os&1

2

-

sin02 0

cos02

thus

0

cos02 - sinBz [Tx oTz] =ITt} !Tz) = coslh sin Bz cosB1 cos82 [ sin lh sin Bz sin B1 cos B2

Ozl

- cos 81 sin Bz sin 81 sin cosfh cosBz - sin 81 cos Bz sin 8 1 COSO) and it follows tha t T1 o Tz 24 .

(a)

i.ha.t T 1 oT2 =-I :;: TzoT1 . This l::llso follows from the fact that

It is ea..<;y t.o see

[Td [Tz!=

=

=

(c)

and lTz]ITJ] =

and [T:d

(b) We have [Tl] = [T2 o T1 ]

f Tz o T1 .

[T2} [T1 )

=

= [:::

=

IIi oTz]

°l-1= [T o 12] . lt fo llows that T j 1

0

S in

1

= jTJ} [T2 ] o 1'2

=

and

f Tz oTt .

=kl ; t.hus fTI] [Tz] === (kJ)[TzJ=kl12] and [TzJ ITl i = ITz]{kl} =

We have 0 0 k

k [T2 }. Jt follows that 1'1 o T 2 = T 2 o T 1 = kT2. 25.

Wt:;

have JJ, 13 =

[

_.\ 'l

and J-1, 16

f _,;

--

2 '

- 1

'\!jl 2

.

T int: rel="nofollow"> the standard matrix for the composition

.ill [ 2 1

2

26. We have H:ri-i =

and f/, 18 =

lirr /8

H.

7ft4

=

-

!2 viJ

-z-

T hus t.he standard matrix for the composition is

..!_

V2

_

7i][0 1] _[ 72 1

72

l O -

_

_!_

v'2

27. The image of the unit square is the parallelogram having the vectors T( e!)

= U]

and T( ez)

= [-

of this parallelogram is jdet(A}I = ldet

as adjacent sides. The area

= 12 +

II= 3.

ChapterS

188

28. The image of the unit square is the parallelogram having the vectors T(e1 )

=

g]

T{e 2 ) =

[!]

y 7f-·

as adjacent sides. The ar(;a

of this parallelogram is ldct(.A)I = \det

!] I

,

J

{5. 7)

J

= 18 - 91 = 1.

"---· ( • 3)

--···

1/

") I

(3.

)

-

'f .1:

5

DISCUSSION AND DISCOVERY D l . (a) 'l'rue. If 1't(X) = 0 with x # 0 , then T2(T1 (x)) = T2 (0) = 0 ; t hus T2 o is not one-to-one. ( b ) False. If T 2 (ran(T1 )) = R" then T2 o T,, is onto. (c) False. If xis in Rn, then Tz(T1 (x)) = 0 if and only if T,(x) belongs to the kernel of Tz; thus if T t is one-to-one and ra.n(TI) n ker(Ti) = {OL then the transformation 72 oTt will be one-to-one. ran(T2 ) #- Rk . (d ) True. If ran{T2 ) :F Rk then ra.n(T2 oT1)

02. We have RfJ

=

sm ""

R H R _1 0

0

1 0 ], 0 - 1

- sinf{J'], H0 = ( cos

2

and R;; 1 "

= R-p = [-

sm

cos"'

Thus

2

[cos {3 - sin {3 2sinf1cosf3 l - [cos2/3 sin2/3] - H f.i ··2 sin {3 cos f3 sin 2 f3 - cos2 f3J - sin 2/3 -cos 2/3 i3 _

and su multiplication by the matrix

1

corresponds to reflection about the line L.

D3. From Example 2, we have He, Hez = R 2ctJ 2 - Ba) · Since {)= 2(0it follows t hat Re Thus every rotation can be expressed as a composition of two reflections.

=

CHAPTER 7

Dimension and Structure EXERCISE SET 7.1 1.

(a)

(b)

(c)

2.

(a)

{b)

(c)

5.

(a) Any one of the vectors (1 2), ,1), ( - 1, - 2) forms a basis for the line y = 2x. (b) Any two of the vectors (1. - 1, 0), (2, 0, - 1), (0, 2, -1) form a basis for x + y + 2z

6.

(!

1

= 0.

(a) Any one of the vectors (1, 3), (2, 6) , -3) forms a basis for the line x = t y = 3t . (b) Any two of the vectors (1, 1,3), (1,-1 2}, ('2, 0, 5) form n. basis for the plane x=t1 +t y =- tl - l:t, z 3t, + 2t2. 1

1

7.

1 1 0 : ] I - 1 1 0

3 5 -· 1

The augmented matrix of the system is [ 1 0 l 0 I 0] : . [0 1 4 1 10

this matrix is

21

and the reduced row echelon form of

Thus a general solution of the system is

x-- (-l" 4

VJ

_!s 4 ·-· t J s ) t) --

"(-.!4 - !4 1 O)+t(O - 1 0 1)

tJI

t

I

l

'

l

\

where -oo < s, t < oo. T he solution space is 2-dimensional wil.h canonical basis {v 1 ,v2_} where V J :.; ( -

1 -

!

1

1) 0)

(l.J1

d

V 2 """ ( 0 > -

1, 0 l) • I

8. Tho augmented mat,;x of tho system i' [: matrix is

I () :1: 0] [

0 1 2 •o . I 0 0 O tO

=: ,;

nnd t ho ceduced row echelon fo,m of th ;s

Thus the general solution is

X= (

-3t, - 2t , t)

t( -3 , -2, 1)

where -co< l < oc. The solution space is 1-dimcnsional with cnnonic..al basis {( -3, -2, 1)} .

0

9.

The augmented matrix of tl1e ::;ystcm is [-

l

0

-

-

1 - 2

0 -1

0

l

I

]

I

°] 0

and the reduced row echelon form

lt' O

I 1 0 IJ I :0

of this matrix is

. Thus the general solution is [

0 0 0 0

o:o

x = (- s - t , s,- t,O,t) = s(-1, 1,0,0,0) +t(-1,0, - 1,0,1) where -oo < .'l, t < oo. The solution space is 2-dimensional with canonical basis {v 1 , v'2} where Vt = (-1, 1,0,0,0) and v 2 = (- J, 0, - 1,0, 1). 195

t96

Chapter 7

10. The augmented matrix of the system is

01 00 - 31 - 31 : 0]0

1

.

-: -!i

r: 0

1

0 -1

i

I

and the reduced row e-chelon form

0

_l .

.

. 0 o f thJs maLnx 1s

I

Thus t he general solution is

[

x = ( -3s +

t, !t,s,t)

= s(-3,1 , 0,1,0) + t( ! , -1, j, o, 1)

where - oo < 8 , t < oo. The solution space is 2-dimensional with canonical basis {vh v 2} where Yt = (- 3,1, 0,1, 0) and Y2 = (3, - 1,1,0, 1). 11.

(a)

T he hyperplane (1 ,2, -3).1 consists of all vectors x = (x, y,z) in R 3 satisfying.:the equation x + 2y- 3z = 0. Using ·y =sand z = t as free variables, the solutions of this·equation can be written in the form

= (-2s + 3t, s , t) = s( - 2, l, 0) + t(3, 0, 1) T hus the vectors v 1 = (- 2, 1, 0) and v 2 = (3, 0, 1) x

where -oo < s, t < oo. form a basis for the hyperplane. (b ) T he hyperplanC! (2, -1, 4, 1)..!. consists of all vectors x= (x 1 , x 2 , xa, x 4 ) in R 4 satisfying t he equation 2xt - xz + 4x3 + x 4 = 0. Using x1 = r, x3 s and x 4 = t as free variables, the solutions of this equation can be written in the form

=

x

= (r, 2r + 4s + t, s, t ) =

r(l, 2, 0, 0)

where -oo < r , s, t < oo. T hus thevcctors v 1 form a basis for the hyperplane. 1 2.

=

+ s(O, 4, 1, 0) 't t(O, 1, 0, 1)

(1, 2, 0, 0), v 2

"'

(0, 4, 1, 0), and v 3

=

(0, 1, 0, 1

(a) The h yperplane ( - 2, 1, 4).L consists of all vectors x = (x , y, z) in R 3 satisfying the equation ·- 2x + y + 4z = 0. Using x = s and z = t as free variables, the solutions of this equation can he written in t he form

x

= (s, 2s -

4t, t ) = s(l , 2,0) + t(O , -4 ,1)

where - oo < s, t < oo. Thus the vect.o-rs v 1 = (1, 2, 0) and v 2 = {0, - 4, 1) form a. basis for t.he hyperplane. (b ) The hyperplane (0, - 3, 5, 7)J. consists of all vectors x = {x 1 , x2 , x 3, X.t) in R4:satisfy ing t he equation -3x2 + 5x3 + 7x4 = 0. Using x 1 = r, X3 = s and x 4 = t as free variables, the solutions of this equation can b e written in the form x = (r,

+

s, t) = r(l , 0, 0, 0) + s(O, j ,1, 0) + t(O,

where - oo < r, 8, t < oo. Thus the vectors v 1 form a basis for the hyperplane.

= (1, 0, 0, 0), v 2 = {0,

0, 1) 1, 0), and

v3

= (0,

0,


(a) (b) (c) (d)

Assuming A :1 0 , the dimension of t he solution sp ace of Ax = 0 is at most n- 1. A hyperplane in R!' has dimension 5. T he subspaces of R5 ha.ve dimensions 0, 1, 2, 3, 4, or 5. The vectors Yt = (1 ;0,1 ,0), v 2 =::: (1 ,1, 0,0), and v 3 = (1,1,1,0) are linearly independent, t hus they span a 3-dimensional subs pace of R 4 •

F.XERCISE SET 7.2

197

D2. Yes, they are linearly independent. If we write the vectors in the order v 4 = (0,0, 0,0,

= (0,0,0,1, "', '),v2 = (0,0, 1, ·, • , •), v1 = (l, ·, • , •, • , •)

then, because of positions of the leading ls, it is clear that none of these vectors can be written as a linear combination of the preceding ones in t.l!c list. D3. False. Such a set is linearly dependent is a linear combination of its predecessors.

when

set is wrltt.eH in reverse order, some vector

D4. The solution space of Ax = 0 has positive dimension if and only H the system has nontrivial solutions, this occurs if and only if det(A) = 0. Since det(A) = t 2 + 7t + 12 = (t + 3)(t + 4), it foll ows t hat the solution space has positive dimension if and only if t = -3 or t = -4. The solution space bas dimension 1 in each case.

WORKING WITH PROOFS Pl.

(a) If S = {vh v2, ... , vk} is a linearly dependent set in R" then t here C\re scalars Ct , cz, ... ,ck

(not all zero) such that c1v 1 + c 2v 2 + · · · +

= 0.

lt follows that the setS'= {v 1, v 2, .. . ,

vk, w1, .. . , wr} is lillearly dependent since c 1v 1 + c2v2 i::; nuut1 i v i<\l Jepcndeucy rela.tio n i:l.lllung its demen ts.

I··· · + C>Yk + Owl+ ·· · + Owr

-=0

(b) If S= {v t, v 2 , . . . ,vJc} is a linearly independent set in R"' then there is no nont rivial dependency relation among its elements, i.e ., if Ct VJ + c2v2 I····+ = 0 then Ct = c2 = ··· = == 0. It follows t hat if S' is a ny nonempty subs et of S then S' must. aho be linearly independent since a nontrivial dependency among its elements would also be a nontrivial among t he elements of S . P2./ If k #·0, then (ka) · x ' - k( a- x) = (l if and only if a· x

=

0: t.lm::; ( k:a )J

= a.l

EXERCISE SET 7.2 1.

(a)

(b) (c)

A bMis for R 2 must contain exactly two vectors (three are t.oo many); any of three vectors in R 2 is linearly dependent. A basis for R3 must centum three vect ors (two nre not eno11gh); s<>t. 0 f (.wo V<'('t.01'S cannot span R:1 Thf: v 1 and v 2 are linc
2.

(a) A basis for R2 must contain exactly two vectQrS (one is not enough). (b) A basis for R 3 must contain exact.ly three vect.ors (four are t.oo many) (c) These vectors are linearly (any set of vectors cont<.>ining t.he zero vector is de pendent.).

3.

(a)

The vectors v 1 = (2, 1) and v 2 ""'(0,3) are linearly independent of the other; thus these two \'ect.ors form a hasis for R 2 . (b) The vector Vz = (-7,8,0) is not a scalar multiple of vl = (4, l,O), linear combination of v 1 and v 2 since a ny such linear combination component. Thus v 1 , v 2 , and v 3 are Iincarly independent, anJ it from a basis for R 3 .

neither is a scalar m1.tltiplc and V:J:::: (1, 1, 1) is not a would have 0 in the third fo llows that t.hese vectors

The vectors v 1 = (7, 5) and v 2 = (4 , 8) arc linearly independent since neither is a scalar multiple of the other; thus these two vectors form a for R 2 . (b) The vector v2 (0 1 8,0) is not a. scalar multiple of v 1 = (0, 1, 3), and v 3 = (1,6,0) is not a linear combi1iation of v 1 and v2 since any such linear combim.t.ioJl would h ave 0 in the first component. Thus v 1 , v 2 , and v 3 are linearly il.dependcnt, a nd it follows that these vecto rs from a basis for R3 .

4 . (a)

=

Chapter 7

198 4

5.

(a)

(b )

(a)

Since de t(.l.)

= 0,

Rl.

T he matrix having v 1 , v 2 , a nd v3 as its column vectors is A= [ :

:=

8

t he column vectors a re linearly dependent and hence do not form a ba.si& for

Jl3 .

The matrix having v 1 , v 2 , and v 3 as its column vectors is A= [

Since det(A) = 0, -8

5 -3

the column vectors are linearly dependent and hence do not form a basis for R 3 .

=

for a.ny value of t . In pa r ticular, corresponding to t = 0, t = 1, and t = -6, we have v v , + v 2 + v 3, v = 4v 2 + + v4, and v = 7v 1 + 4v2 + 3v 3 - 6v4 .

=

l

4

7

(a)

-!

det(A)

(b) The matrix having v 1 , v 2 , and v 3 as its column vectors is A =

=f 0, t he column vec tors are linearly independent

An arl•itmry vector x

= (:r., y , z:)

l

-1

a..nd hence form a basis for R3 .

in R 1 can be written as

t hus the vectors v 1 , v 2 , v 3 , and v 4 span R 3 . Oo tbe other hand , since

the vecl ors v 1,

v2,

v 3, and v 4 are linea rly dependent. and do not form a basis for R 3 .

(b ) T hP ,·ector equa t ion v

= c 1 v 1 + c2 v 2 + c 3 v 3 + c4 v 4 1 0

=··'··-. . where - oo

(a)

3

:J

=

m

= t, n general solution of this system is given by

a nd,

8.

is equivalent to the linear system

0

0 2 0

[0 0

-· . . .

l

. ·......

.t

= .•

...

< t < oo. Thus

An nrhitra ry \'ector x

= (x , y, z} can be written as x = zv 1 + (y- z)v, + (x - y)v3 + Ov4

.". '

r} I

:f: 0,

Sir•.cc det{A)

-4 -2 -3 the column vectors a re linearly independent and bence form a basis for

- 5 -2

6.

-5

The ma t r ix having v 1 , v2 , v 3 as its column vectors is A = [

il- \

.,

t.hus the vectors v , , v2, v 3 , and v 4 span R3 • On the other hand, since v 4 = 2v 2 vt:ct.ors v 1, v 2 , v 3 , v 4 are linearly dependent and do not form a basis for R 3 .

J ·

.

(l ' I

..

. .,.),

. . V1 . ., ·.) I

\1.,

+ v3 ,

the

199

EXERCISE SET 7.2

{b) The vector equation v

= c1v 1 + c2v 2 + c3v 3 + c4v4 is equivalent to the linear system

The augmented matrix·of this system is

1 3 1] I

1 0 2 :2

o o ol3 and the reduced row echelon form of this matrix is

1 0 0 0: 3] [0Q 01 01 2:-1 1 -1 I

Thm:,

c.1 = t , a general solut io n of the sys tem i:; t;i•:cn by C1 =

where -oo < t

3, cz

=:

-1-

2t , C3 =

- 1 - - t,c4

=t

< oo. Thu:; v

= 3v1 ·- (1 + 2t)v-z -

(1 -1- t)v3 + LV4

for any value of t. For example, correspond ing _t9 t 3v 1 - v7 - v3, v = 3v 1 + v2- v 4 , and v = 3v , -

=0,

t = - 1, and t = -

we have v =

9. The vector v 2 = {1, -2, - 2) is not a scalar multiple of v 1 = ( - 1, 2, 3) ; thus S == {v 1 , v 2 } is a linearly independent set. Let A be the matrix having v 11 v 2 , and e 1 as its columns, i.e., A :::

-1 1 1] [ 2 3

-2 - 2

0 0

Then det( A) = 2 ;6 0, and it follow:; from this that S' = { v 1, Vz, e1} is a linearly indepew.lent set and hence a basis for R 3 . [Similarly, it can he $hown that {v 1 ,vz, e2} is a b asis for R3 . On t.he other hand, set. {vl> v 2 ,e3 } is linearly dependent and thus not t':. basis for R 3 .J 10. The vector v2 = (3, 1, - 2) is not a scalar multiple of Vt = (1 , -1 , 0) ; thns S = {v1 , v 2} is a linearly independent set . L€t A be the matrix having v 11 v 2, and c 1 as columns, i.e.,

.4

= [-

0

:

- 2

0

Then det(A) = 2 f 0, and it follows ftom this that S' = { vb v2 , e1} is a linearly independent set and hence a basis for R 3 . [Similarly, it can be shown t hat {v 1,v 2 , e2} and {v 1, v 2,e3} are bases for R 3 .] 11.

Chapter7

200

12.

vector v 2 from the set S then the remaining vectors a.re

We h ave v 2 = 2v 1, a.nd if we linearly independent since· det

11 I] = [

13. Since dct o

1 1

1

2

2

l

2 -3

4

f 0, the vectors

0 0 1

The vector equation v

VJ.

= 27 =I= 0. Thus S' = V3

{ Vt, v 3 , v 4 } is a basis for R3 •

(column vectors of the matrix) form a basis for R3 .

= c1v1 + c2v2 +c3v3 is equivalent to

which, from back substitution, has the solution c3 v = -3vl + 4v2 + v3.

= 1, c2 =

the linear system

5- c3

= 4, c1

= 2- c2 -

c:v=- -3. Thus

= - 2 =f. 0, the vectors v1r v2, va (column vectors of the matrix) form a basis for

14. Since del 0

I

J

R3 . The \'P.ctor cquatioH v ::: c1v 1 + c2v 2 + c3v3 is equivalent to the linear system

15.

Since u · n . (1)(2) + (2)(0) ( - 1)(1) = 1 i: 0, the vector u is not orthogonal ton. Thus Vis not contained in W. (b) The line Vis parallel to u = (2, - 1, 1), and the plane W has normal vector n = (1,3, 1). Since u·n:o::: (2)0)+( -1 )(3)+(1}(1) :::0, the vector u is orthogonal ton. Thus Vis contained (a)

il'

w.

Sim:e u · n = (1) (2) + (1)( l) + (3)( - J) = 0, the vector u is ton. Thus Vis contained in IV. (b) T he line V is parallel to u = (J, 2, - 5), a.nd the plane W has normal vector n = (3,2, 1). Since u· n = (1) (3) + (2)(2) + (-5)(1) = 2 i: 0, the vector u is not orthogonal ton. Thus. Vis not contained in W .

16.

(a)

17-

(a)

T he vector equation c1(1, 1, 0) + cz(O, 1, 2) ::;ystem with augmented mo.trix

+ CJ(2, 1, 3) = (3, 2 , -1)

is equivalent to the linear

0 1 2

T he reduced row echelon form of this rna.t.rix is

1 hus (3, 2, - 1) =

13 5

(1, 1, 0) -

t

l;l

1

0! Q l _i

0

1

0

I I

I I

1i

1

S

(0, 1, 2) + {2, 1, 3) and, by linearity, it follows that

... 1)= \ 3 (2,1 ,-1)-

= !(26, 13, -20)

201

EXERCISE SET 7.2

(b) The vector equation CHI, 1,0} + c 2 (0,1, 2) + c 3 (2, 1,3) te m with augmented matrix

= (a,b,c)

is equivalent to the linear sys-

1

..o

2

3 : c

The reduced row echelon form of this matrix is

lb + tc){O, 1,2) +

0) +

Thus (a, b, c)

1, -1) +

T(a , b, c) = (! a +

ic)(2, 1, 3) and

+

+ !c)(l , 0, 2)

1, 0)

3b + ISC> ga 1 + [; 4b - -gC, 2 7 , !) = ( gaT -a + C)

( c)

From lhe formula above, we have {T I

!

=[

-1

18.

(a)

'fhe vector equation c 1 (1, l , l) + c2(l , l, 0} + c3(l, 0, 0) = ( 4, 3, 0) is equivalent to t he linear systern

has solution c1 hy Jineari t.)',

\V€

= 0, c 2 = 3,

= 1. Thus (4, 3,0) = O(l, l.Jlj-.-3.(Ll 0) +

nave

-

.

·.

/

T('t 3, o) = o(3, o, I)+ 3(2, 1, 3, - 1) + 1(5, - 2, 1, ( b)

T he vector eqnatiun c, (1, l, 1) I· c2 ( l, l, 0)

+ c:l( 1, 0, 0) = (a , b, c)

.(1 1, 1, 10, -3)/ - ...• •.

-_..? ·

ill equivaleht .to the linear syil-

rl I 1] [:t1] [a]

::

which ha.:; s o!ution

x1 =

c,

x2

=b -

c,

X3

=

=a - b. Thus

(a, b, c) = c(l, 1,1) + (b - c)(l, 1,0) + (a-- b)(l , O, 0) a nd so, by lmearity, we have

T(a, b, c)

= c(3, 2, 0 , 1) + (b- c)(2, 1, 3, -1) + (a -

b)(5, ·-2, 1, 0)

= (5a - 3b+ c, - 2a +3b+ c,a + 2b - 3c,-b + 2c)

(c)

F\-om the lo<mula above, we have

ITI =

[-j -3 1]

3 l 2 -3 .

- 1

2

202

Chapter 7

19. Since det -5

-! -:] 8

:=

5

- 100

0, the vectors

v 1

= (1 , -7, - 5), v

2

= (3, -2, 8),

V3

= (4, :-3, 5) form

-

a b asis for R 3 . The vector equation c 1v 1 + c 2 v 2 augmented matrix

+

= x is equivalent to the linear system with

1 - 32 - 34:x] ly

-7

[- 5

8

5 : z


[

1 0 0 0 1 0 0

i- iox - doY + 1 1

1 I

0 1 1:

-

!x - !., + 1z 2 4" 4

33 50 x

23 19 + 10oYImiz

Thus a general vector x = (x, y, z) can be expressed in of the basis vectors v 1 , X

= (- ;0 X

-

y+

160 Z j V l

+ (-

-

h+

Z) V2

+

+

'!/ -

as follows: Z )v 3

DISCUSSION AND DISCOVERY 01.

(a) True. Any set of more than n vectors in Rn is linearly depe ndent. (b) True. Any set of less than n vectors cannot be spanning set for Rn. ( c) True. 1f every vector in Rn can be expressed in exactly one way as a linear combina tion of the vectors in S , then S is a busis for R n and t hus must contain exact ly n vectors. (d) True. If Ax = 0 Iws infin itely many solutions. then det(A) ::: 0 and so the row vectors of r1 are linearly dependent. (e) True. If V £:; W (or W V) and if dim( V) = d im ( W), then V = W .

02. No. If s = {vl , Y2, - . . 'Vn} is a linearly dependent set in Rn, then s is not a spanning set for n n; r.hus it. is not possible to create a basis by formiug linear combinat ions of the vectors in S . 03. E ach such operat()r corresponds to (and is determined by) a permutation of the vectors in t he basis B . Thus t here a rc a t.otal of n! s uch operators. 04. Let A be the matrix having the vectors v 1 and dct (A) ""' (sin 2 o- sin 2 /3)

v2

as its columns. Then

- (cos 2 a - cos2 {3) = - cos 2o + cos 2{3

and det(A) =/= 0 if and only if cos 2a j cos 2/3 , i.e., if and only if o =/= ±/3 + br where k = 0, ± I , ± 2, .. . . For these values of a and {3, the vectors v 1 and v 2 form a basis for R 2 . D5. Suppose W is a. s ubspace of Rn and dim( W) = k. If S = {w1 , w2 , .. . ,wj } is a spanning set for W, then eit her S is a basis for W (in which caseS contains exactly k ve<::tors) or, from Theorem 7.2.2, a. b asis for W can be obtained by removing appropriate vectors from S. Thus the number of clements in a spanning set must be at least k , and the smallest possible number is k.

WORKING WITH PROOFS Pl. Let V b e a nonzero subspace uf R" , and let v 1 be a nonzero vector in V. If dim(V) = 1, then S = {v 1 } is a ba..<;is for V . Otherwise, from Theorem 7.2.2(bJ, a busis for V can be obtained by adding appropriate vedors from V to the set S.

WORKING WITH PROOFS

203

P2. Let {v1, Vz, ... 1 vn} be a basis for Rn and, fork any integer between 1 and n, let V = spa.n{v1, v2, • • • 1 vk}· Then S = {vt. v 2, ... , vk} is a basis for V and so dim{V) = k. The subspace V = {0} has dimension 0. ·

P3. LetS= {v1o v2, ... , vn}· Since every vector in Rn can be written as a Hnearcombinationofvectors in S, we have span(S) = Rn. Moreover, from the uniqueness, if Ct v1 + cz v2 + ·· · +env n = 0 then Ct = Cz =···=en= 0. Thus the vectors Vt. vz, ... , Yn span Rn and are linearly independent, i.e., S = {vl, Vz, ... 1 vn} is a basis for Rn.

P4. Since we know that dim(R"} = n, it suffices to show that the vectors T(vt) 1 T(v2), ... ,T(vn) are )inearly independent. This follows from the fact that if ctT(vl) + CzT(vz) + · · · + c 11T(v,) = 0 then

T{c1v1 + CzVz + · · · + CnVn) = CtT(vt} + CzT(vz) + · · · + CnT(vn) = 0 and so, since Tis one-to-one, we must have c1v1 + czvz + · · · + CnVn = 0. Since VJt v2, ... , Vn are linearly independent it follows from this that c 1 = c2 = ···=en= 0. Thus T(v1 ), T(vz), ... , T(v,) are linearly independent. P 5. Since B = {v 1 , v2, ... , vn} is a basis for R"', every vector x in Rn can he expressed as 1i linear combination x = c1 v 1 + czv 2 + · · · + c,. Vn for exactly one choice of scalars c 1 , ... , Cn. Thus it makes sense to define a transformation T: Rn --1 Rn by setting

T(x) = T{ctVl + CzVz

+ ... + CnYn) =clwl + c2w2 + ... + CnWn n

It is easy to check that Tis linear. For example, if x

n

= I: CjVj andy= E d;v1 , then j=l

n

T(x + y)

= T(l= (c; + d;)vJ) j=l

=

j=l

n

n

j=l

j;l

L (cj + dj)wj = L

n

CjWj

+L

d;w;

= Tx + Ty

j=l

and so T is additive. Finally, the transformation T has the property that Tv; = w; for each = 1, ... , n, and it is clear from the defining formula that T is uniquely determined by this property.

j

P6. (a)

Since { u 1 1 Uz 1 U3} has the correct number of elements, we need only show that the vectors are linearly independent. Suppose c 1, C:lJ c3 are scalars such that c1u 1 + CzUz + C3U3 = 0. Then (c1 + Cz + c3)v1 + (cz + c3)v2 + c3v3 = 0 and, since {Yt. Vz, v3} is a linearly independent set, we must have c 1 + c2 + c3 = c2 + c 3 = c3 = 0. It follows that c 1 = c2 = C3 = 0, and this shows that Ut, uz, us are linearly independent. (b) If { v1, v2, ... , Vn} is a basis for R 1", then so is { u 1, u2, ... , un} where

... ' P7. Suppose x is an eigenvector of A. Then x :f. 0, and Ax = >.x for some scalar >.. It follows that span{x,Ax} = spa.n{x,>.x} = span{x} and so span{x,Ax} has dimension 1. Conv\...sely, suppose tha.t span{x, Ax} has dimension 1. Then the vectors x f. 0 and Ax are linearly dependent; thus there exist scaJars CJ a:.d c2 , not both zero, such that c 1 x + c2 Ax = 0. We note further that c2 =I= 0, for if c2 = 0 then since x =I= 0 we would have c1 = 0 also. Thus Ax = >.x where

>.::;;; -cdc2.

P8. SupposeS= {vt, v2, ... , vk} is a basis for V, where V C Wand dim(V) = dim(W). Then Sis a linearly independent set in W, and it follows that must be basis for W. Otherwise, from Theorem 7.2.2, a basis for W could be obtained by adding additional vectors from W to Sand this would violate the assumption that dim(V) = dim(W). Finally, since S is a. basis for W and S C V, we must have W = span(S) C V and soW= V.

Chapter 7

204

EXERCISE SET 7.3 1 . The orthogonal complement of S

= {v 1 , v 2 } is the solution set of the .system

-Jt,

!t,

A general solution of this system is given by x = y= z =tor (.x, y,z) S.l is the line through the origin that is parallel to the vector (1}.

A vector that is orthogonal to both v 1 and v 2 is w

!,

= v1 x Vz = [ 1i 0

j 1

I<] .3

=

Thus

t(

= -7i + j

+ 21<: =

( -7, 1, 2).

2 -1

Note the vector w is parallel to the one obtained in our first solution. Thus SJ. is the line through the origin that is parallel to {-7, 1, 2) or, equivalently, parallel ,1). 2. The orthogonal complement of S = {v 1 , v 2 } is the solution set of the system

A general solution of this system is given by x = y =-1ft, z =tor (x,y,z} = S.l. is the line tluough the origin that is parallel to the vector (!, - ¥ , 1).

..

A vector that is orthogonal to both v 1 and v 2 is w = v 1 x v 2

=

[I 2

j 0

- 1

1

1

5

kl = i -

ll j

-

i

1

1

1). Thus

+ 2k = (1, - 11 , 2).

Note the vector w is parallel to the one obtained In our first solution. Thus 51.. is the line through the origin that is parallel to (1, -11,2) or, equivalently, parallel to (j,3. We have u · v 1 = ( -1)(6) + (1)(2) + (0)(7) ·1- (2)(2) = 0. Similarly, u · v 2 = 0 and u · v3 = 0. Thus u is orthogonal to any linear combination of the vectors v 1 , v 2 , and v 3 , i,e., u bek>,ngs to the orthogonal complement of W = spa.n{vl> v2, v3}. " 4. We have u · v 1 = (0)(4) + (2)(1} + (1)(2) + (-2)(2) = 0, u · v 2 = 0, and u · v 3 = 0. Thus u is orthogonal to any linear combination of the vectors vlt v 2 , and v 3 , i.e., u is in the orthogonal complement ofW. 5; The line y = 2.x corresponds to vectors of the form u = t{1,2), Le., W = span{(l, 2)} . Thus WJ. corresponds to the line y = or, equivalently, to vectors of the form w = s(2, -1). 6. Here W J. is the line which is normal to the plane x - 2y - 3z = 0 8Jld pa:3Sea through the origin. A normal vector to this plane is n = (1, - 2, -3); thus parametric equations for WJ. are given by: X =

t,

y

= -2t,

Z

= - 3t

7. The line W corresponds to S<:alar multiples of the vedor u = (2, -5, 4); thus a vector w = (x, y, z) is in WJ. if and only if u · w 2x- 5y + 4z = 0. Parametric equations for this plane are:

=

X

=

-

2t,

y

= s,

Z

=t

q

EXERCISE SET 7.3

205

't \

8.

I

Here W is the line of intersection of the planes x + y + z = 0 and x - y + z = 0. Parametric equatip_ns for this line are x = W corresponds to vectors of the follows that w (x,y, z) 1s m w1. if and only if x = z, i.e. if and only if w is of the Torm

- tli==-9.!! =

= (r, s , r) = r(l , ····---···0, 1) + s{O,l, 0) Parametric equations for this plane are given by x = r, y == s, z = r . w

-19 . 0

Thus the vectors w 1 = (1,0, -16) and w2 == (0, 1, - 19) form a basis for the row space o( A or, equivalently, for the s ubspace W s panned by the given vectors. We have W..l. = n.1w(A).l = nuii(.A) = null(U}. Thus, from the above, we nsists of all vectors of the i.e., the vector u = (16, 19, 1) forms a basis for w.l..

Let A bt!

•w•L• ix having lhe given vect ors as ils rows. This matrix can be row reduced to

!

u

Thus the vect orl; w 1 = (2. 0, - 1) and w 2 = (0, 1, 0) form a. basis for the row space of A o r, equivalently, for the subspace W spanned by the given vectors. We have w.t = row(A) .L = null(A) = null(U). Thus , from the above, we conclude that consists of all vectors of the form x = ( 0, t), i.e., the vector u = (!, 0, 1) forms a basis for W .1..

w.1.

11. Let A be t he matr ix having the given vectors as its rows. The reduced row echelon form of A is 1

· .-:"

. ,. '·

y •• •

R=

' \._--, ';- l

l

0 1 0 0

0 0 1

0

Thus the vectors w1 = (1, 0, 0, 0), w 2 = (0,1, 0, 0), and w 3 = (0, 0, 1,1) form a basis for the row space of A or; equivalently, for the space W spanned by the given vectors. We have w.t = row(A).l = null(A) = null(R). Thus, from the above, we conclude that W.l consists of all vect OrS Of the form X = (Q, 0, -t, t), i.e., the vector U = (0, 0, -1, 1) formS a basis for W.l. 12.

Let A be the matrix having the given vectors as its rows. The reduced row echelon form of A is

R= 01 01 00 [0 0 0 0

1 0

i 0 0

Thus the vectors w 1 = (1, 0, 0, 0), = (0, 1, 0, i), and w3 = (0, 0, 1, 0) form a. basis for the row space of A or, equivalently, for the space W spanned by the given vectors. We have Wl. = rvw(A).l = null(A) = null(R). Thus, from the above, we conclude that consists of all vectors of the form x = (0, 0, t) , i.e., tho vector u = (0, 0, 1) foriDB a basis for W .1. •

w.t

.\

. ···. I

"".

-..i...· ',

\

. .1

,

0 -16]

l 0.

'\

t'

Let A be the ma trix having the given vectors as its rows. The reduced row echelon form of A is

1 0

\

Chapter7

206

13. Let A be the matrix having the given vectors s.s its rows. The reduced row echelon form of A is

R=

;

=

!

Thus the vectors w 1 = (1,0, 0,2,0), w2 (0, 1,0,3,0), W3 = (0, 0,1,4, 0) and W4 = (0,0, 0,0,1) form a basis for the row space of A or, equivalently, for the space W spanned by the given vectoiS. We have W.J. = row(A)..L = nuli(A) null(R). Thus, from the above, we conclude that W.l consists of all vectors of the form x = ( -2t, -3t, -4t, t, 0), i.e., the vector u = (-2, -3, -4, 1, 0) forms a basis for W i.

=

14. Let A be the matrix having the given vectors as· its rows. The reduced row echelon form of A is

R=

1 0 0 0 0

0 1 0 0

0

0 0

1 0 0

0 -243 0 57 0 -7 1 2 0 0

Thus the vectors w1 = (1, 0, 0, 0, -243), w2 = (0, 1, 0, 0, 57), W 3 = (0, 0, 1, O, -7), W4 = (0, 0, 0, 1, 2) form a basis for the row space of A or, equivalently, for the space W spanned by the given vectors. We h ave Wi = row(A).l = null(A) = null(R). Thus, from the above, we conclude that W.J. consists of all vectors of the form x = (243t, -57t, 7t, - 2t, t}, i.e., the vector u = (243, - 57, 7, -2, 1) forms a. basis for W J..

=

15. In Exercise 9 we found that the vectoru = (16, 19, 1) forms a basis for W.l; thus W w.t..t consists of the set of all vectors which are orthogonal to u, i.e., vectors x = (x, y, z} satisfying 16x + 19y t z = 0.

=

16. In Exercise 10 we found that the vector u = (!, 0, 1) forms a basis for W J.; thus W W .J..J. consists of the set of all vectors which are orthogonal to u, i.e., vectors x = (x, y, z) satisfying x + 2z = 0. 17. In Exercise 11 we found that the vector u = (0, 0, -1, 1) forms a basis for W .l; thus W = W l..l consists of the set of all vectors which are orthogonal to u, i.e., vectors x = (x 1, x2, X3, X4) satisfying -X3 + X4 = 0. 18. In Exercise 12 we found that the vector u = 1) forms a basis for w.t; thus W = w.t.L. consists of the set of all vectors which are orthogonal to u , i.e., vectors x = (x1, x2, x3, x4) satisfying 3x-z- 2x4 = 0. 19.

Solution 1. Let A be the matrix having the given as its columns. Then a (column) vector b is a linear combination of v1, v2, v3 if and only if the linear system Ax= b is consistent. The augmented matrix of this system is

1 5 7l bl]

-1 -4 -6! 02 3 - 4 21 03

[ and a row echelon form for this matrix is

207

EXERCISE SET 7.3

From this we conclude that b = ( o1 if 16b1 + 19bz + b3 = 0.

Solution 2. The matrix

1

1

b3 ) lies in the space spanned by the given vectors if and only

-1 3]

VI] [ 5\ -4 -4 v2 = , _1_::!! __ 2_ [ b

bt

b2


b3

l 0-16]

0

1

- 19

b1

b2

ba

_o__ .s'___ , and

[

then further

reduced to

From this we. conclude that W = span{ v 1, v 2, v3) has dimension 2, and that b = (bt, b2, b3) is in lV if and only if 16b1 + + b3 = 0.

Solution 3. In Exercise 9 we found that the vector u = (16, 19, 1) forms a basis for w1.. Thus b= is in W ::-: W l.l. if and only if u · b = 0, i.e., if and only if Hib1 + 19b2 + b3 = 0. 20.

Solution l. Let .4 be the ma.trix having the given vect ors ns its columns. Then a (column) vector b is a .linear combination of v 1 , v 2 , v 3 if and only if the linear system Ax = b is consistent. The a ugmented matrix of this system is ·· ·

and a row echelon form for this matrix is

T hus a vector b = (b 1 , b2, b3) lies in the space spanned by the given vectors iJ and only if b1

Solution 2.


reduced to 1 0 [

_!

+ 2b3 =

0.

and then further

l b;

From this we conclude that W = spau{v 1, v2 , v3} has dimension 2, and that b if and only if b1 + 2b3 = 0.

n

= (b t , b2, b3}

Solution 3. In Exercise 10 we fOIJnd t hat the vector u = I ul 1) forms a basis for b = (bt , b2, b3 ) is in W = W.l..J.. if and only if u · b =0, i.e., if and only if bt + 2b3 = 0.

is in W

wl..

Thus

208

21.

Chap1er 7

Solution 1. Let A be the matrix having the given vectors as its columns. Then a (column) vector b is a linear combination <>f. v 1 , v 2 , v:l, v 4 if and only if the linear system Ax ::::: b is consistent. The augmented matrix of this system is 0

-2

4

0 1

0 2

1

2

2 -1 -1

b,]

b2

b3 b4

and a row reduced form for this matrix is

z

1

0

-1

b,

I

-2:0:

2' 0

0

Thus a vector b

4:

-2

0

]

ba

1

-bt

+ b2 + b4

= (b1 , bz , b3) lies in the space spanned by the given vectors if and only if -b3 + b4 = 0.

8olution 2. The matrix v3

! ----------4

b

further reduced to

bl

can be row reduced to [

02

b4

b3

l

0

0 0

()

0 0

0

0

:

and then

0 0 0 -·--------0 0! b2 ba b4

2 -1 - 1

0 0 1 0

0 0 1

0

From t his we conclude t hat W = span{vlo v z, v 3 , v 4 } has dimension 3, and t.hat b = (b1.bz,b3) is in W if and only if - b3 + b." = 0.

Solutwn 3. Tn E..xcrcise 1J we found t.hat vector u = {0, 0, -1, l ) forms a basis for W .L. Thus b = (b 1 , b2 , is in IV =- WL1 if and o nly if u · b = 0, i.e., if and only if - b3 + b4 = 0. 22.

Solution 1. Let A be the matrix having the gi ven vectors as its columns. Then a (column) vector b is a linear combination of these vectors if a.nd only if the linear system Ax = b is consistent. The augmented mat.rix of this system is ]

[

2

-2 3

3 5 6 4 -3 -1 9 6

I I

I I

I I

b,] b2 03

04

and a row echelon form for this matrix is

Thus b

= (b;.

i !! 5bl -b32b:l

23

l

1

9

0

0

t:-!bl+ i b2

0

0

0 :I !3 b4 - !2 b2

b3 , b4 ) lies in the space spanned by t he given vectors i£ and only )f

2b4 = 0.

209

EXERCISE SET 7.3

vl] rl}2 4 -s 6] [ Solution 2. The matrix ·= can be row reduced to 2 - 2

Y2

b

In b

bt

J

l

0

0

0

0

1

0

a2

o o

1

o , and then further

0 0 0 0 ---------

b1

reduced to

1

0

0

0

0

1

0

2

0 0

0 0

1 0

0

3

0

-----------------0 0 0 +b4 From this we conclude that W = span{ v 1 , vz, v3, v 4 } has dimension 3, a nd that b = (bl> bz, b3, b4 ) is in W if and only if 3b2 - 2b4 0.

=

Solution 3. In Exercise 12 we foun d that t he vector u = (0, - 0, 1) forms a basis for W .L. Thus b = {bJ , b2. b3 , is in W ;; w .t.t if and only if u · b = + b4 0, i.e., if and only if 3b2 -- 2b4 0.

=

23.

The aug mented matrix lv 1

v2

V:;

=

v 4 I ht l b z l b "J is

-··2

- ]

0 1 1 3

1

l

0 -1 2

2 1 - 1

0

I I

2:

-2

I

-2

2:' -1 5: 5

I I I

'

1

0

2 4: - 2 2: - 3 - 1

I

- 1 1

I I

I

1

0

and the reduced row echelon form of this matrix is 0

0

l

0

0

1 0 0

0 0

matrix jv 1

Vz

'13

0 1

0 0

l: -2 0 0

0 0

1 I' 1:

0

l ''t

1

o:

0

b :t lie in span{v 1 , v 2, v 3 , v 4 }, but b s does not.

From this we conclude Uwt the vectors h 1

2 4. The

0 0

v 4 I b 1 I b2 I b s l is 1

- L

1

0

3 2 1 0 -l

2

0 1 0

3 •t

-2 : -1: o:2 I

1 I'

0 ' 1

'

1 :

7 : -1 : 6 I I 2 I' 2 : 4 1: 2:1

and the reduced row echelon form of this matrix is 0

1

0 0

0

0

1

1

0 0

0

0 0

0 0

0: _! : 0

I

1)

1

I I

§

I-} I

0 0

1

0

0 0

0

o: 1

t

t t

5 6 t - t 5 ' 1t

o: o:'

1

From this we conclude t ha t t he vectors b 1 and b 2 lie in span{v1 , vz , v 3 , v 4 } , but b 3 does not.

Chapter7

210

25.

The reduced row echelon form of the matrix A is

0 0 0 0 1 0 0 0 0 0

0 1

=

Thus the vectors r 1 = (1,3, 0,4, 0,0), r2 = (0,0,1,2,0,0), r 3 = (O, O,O,O, l,O), r4 (0, 0,0,0,0,1), form a basis for the row space of A. We also conclude from a.n inspection of R that the null space of A (solutions of Ax= 0) consists of vectors of the form

= s( - 3, 1, 0, 0,0,0) + t{-4, 0, -2, 1, 0, 0)

x

Thus the vectors n 1 = ( - 3, 1, 0, 0, 0, 0) and n2 = (-4, 0, -2, 1, 0, 0) form a basis for the null space of A. It is easy to check that r; · Dj = 0 for a.ll i, j. Thus row( A) and null(A) are orthogonaf.subspat:es of R 6 • 26 . The

row (>Chf'lon form o f the m11.trix .4 is

R=

1

1

0

0

4

0

1

0

2

0

0

1

t ;j

0

0

0

0

n.

-!]

1

5 3

5 0

4.

Thus the vectors f ) = (1, 0, 0, r 2 = (0, 1, 0, a.nd TJ = (0, 0, 1, } , form a basis for the row of A \Ve also conclude from nn inspection of R that the null space of A (solutions of Ax -::: O) consists of vectors of the form

Thus the vectors n 1 = ( · It is easy to che<.:k t.hat r, · n 1

1, 0), n 2 =: ( ·0, 1) form a basis for the null s pace of A. j . T hus row(A) and onll(A) are orthogonal subspaces of

= 0 for all i,

R" .

27. The vectors r1 = (1,3 ,0,4,0, 0), r2 = (0,0, 1, 2,0 ,0), basis for t he row space of A (see Exercise 25).

!. - ),

V. a..nd

28.

The vectors r, = ( l , 0, 0, space of A (see Exercise 26).

29.

The reduced row echelon forms of A and B are

r2 = (0, l, 0,

r3 =

(0,0,0,0, 1, 0),

r3

r4

= (0, 0, 1, ! ,

and

= (0,0,0,0,0,1)

f<'l rm a.

for m a. basis for t he row

respectively. Thus the

0 0 0 1

vectors r l :::: (1,0,3, 0), r2 = (0, 1, 2,0) , r 3 = (0,0,0, follows t.hnt row(A) = row(B).

30. The reduced row

forms of A and B are

n form

00 13

a basis for both of the row spaces. It

and

OQI J


0 0 0 0

vectors r 1 :::;: (1,0,0,5), r 2 = follows that ruw(A) = row(B).

= (0,0,1 ,3) form a basis for

of the row spaces. It

211


31.

Let B be the matrix having the given vectors as its rows. The reduced row echelon form of B is

0 -1 -4

R=

a nd from this it follows that a general solution of Bx= 0 is given by x = s (1 , 4, 1,0) 1), where -oo < s, t < oo. T hus the vectors w 1 = (1, 4, 1, 0) and w2 = (-2, 0, 0, 1) form a ha.sis for the null space of B , and so the matrix

A = [ 1 4 1 0] -2 0 0 1 has the property that null (A) = row{A)J.

32. (a) If A=

...

= null( B).!. = row(B).l.J. =row(B).

and x = [:], then Ax = 0 if and only if x = y

=

0. Thus the null sp•ce of A

corresponds t o point.s on t he z-axis. On the othe:- hanrl, t he cokm n sp:.:.cc cf A t.'Onsists of all vectors tha t are of the form

which corresponds t.o points on lht} xy-plane. ( b ) The matrix B

=

has the specified n ull space n.nd column space. 0 0 I


(a) False. T his statement is t.rue if and only if the nonzero rows o f A are linear.ly inde pendent, i.e. , if Lhcre a.re no additional zero rows in i.tll echelo n form. " ( b) T rue. 1f E is a11 elementary matrix , t hen E is invertible and so EA x = 0 if a nd only if Ax= 0 . (c) True. If A ha.s ru.nk n , then there can be no zero rows in an echelon form for A; thus the red11ced row echelon form is t he ident.it.y matrix. (d) False. For example, if m = n and A is invert ible, then row(A) -= Rn and null(A) = {0}. (e) True. T his follows from the fact (T heorem 7.3.3) that SJ. is a subspace of Rn .

D2.

(a) True. 1f A is invertible, then the rows of A a nd the columns of A each form a basis for W'; thus row(A ) = col(A) = R" . ( b ) Fah>e. ln fact, t he opposite is true: lf W is a subspace of V, then V J. is a subspace of W .l. since every vector that is orthogona l to V will also be orthogonal to W . (c) False. The specified condition implies that row(A) row( B ) fro m which it follows that mJll(A) = row(A)J.

2 row(B)L =

null(B)

but in general the inclus ion can be proper.

(d) False. F.or example A= spaces.

and B =

have the same row space but

column

212

Chapter 7

(e)

True. This is in fact true for any invertible matrix E. The rows of EA are linear combinations of t.he rows of A; thus it is always true that row(EA) row(A) . If E i:; invertible, then A = E- 1 (EA) and so we also have row{A) row(EA).

D3. If null{ A) is the line 3x- 5y

= 0, then row(A) = null(A)l.

is the line 5x + 3y = 0 . Thus each row

of A must be a scalar multiple of the vector {3,-5), i.e .• A i'i of the form A= [:ls

-!Js].

3t - 5t

D4. The null space of A corresponds to the kernel of TA, and the column space of A corresponds to the range of T A . D5. lf W D6.

= a..t. , then w..t. = a.L.L =span{a}

is the !-dimensional subspace spanned by the vector a.

(a) If nuU( A) is a line through the origin, then row(A) = null( A )..I. is the plane through the origin that is perpendicular to that line. (b) If col(A) is a line through the origin, then null( AT)= col(A)..l is the plane through the origin thnt is perpendicular to that line.

D7. The first. two matrices arc innrtiblc; thus in each cnse the null spuce is {0}. The mill matri..x A

is the line 3x + y = 0 , and the null space of A =

=

is all of R

2

of the

.

DB. (a) Since S has equation y = 3x, Sl. has equation y = -and Sl.J. = S has equation y = 3x. (b) If S == { ( 1, 2)}, then spa.n(S) has equation y = 2x; thus Sl. has equation y = - x ann Sl. l = span(S) has equation y = 2x . 09.

·No, this is not

possible. The row space of an invertible n x n matrix is all of Rn since its rows form a ba.sis for R ... On the other hancl , the row space of a singular matrix is a proper subspace of Rn since its rows are linearly dependent and do not s pan all of Rn.


lt is clear, from the definition of row space, that (a) implies (c). Conversely, suppose that (c) holds. Then, since <::ach row of A is a linear combination of the rows of B, it fo llows that any li ne..a.r c.ombination of the rows of A can be expressed as a linear combination of the rows of B . This shows that row(A) row{B), and a si1uilar argument shows that row(B) row(A). Thus (a) holds.

P2. The row vectors of an invertible matrix are linearly independent and so, since there are exactly n of them, they form a basis for R" . P3. If Pis invertible, then (PA)x = P(Ax) = 0 if and only if A x= 0 ; thus the matrices PA and A have the same null space, and so nullity{PA) = nullity( A). From Theorem 7.3.8, it follows that PA and A. have also have the same row space. Thus rank(PA) = dim(row(PA)) = dirn(row(A)) =rank( A). P4.

From T heorem 7.3.4 we have SJ. = span(S)l., and (span(S)J.)..l subspace. Thus (S l. )l. = (span{S)l. )l. = span(S) 1

P5. WehaveAA- 1

rJ(A)·cl(A- ) rl(A)·c:J(Ar'l(A)·ct(A- 1 ) r 2{A)·c2(A-

= [

...

..

1

) 1)

.

r 11 (A) · C! (A- 1 ) r n{A) • C2{A -

1

)

•• • ••

..

· •·

= span{S)

since spa.n(S) is a

1

rt(A)·cn(A- ) ] [..1 0 ··· 0] r 2(A)·cn(A- 1 ) 0 1 ·· · 0 = . . . .

.

r n(A) · Cn(A - 1)

..

..

0 0 ·· · l

= [Oijl·

In par-

ticular, if it= {1, 2, . . . ,k} and j E {k + 1, k + 2,. _., n}, i < j and so ": ( A)· c;(A - 1) This shows that the first k rows of A and the last n - k columns of A- 1 are orthogonal.

= 0.

EXERCISE SET 7.4

213

EXERCISE SET 7.4 1.

The reduced row echelon form for A is

0 -16] 1

-19

0

0

Thus rank(A) = 2, and a general solution of Ax= 0 is given by x = ( 16t, 19t, t) = t(16, 19, 1). It foll ows that nullity(A) = 1. Thus rank( A) + nullity(A) = 2 + 1 = 3 the number of columns of A. 2.

T he reduced row echelon form for A .is

0

0

Thus rank(A) = 1, a nd a general solution of Ax= 0 is given by x = (1t, s, t) = s(O, l ,O) + It. follows t hat. nullity(A) = 2. Thus rank(A) + nullity(A) = 1 + 2 = 3 t he number of columns of A .

3. The reduced row echelon form for A is 1

ll

-tl

0

0

0

0

:!7

T hus rank(A) = 2, and a general solution of Ax= 0 iR given by x <= s(- 1,-- 1, 1, 0) 1). It follows t hat nullit.y(A) = 2. Thus rank (A) + nullity(A) = 2 + 2 = 4 the number of columns of A. 4.

form for A is

The reduced r ow

] l 0

0 1

1 2 ]} 1-..J 2

..

·-...._______ - .- ·

Thus rallk(A) = 2, and a general solution of Ax= 0 is given by

x

= r( -1 , -

It fo llows that n111lity(A)

5. The reduced row

= 3.

1, 1, 0, 0) + s(- 2, - 1, 0, l, 0)

Thus rank(A)

+ t( -1 , -· 2, 0 , 0, 1)

+ nullity(A) = 2 + 3 =

5.

form for A is 1

0

0

4

0

2

3

0

0

-- 6

- n5

0

0

1

0

0 0

0 0

0

0

0

0

I

0 0

Thus rank(A) = 3, and a general solution of Ax= 0 is given by

fi,o, 1) nul;ti.y(A) = 2. Thus rank(A} + nullHy(A) ""'3 + 2 = 5. x ""' s(-2,0,0,1,0} +

It follows that

214

Chapter 7

6. The

row echelon form for A is 1

0

3

1

3

0

0 0

0

1

2

2

3

3

7

4

-3 -3 0 0 0 0

4

3 l

3 0 0

Thus rank(A) = 2, and a gene1al solution of Ax = 0 is given by X =

r(

-s,

+

0,0, 0,0} + s( -j,

-A. 0,0,0,1,0) +

0, 1, 0, 0, 0) + t( -j, 0, 0, 0, 0,1)

ll follows that nullity( A)= 5. Thus rank{A) + nullity(A)

7.

t,0, 0, 1, 0,0)

=2+ 5 =

7.

(a) lf A is a 5 x 8 matrix having rank 3, then its nulHty must be 8- 3 = 5. Thus there a.re 3 pivot variables and 5 free parameters in a general solution of Ax = 0 .

(b) Tf A is a 7 x 4 matrix having nullity 2, then its rank must be 4 - 2 = 2. Thus there are 2 pivot variables and 2 free parameters in a general solution of Ax = 0. (c) lf A is a 6 x G matrix whose row echelon forms have 2 nonz!'rO rows, then A hn.s rank 2 and nullity 6- 2 = 4 Thus there are 2 pivot variables and ..1 free parameters in a general solution of Ax - 0

8.

If A is a 7 x 9 matrix having rank 5, then its nullity must be 9 - 5 = 4. Thus there are 5 pivot variables and t free parameters in a general solution of Ax = 0 . (b) If A is an 8 x 6 matrix having nullity 3, then its rank must be 6-3 = 3 Thu:; there are 3 pivot variables and 3 free parameters in a general.solution of Ax = 0 (c) If A is a i x 7 matriX ''hose row echelon forms ha\f' 3 nonzero rows then ..1 has rank 3 and nulhty 7 - 3 = ·1 Thus there are 3 pivot variables and 4 free paramett-rs in a genera] solution of Ax - 0

9.

(a)

(a)

(b) (c) 10.

[fA a 5 x 3 rnatnx. then the largest possible value for rank( ...l) is 3 and the smallest possible value for nulllty(A) is 0 If A is a 3 x 5 :natrix, then the largest possible value for rank(A) is 3 and the smallest possible valur for null1ty( '\) is 2 If A is a 4 x 4 mn.trix, t hen the largest possible value ior ranK( A) IS 4 and the smallest possible value for nulltty(A.) is 0.

If 4 1s a 6 x I then lhe largest possible value for rank(.4) is 4 and the smallest possible value fm nulhty(A) is 0. (b) If A is a. 2 x 6 matrix, then the largest possible value for rank(A) is 2 and the smallest possible value for nullity(A) is 4 (c) If A 1s a 5 x 5 matrix, then the largest posstble value for rank(A) ts 5 and the smaiJest possible value for nulltty(A) is 0

(a)

11. Let A be the 2 x 4 matnx havmg v 1 and v 2 as its rows Then the reduced row echelon fonn of A is

0

"

3

and so a g<>n('ral solulton of Ax= 0 is given by x = Vt= (·! , l ,O,O), form a basts for R 4

v2=(0,3,.t,5),

0)

+

tn,-

0, 1). Thus the vectors

215

EXERCISE SET 7.4

12. Let A be the 3 x 5 matrix having v 1 , v 2 , and v:l as its row:>. T hen the reduced row echelon form of A is 0 0 -- 819 2:l

0 0

and so a general solution of Ax= 0 isx Vt

W3

v2

15

- 8

43 - 16

8

= 1i ,-

= (1, 0, -2, 3, -5),

¥]

15

l, 0)

43

+ t (- 21, 1i, -

= ( - 2, 3, 3, }, -1) ,

=

1,0),

W4

V3 15 8

=

= (4

1

0, 1). Thus the vectors

1, -3,01 5)

,-¥,0, 1)

form a basis for R5 • 13.

(a) This matrix is of re.nk 1 with l1 =

(h) T his matrix is of rank 2. (c) T his matdx i< oh ank

1 4.

(a)

1

with A =

u vT.

H -:-: _:;] +:1[1

This matrix is of ra.nk 2.

_ ,: : ]

of '""k 1 with A=

(b) This

-7] =

=

= [

6

-3[

-3

5

1 3

3

= uv'''.

(c) This matrix is of rank 1 with

;\ = [

-6 1010

lJ =

3

- .)

8]8 [ 22] - i·Z - ·l

3 -1

15.

12 - 6 12 -6

-6

-6

3 - 1

I1

6

- 3

4) =

UVT

69 32 32] .ts a sv mmetnc. rnat nx. . 3 l

1

•

3 1 1

0

16.

0

16 -20 - 20 25 28 -35

is a symmetric matrix.

l [1

1 7. The matrix A can be row reduced to upper triangular form as follows:

t 1 -t--+ 0 2 1- t 0

1

t- 1 0

t t 1-

]

(2+ t)(l - t)

If t = 1, then the la.tter has only one nonzero row and so = 1. If t = -2, then there are two nonzero rows and ra.nk(A) = 2. If t :f: lor - 2, then there are three nonzero rows and rank(A) = 3.

216

18.

Chapter 7

_, l r -tl r

The matrix A can be row red11ced as follows:

A=

[

-1

3

-1]

6 -2 -+ 3 t

3 6

[i

3

-2 -+ 0 0 3 -1

-3

-2+3t -1 + t 2

3- 3t

-+

3

-t ] -2+ 3t -(t- l){2t - 3)

0 -3 0 0

If t = 1 or t = then t he third row of the latter matrix is a zero row and rank( A) values of t there are no zero rows, and rank(A) = 3. 19.

If lhe matrix

Thus (x.y,z) 20.

[:r1

= 2.

For all other

z] has rank 1 then the first row must be a scalar multiple of the second row.

Y

l/

X

= t(l,x,y), and sox= t, y =

Let A be the 3 x 3 matrix having

VI.

tx

= t2 , z =

ty

= t3 .

v 2 , v 3 as ils rows. The reduced row echelon form.of A is

...

thus the vt•ctors Ax

=0

-=

w1

(1, 0, 5) and

w2

= (0, 1, -4) form a basis for W

=row( A)

= row(R ).

tf and only if Rx = 0 , and a general solution of the latter is given by x

Lhus the vf'r.tor x 1

-

(-5, 4,1) forms a basis for W.l.

'>+I== 3

= row(A).l..

We have

= [-Stl 4: = t [-5] ;

Finally, dim(W)+dim(W).l. = .

21. The suh.spare W, consisting of all vector!'J of the form x = t(2, -1, - 3} , has dimension 1. The subspace n·.l is the hyperplane cons1$tmg of all vectors x = (x,y,z) whlch are orthogonal to (2,-1,3), i.e., which satisfy the equation 2x- y- 3z = 0 A gt•neral !iolution of the latter ts given by

x

= (s. 2s- 3t, t) = s(1. 2, O) + t(O, -3, 1)

wtwre - oo < t < oo. fhus the vectors (1, 2, 0) and (0, -3, 1) form dim(W)

l\

basis for w.t' and we have

+ dim(lV.l.) = I + 2 = 3

=

=

22 .

If u and v nre nonzc>ro column vectors and A = uvr, then Ax = (uvT)x u (vrx) (v. x ) u . Thus Ax= 0 if Bnd only if v · x = 0. i.e.• if and only if xis orthogonal to v . This shows that ker (T) = v.l. . Similar!_\, I he range ofT conststs of all vectors of the form Ax= (v · x }u , and so ran(T) =span{ u }.

23.

If B 1<:: obtainerl from A by changing only one entry, then A - B has only one nonzero entry (hence only one nonzero row), and so rank(A- B) = 1. (b) If B is obtained from A by changing only one column (or one row), then A - B has only one nonzero column (or row) , and so rank{A - B) = 1. (c) 11 B 1s obtamed from A in the specified manner. Then B - A has rank 1. Thus B - A is of the form uvT and 8 =A+ uvT.

24 .

(a)

(a)

If A= uvr, lhcn A 2 = ( uvT)( uvT) == u (vTu)vT = (v · u ) uvT

= (u. v )A.

( b ) If Ax= >.x for some nonzero vector x, then A x = >. x. On the other hand, since A 2 = (u · v )A, we have A2 x = (u · v )Ax = (u · v)>.x . Thus >.2 = (u · v )>., a.nd if>. :f nit. follows that>.= u · v . 2

2

217


(c)

If A is any square matrix, then I - A fails to be jnvertible if and only if there is a nonzero .vector x such tha.t (I - A)x = 0; this is equivalent to saying t hat>. = 1 is an eigenvalue of A. Thus if A ::::: uvT is a nmk 1 r£tatrix for which I - A is invertible, we must have u · v =/= 1 and .4 2 :fA.

DISCUSSION AND DISCOVERY Dl. (a) True. For example, if A ism x n where m > n (more rows than columns) then the rows of A form a set of m vectors in Rn and must therefore be linearly dependent. On t.he other hand, if m < n then the columns of A must be linearly dependent. (b) False. If the additional row is a linear combination of the existing rows, then the rank will not be increased . (c) False. For example, if m = 1 then rank( A)= 1 and nullity( A) = n - 1. (d) True. s·uch a matrix must have rank less than n; thus nullity(A):::::: n - rank(A) == L (e) False. U Ax= b is inconsistent for some b then A is not invertible and so Ax = 0 hM nontrivial solutions; thus nulHty(A) ;::: 1. (f) 11-uP.. We must have rank(A) + nullity(A) = 3; thus it is not possible to have rank( A) and nullity(A) both C'qual t.o 1. D2. If A is m x n, then AT is n x m and so, by Theorem 7.4.1, we have rank(AT) + nullity( AT )= m . D3. If A is a 3 x 5 matri."<, then the number of leading l's in the reduced row echelon form is at most 3, und (assuming A =/= 0) the number of parameters in a general solution of Ax= 0 is a.t most 4. D4. If A is a 5 x 3 matrix, thP-n ·rank(A) :::; 3 and so the n umber of leading l's in the red uced rvw echelon form of A is at most 3. Assuming A =I 0, we have rank(A) :;::: 1 and so the number of free parameters in a general solution of Ax = 0 is at most 2. D5. If A is a 3 x 5 matrix, then the possible values for rank(A) are 0 (if A= 0) , 1, 2, or 3, and the corresponding values for If A is a 5 X 3 matrix, then the possible values for rank(/\) are 0, 1, 2, or 3, and the corresponding values for nullity(A) are 3, 2, 1, or 0. If A is a 5 x 5 matrix, then the possible values for rank(A) are 0, 1, 2, 3, 4, or 5, and t.he corresponding values for nullity(A) are 5, 4, 3, 2, 1, or 0. D6.

Assuming u and v are nonzero vect.ors, the rank of A= u vT is 1; thus t.he nullity is n - 1.

D7. Let A be the s tandard matrix ofT. If ker(T) is a. line through the origin, then nullity(A) = 1 and so rank(A) ::: n - 1. Tt follows that ran(T) = col (A) has dimension n- 1 and thus js a hyperplane in R" . D8. If r == 2 a.nd s = 1, then rank(A) = 2. Otherwise, either r - 2 or s- l (or both) is 0 , and rank(A) = 3. Note thnt, since the first and fourt h rows are linearly independent, rank(A) can never be 1 (or 0).

D9. If,\¥ 0, then the reduced row echelon from of A is

0 L.

0

0

1

0

0

1

0

0

-{]

Chapter 7

218

and so rank(A)

= 3.

On the other hand, 1f >. = 0, then the reduced row echelon form is 0 1

0 0

5

2 0

0

il

and so rank(A) = 2. Thus>. = 0 is the value for which the matrix A has lowest rank.

DlO.

and B

Let .4 =

B2

=

=

Then rank(A)

= rank(B) =

1, whereas A 2

=

bas rank 1 and

has rank 0.

= 0 then rank( A) + rank(B) - n = rank(AB) = 0, and so rank(A) + rank(B} 5 n. Since rank( B) - n- nulHty(B), it follows that rank(A) + n - nullity( B) 5 n ; thus rank{A) B). S1milarly, rank(B) 5 nullity(A).

Dll. If AB

WORKING WITH PROOFS 011

012

021

an

Pl. First. we note that. the matrix A= (

013 ] 023

fails to be of rank 2 if and only if one of the rows

is a scalar multiple of the other (which includes the case where one 01 both is a row of zeros). Thus it is sufficient to prove that the latter is equivalent to the condition that

Suppose that one of the rows of A is a scalar multiple of the other. Then the same 1s true of each of the 2 >< 2 matrices that appear in (#) and so each of these det erminants IS equal to zero. Suppose, conversely, that the cond1t10n ( #) holds. Tlten we have

\\'ithout loss of gPnerality ,,.e may assume that the first row of A is not a row of zeros, and further t Interchange columns if neces!lary) that. a11 =1: 0 We then have a22 = (l!.U )at 2 and = ( l!.U )a.l3 an an Thus and so the second row is a scalar multiple o f the first ••II row

P2. lf A is of rank 1, then A = xyT for some (nonzero) column vect ors x andy. If, in addition, A is symmetric, Lhen we also have A = AT= (xyT)T = pr . From this it follows that

x(yr x)

= (xyr )x = (yxT )x = y(xr x ) = yllxll 2

Since x andy are nonzero we have xry that A= yxr

= y(gy)T ==-

=

yTx = y · x :/; 0 and x

=

respondmg formula 1s A = - uur where u

=

If xTy > 0, it follows

= uuT IfxTy

< 0, then thecor-

+ c2(A)r 2(8) + · · · +

and

=

(B) r 2(B}

r l

P3. AB

= lc.CA )

c1(A)

= Ct(A )r t(B ) [

each of the

c1 ( A)r1 (B) is a rank 1 matrix.

EXERCISE SET 7.5

219

P4 . Since the set V U W contains n vectors, it. suffices to show that V U W is a linearly independent set. Supp?se then tha.t c1 , c2 , . •. , and d 1 , d 2 , ••• , dn- k are scalars with the property that

+ CzVz + · · · + CJ.:Vk + d1W1 + dzWz + · ·· + d-n-kWn-k = 0 c1v1 + czvz + · · · + c,.vk = -(d1w1 + dzW?. + · · · + dn - JcWn-k) belongs both

CtVJ

Then the vector n = to V = row(A) and to W = null(A) = It follows that n · n = llnll 2 = 0 and son= 0. Thus we simultaneous ly have c1 v1 + czVz + · · · + q v,. = 0 and d1 w1 + d2w2 + · · · + dn -k Wn-k = 0. Since the vectors v1, Vz, . .. , Yk are linearly indepenjent it follows that c1 = cz = · · · = ck = 0, and since w1, Wz, ... , Wn-k are linearly independent it follows that d1 = dz = · · · = dn-k = 0. This shows that V U W is a linearly independent set. P5.

From the inequality rank(A) + rank(B) - n

n- rank(A)

A) it follows that

rank{AB}

n- rank(AB}

2n - rank{A) - ra.n.k(B)

which, using Theorem 7.4.1 , is the same as

nullity(AB) S nullity(A) +nullity( B)

nullity( A) Similarly, nullity(B) S nullity(AB)

P6. Suppose .A= 1 + vr ; t ' 1u Bx

.=:

T) (A + uv

# 0, and -

. =l+uv rA - 1

-

nullity(A)

+ nullity(B).

let X= A- 1

A- l uv1'A-t)

A

=

I

-

A-'.

+ uv

T A-l

-·

Then

uvTA-1

uvTA- 1 + u (vTA- 1 u )vTA - 1 = l + uvT A- 1 ). TA-t

-1

and so B is invertible with B -· 1 = X= A- 1 - A uvA Note. If vTA- 1 u = - 1, then nA- lu =(A+ uvT)ABA- 1 is singular and so B is singular.

=A - l -

1u

=u+

+ uvT A-luvr AA

-

A-1

t

l +vTA- 1 u uvTA - 1 A

;;;;;[

TA - 1

uv

l+vT.'I. - 1u ' u(vT A- 1 u) =

u-

u

= 0; thus

....··

,

EXERCISE SET 7.5 1.

The <edueed cow .ehelon fonn of A is 0 - 1

2

0

l

1

AT is 0

0 0 0

0

J 0 0 0

0 0

0

0

[j

-5] and the reduced row echelon form of

0 -7 l -9

4. 5 -6

0 0

0 0

0 0

0 0

T hus dim(row(A)) = 2 and dim(col(A) = dim(row(AT))

diro(null(A)) =5-2 = 3, and dim(null(AT)) parameters in a general solution of Ax = 0.

2. The reduced row echelon form of A is l 0 0 1 0 I 0 l

of AT is 0 o 1 0 . Thus dim(row(A)) 0 0 0 0

= 4- '2 = 0

0

1

0 _ .!§. 2

0

0

1

-2

0

0

0

0

7

--1]

3 a nd di.m(col(A)

dim(null(A)) = 5-3 == 2, and dim(null(AT)) == 4-3 parameters in a general solution of Ax = 0.

It follows that

2. Sjnce dim(null(A)) = 3, t here are 3 free

0

0 0 0 0

= 2.

= 1.

and the reduced row echelon form

= dim{row(AT)) =

3. It follows t hat

Since dim(nuli(A)) = 2, there are 2 free

220

Chapter 7

!]

3. The reduce
3 and rnnk(AT) = 3.

4. The reduced •ow echelon form of A is

0

and the reduced row echelon form of AT is

0

:-,! -:]

2

1 - 1

0

and the reduced row echelon form of AT is

0

] . Thusrank(A)

=

= 2 and rank(A;) = 2

'o

=

5.

dim(row(A)) dim(col(A)) rank(A) = 3, dim(rlllJI(A)) = 3 - 3 = 0, dim(nuli(Ar)) = 3-3 = 0. (b) dim(row(A)) = dim(col(A)) = rank(A) = 2, dim(null(A)) = 3 - 2 = 1, dim(null(AT)) = 3- 2 = 1. (c) dim (row(A))- dim(coi(A)) = rank{A) = 1, dim(null(A)) = 3 - I - 2, dim(null(AT)) = 3 - 1 = 2. (d) dim(row(A)) = dim(coi(A)) = rank(A) = 2, dim(nuii{A)) = 9 - 2 = 7, dim(nuii(AT)) = 5-2 = 3. (e)
6.

(a) dim(row(A)) = d im(col(A)) = rank(A) = 3. dim(nuii(A)) = ot - 3 = 1,
(a)

7. (a) Since rank(A)

= rnnk!A I bJ=

solution is n - r (b) Since rank(A ) t(c) Since rank{ A) = solution is n - r {d) Since rank(A) soluLion JS n - r

=

3 the system is consis tent. The number of parameters in a general

= 3- 3 = 0; i.e., the system has a unique solution .

rank[A I bJ the system is inconsistent. rank[A I bJ = 1 the system is consistent. The number of parameters in a general = 3 - 1 = 2. rank[ A I bj = 2 the system is consistent. The number of parameters in a general = 9- 2 = 7.

t- rank[A I b) the system is inconsistent. (b) Since rank( A) = rank[ A I bl = 2 the system is consistent. The number of parameters in a. general solution is n - r = 4 - 2 = 2. (c) Since rank( A) = rank[A I bJ = 3 the system is consistent. The number of parameters in a general solution is n - r = 3 - 3 = 0, i.e. , tLc system has a unique solution. (d) Since rank( A) = rank[A I bJ = 4 the system is consistent. The number of parh.weters in a. general solution 1s n - r = 7 - 4 = 3.

8. (a) Since rank{A)

221

EXERCISE SET 7.5

9.

(a) This matrix bas full column rank because the two column vectors are not scalar multiples of each other; it does not full row rank because any three vectors in R 2 ar..; linearly (b) This matrix does not have full row rank because the 2nd row is a scalar mul'.iple of the 1st row; it does not have full column rank since any three vectors in R2 are linearly dependent. (c) This matrix has full row rank because the two row vectors arP. not scalar mult iples of each other, it. does not have full column rank beeause any three vectors iu R 2 are linearly (d) This (square) matrix is invertible, it has full row rank and full column rank.

10.

(a) Thls matrix has full column rank because the two column vectors are not scalar multiples of each other; it does not have full row rank because any three vectors in R 2 are linearly dependent. (b) This matrix has full row rank because the two row vectors are not scalar multiples of each other; it does not have full column rank because any three vectors in R2 are linearly dependent. (c) Thls matrix does not have full row rank because the 2nd row is a scalar multiple of the lst row; it does not have full column rank because any three vectors in R 2 are linearly dependent. (d) This (square) matrix is invertible; it has full row rank and full column rank.

11.

(a) det(ATA) = det

l-

6 1

2

.

="-"

149

=f 0

and det(AAT)

= dct [102

··

24 - 1]2 = 0; t hus ATA is invE>rti

ll - 2

17

ible and AAT is not invertible. This corresponds to the fact (see Exercise 9a) that A has fuJl column rank but not full row rank. (b) det(AT A) =- det [

= o; thus neither AT A nor AAT

= o and d et(AAT) = det.

: -!0 -40

20

invertible. This corresponds t.o the fact. that A doeR not have full column rank nor full row rank.

IS

(c)

det(AT A)

det [:: ; __;_;]:

'l-'"1 det(AAT)

:]

66,; 0; thus AT A Is not lnmtlble

but A A1 -i&inveitll)fe. This corresponds to the fact that A hu.s does not have full column rank but. does have full row rank . (d)

1 ] = 4 ]{)

"" d el(!'>

4 2 ] 2 17

64 -:f= 0 and det(AA T) "" det (

hoth invertible. This 12.

(a)

det(AT A) ...,. det [

to the fact that A has full column rank and full row rank. = 45

: 1

= 61 =f 0; thus AT A and AAT are

=f 0 and det(AAT)

= det [

= 0; thus AT A is invertible ·- 1 - 2

I

and AAT is not invertible. Th1s corresponds to the fact (see Exercise lOa) that A has full column rank but not full row rank (b) ciet(AT.1)..,., rlct

4 10 -2] = 0 and det(AAT) = del

[-2

10

34

13

g

= 1269 :j: 0; thus AT A is not

13 37

invertible but AAT is invertible. This corresponds to t he fact that A does not have full column rank but does have full row rank . (c)

= 0 and det(AAT) = det

det(AT A)::; det 60 -15

= 0;

thus neither AT A nor

45

AAT is invertible. This corresponds to the fact. that A has does not have full column. rank nor

full row rank.

222

Chapter 1 61 41

(d) det(AT A)= dct [

41

soj1 = 1369 =/= 0 and det(AAT ) =

37 ] 31

det

1369 =/= 0; thus AT A and AAT

are both invertible This corresponds to the fact that A has both full column r
!4b

I 0:

1 3. The augmented matnx of Ax= b can be row reduced to o [

1

b3 1 -

0 01 bt

l

b3 •

Thus lhe system Ax = b

+ b'J

is either inconsistent {if b1 + :f:. 0}, or bas exactly one solution {if b1 + = 0). Note Lhat the latter includes the case b1 = b2 = b3 = 0; thus the system Ax = 0 h as only the t rivial solut ion.

l

4: bt -3: i>l- 2bt . Thus the system [ 0 0 1 b1 - 2i>l + b3 1

o

1 4. The augmented matrix of Ax = b can be reduced to

Ax = b

is t!ither inconsistent (if bt - 2b2 + b3 =I= 0), or has exactly o ne solution (if b1 - 2b2 + 63 = 0). The latter includes t he case b1 = b2 = b3 = 0; thus the system Ax = 0 has only the trivial sdl'i.ition.

15.

If A

= [- )

then A r A 2

[

I

6 12 12 24

It is clear from inspection t hnt the rows of A and .4.T A

].

are multiples of the single vector u = (1,2). Thus row(A) = row(Ar A) is the !-dimensional space consisting of all scalar multiples of u . Similarly, null (A) = null(AT A) is the !-dimensional space consisting of all vectors v in R2 which are orthogonal to u , i e all V('ctors of the form v = s( -2, 1).

16. The reduced ·row echelon from of A = [ 1

2

AT A = [

1

1

3 -4

]

1 0

is [

0

7 ] and

I -6

the reduced row echelon form of

Thusrow(A)=row(ATA)isthe2-dimensionalspaceconsisting -7 - 11

17

0

0

0

of all linear comb1ne.tions of the vect ors u 1 = (1, 0, 7) and u 2 = {0, I , - 6). Thus null(A) =null( AT A) 1s the 1-dimtlnsional space consisting of aJI vectors v in R3 which n.re o rthogonal to both u 1 and u 2, i e, all \'CClor of the form v = s( - 7,6, 1) 17.

augmented ma.trix of the system Ax = b can be rel!\lo.:ed to

-3

)

0 0 0 0

bt b2 - bl

1

0 0 0

b3 - ·lb2

b4

+

+ 3bl

2b, b5- 8b2 + 7bt

thus the system will be inconsistent unless (b 1 , b2 , b3 , b4 , b5 ) the equations b3 = -3bt -1 4b2, b,. = 2btbs = -?b1 + where b1 can assume any values. 18.

The augmented matrix of the system Ax= b can be 1educed to

1 2 3 0 - 7 -8 [ 0 0 0 thus the system is consistent if and only if b3 solutions.

-1 : 5 0 :

!

= b2 + b1 and, in this case, there will be in!'lnitely many

223

WORKJNG WITH PROOFS

DISCUSSION AND DISCOVERY D l. If A is a 7 X 5 matrix wit h rank 3, t hen A'r also h as rank 3; t hus dim( row( AT)) = dim( col( AT ))= 3 and dim (nuli(AT)) = 7 - 3 = 4. 02. If A has rank k the n, from T heorems 7.5.2 and 7.5.9, we h ave dim(row(AT A))= rank(A1'A) rank(A1 ') = rank(A) = k and dim(row(AAT}) = rank(AAT) = rank( A) = k.

=

D3 . If AT x = 0 has only the trivial solution then, from T heorem 7.5.11, A has full row rank. Thus, if A is m >< n, we mus t m and dim(row(A)) = dim(col(A) ) = m .

D4. ·(a ) False. The row space and column space always have t he same dimension. (b) False. It is always true that rank(A) = ra.nk(AT) , whether A is square or not. {c) True. Under these assumptions , the system Ax = b is consistent (for any b) and so the ma t riCes A and [A I b J have the same rank. (d} True. If an m x n matrix A has full row rank u.nd full column rank, then m = dim(row(A )) = rank(.4) = dim(col(A)) = n. (e) True. If A 1'J\ and A A'T a re both invertible then, from 'Theorem 7.5.10 , A has full column (f)

rank ami full row rank; t hus A is square. True. T he rank of a 3 x 3 matrix is 0, 1, 2, or 3 a.nd the corres ponding nullity is 3, 2, 1, ·or 0.

D5 . (a) The solution$ of the system are given by x = (b - s- t, s, t) where - oo < s, t < oo. This docs not violate T heorem 7.5.7(b). (b) The solutions can be expressed as (b, 0, 0) -1- s( - 1,1, O) + t ( - 1, 0,1), whe re (b, 0, 0) is a pa.rt.icula r solution and s( - 1, 1, 0) + t( - 1, 0, l) is a. general solution of the corresponding homogeneous system.

\

n6.

If A is 3 x 5 , t.hen the columns of /1 are a. set of five vectors in R3 and thus linearly dependent. (b) Lf A 5 x 3, t.hen t.he rows of A are a se t of 5 vectors in R3 a nd thus a.re linearly dependent. (c) Jf A is m x n , with m f n, then eit her t he columns of A are line arly dependent or the rows of A arc linearly depende nt (or both). (a)

WO RKING WITH PROOFS Pl.

Frnm T heorem 7.5.8(a) we have nuli(AT A) = nuii(A) . Thus if A .is m >< n , t hen AT A is n x nand so ra nk (ATA) = n - nuility (AT A) = n - nullity(A} = rank( A). Similarly, nuli{AAT) = nuli(AT ) a.nd so rank(A AT) ::= m = m - nullity(AT) ""' ra.nk(Al' ) = ra nk( A) .

P2.

As above, we have rank( AT A) = n -

P3,

(a) Since null(A1'A) = null(A), we have row(A) {b) AT A is symmetric, we have col(AT A )

P4.

A)= n - nullity(A)

=

ra nk(A) .

= nuB{A).l = null(A'

= row(AT A) =

A)l.

row(A)

= row(A T A ).

= col(AT).

If A is m x n where m < n, then the columns of A form a set of n vectors in Rm and thus are linearly dependent . Similarly, if m > n , then the rows of A form a set of m vectors in and thus are linea r ly depender.t

224

Chapter 7

=

P5. If rank(A 2 ) = ran 1'(A) then dim(null(A 2 )) = n- rank(A 2 ) n- rank( A)= dim(nuii(A}) and, since null(A) <;;; null(A 2 ), it follows that null(A) = null(A 2 ) . Suppose now that y belongs to null( A) n col( A) . Then y = Ax for some x in R" and Ay = 0 . Since A 2 x = Ay = 0 , it follows that the vector x belongs to null(A 2 ) = null(A) , and soy = Ax= 0. This shows that null( A) n col( A)= {0}. P6. First we prove that if A is a nonzero matrix with rank k, then A has at least one invertible k x k submatrix, and all submatrices of larger size are singular. The proof is organized as suggested. Step 1. If A is an m x n matrix with rank k, then dim(col(A)) = k and so A has k linearly independent columns. Let B be the m x k submatrix of A having these vectors as tts columns. This matrix also has rank k and thus bas k linearly independent rows. Let C be the k x k submatrix of B having these vectors as its rows. Then Cis an invertible k x k submatrix of A. Step 2. Suppose D is an r x r submatrix of A with r > k. Then, since dim(coi(A}) = k < r, the columns of A which contain those of D must be linearly dependent. It follows that bne columns of D are linearly dependent since a nontrivial linear dependence arx1ong the containing columns results in a nontrivial linear dependence among the columns of D. Thus D is singular Conversely, we prove that if the largest invertible submatrix of A is k x k, then A has rank k Step 1. Let C be an invertible k x k submatrix of A. Then the columns of C are linearly independent and so the columns of A that contain the columns of C are also linearly independent. This shows that rank(A) = dim(coi(A)) k . Step 2. Suppose rank(A) = r > k. Then dim(col(A)) = r, and so A has r linearly independent columns. Let 8 be the m x r submatrix of A having these vectors as its columns. Then B also has rank r and thus has r lmearly independent rows. Let C be the submatrix of B having these vec-torc; as lt'l rm\''l Thf'n (' is a nonsingular r x r submatrix of A Thus the a..c;sumption that rank(A) > k has led to a contradiction. This, together with Step l, shows that rank(A) = k. P7.

If A is invertible then so is AT. Thus, using the cited exercise and Theorem 7.5 2, we have

= rank((C P)T) = rank (Prcr)- rnnk(CT) =rank( C) this 1l also follows that nullity() = n- rank(G' P) = n- rank( C)= nullity(C). rnnk(C P)

and from

EXERCISE SET 7.6 1.

A mw eehelon fmm fo• A ;, 8

=

-: 0

0

-•!];

and these column vectors form a. b asis for dim(row(A))

= 2, and

thus the fi.st two columns of A are the ph•oL columns A row echelon from for AT is C =

the first two rows of A form a basis for row(A)

2. The first column of A forms a basis for col(A), and

'I

3. The ma.tnx A can be row reduced to B =

:l:[r

thus 0 0 0

first row of A forms a basis for row(A). thus the first two columns of A are the pivot

0 0 0 0

::::0:o::::eed ,0

EXERCISE SET 7.6

4. The redur.ed row echelon form of A is B =

is C

=

0 0 0 0

0 0 0

=:

[i

iii]

0 . Thus the first two columns of 0 0

0 0 0

225

and the reduced row echelon form of AT

A form a basis for col(A), and the first two

rows of A form a basis for row(A) . 1

0

2

0

0

0

l

0

0

0 0

0 0

Q

0

5. The reduced row echelon form of A is B =.

;!(;

0 0 0

0

i

3 I

-· 6

and the reduced row echelon form 0 OJ

1

()( Al' is C: =

] .

0

0

()

Thus the fi rst tii <ee column..<> of A form a basis for col(A), and the

0

first three rows of A form a hasis for row( A) . 1 3

6. The reduced row echelon form of A is fJ

0

c=

0 0

1

0

5

-4

0

0 0

t

2

0

:

1

0

-

l1'1 . 0

=

·o o

0 0 1 0 0 0 0 1

0 0 0 0 0 0 0 0 0 0 0 0

and the reduced row echelon form of AT · iS

Thus lhe 1st. 3rd , aud •lt.h columns of A form a basis for co!( A), and

o oJ

the 1st, 2nd, and 4th rows of A for m a basis for row(A). 7 . Proceeding as in Example 2, we first form t he matrix ha.¥ing the given vectors ns its columns:

4 1]

G

10

-6

- 11

2

3

The reduced row echelon form of A is:

From this we conclude that all three columns of A are pivot columns; the vectors v 1 , v 2 , v 3 a.re lineMiy independent and form a basis for the space which they span (the column space of A).

Chapter7

226

8. Let A be the matrix having the given vectors as its columns.

R= 0

0

0

0

From thls we conclude that {v 1, v2} is a basis for W and v 4 == -2Vt + v2.

0 0

reduced row echelon form of A is

0 0

= col(A}, and it also follows that v3 = 2v 1 + v 2

9. Proceeding as in Example 2, we first form the matrix having the given vectors as its columns:

A==

0

0

2

2

3

6

0

3

The reduced row echelon form of A is:

From this we conclude that the 1st and 3rd columns of A are the pivot columns; thus {v 1, v3} is a basis for W col(A). Since c2 (R) == 2c1 (R) and c4(R) = c 1 (R) + c3(R), we also conclude that v2 = 2vt and v 1 v 1 -+- v 3 10. Let A be t he matrix having the given vectors as its columns. The reduced row echelon form of A is

From this we conclude that the vectors and = -v 1 1- 3vz +

11. The m•tnx

lA I 1,1 = [:

v 1,

0 -1 : 1 0 -'2 I 0

0

0

0

0

I

I

2

1

-1

0 0

0 0

0-1] 0

3

1

2

0

0

v 2 , v 4 form a basis fnr W = coi(A), and that v 3 = 2vt- v 2

can be row reduced (and further parmioned) to

I

0

0

= 0 Thus the vectors

v1

0

0 : 0

0

= (1,- 4,0) and v 2 = (0, 0, 1) form a basis for

12. The matrix {A I /3j can be row reduced (and further partitioned) to

= Thus the vector v 1

= (-1, 1, 1) and forms a basis for nuii(AT) .

1 null(AT).

227

EXERCISE SET 7.6

13. The redu-:ed row echelon form of A is R

! .

=

From this

conclude that the first two

lo o o o

columns of A are the pivot columns, and so these two columm; forrn a has is for col(A) . The fi rst two rows of R form a basis for row(A) = row(R). We have Ax = 0 if and only if Rx = 0, und the general solution of the latter is x = (s+4t ,3s+1t, s,t) = s{1,3,1,0) +t(4, 7, 0,1). Thus (1,3,1,0) and (4, 7, 0, 1) fonn a basis for null(A) . The reduced row echelon form of the partitioned matrix [A I !4} is

[ Thus t he vectors ( 1, 0, -

1 0 0

0 1 0

0

0

-

l

- 1 -4 0 - 3 -7 I 0 0 0 : 1 I

0

and (0, 1,

0 : 0 -

t

0 0 0

0

-1 _ .! 4

1

form a basis for null(AT).

14. The •educed ww echelon fmm of A ;, R =

-: 0

0

:]. From this we conclude t hat the lsl. 2nd ,

0

0

and 4th columns of A are the pivot columns, and so these three columns form a basis for col( A). The first three rows of R form a basis for row( A) = row( R). We have Ax = 0 if and only if Rx ""' 0, and the general solution of t he latter is x = s, O) l ,O). l , Orrorms a basis for null( A). The reduced row echelon form of the part itioned mat rix [A I J 1 j is l

0

0 1

:...12

0

:-!

4

0

1 I

s

I

I

Thus t.he vector (0, 1, -£ , 15. {a)

0

8 1 lG

11o

- .l. 40

o

1 80

t---

1

-5

---}

5

5

t) forms a b&Sis for null (A-r).

'The lst, 3rd, and 4th columns of A are the pi vot columns; thus these vectors form a basis for col(A).

(b) The first three rows of A form a basis for row( A). (c) We have Ax = 0 if and only if Rx = 0, i.e., if and only if x is of t he for(ll X =

(d)

(- 4t,t,O,O)

= t(-4,1,0,0)

Thus the vect or u = (-4 , 1, 0, 0) forms a basis for null(A). We have ATx = 0 if and on ly if C x = 0, i.e. , if and only if x is of the form

x=

+ s(O,

=

Thus t he vectors v 1 = (- LO, O, O, 1) and v 2

16. The reduced ww <ehelon fo•m of A = [: - : ticn is

A=CR =

= (0, is

.Ro

1, 0) form a basis for nuli(Ar).

[

t]·

il

Thus the column-row factori>a·

228

Chapter 7

.1nd the ro rrespcmding column-row expansion is

m

[1 0

II+

HJ

1

jO

0 -37

- 4

17. The reduced row echelon fo rm of A

=

[l

1

-9

-4 82

- 2 . "] 18 14

Ro

s -3

0

= ['0

6 -10 -10

0 0

0

0 0

=;].

Thus the column-

row factorization is

A = CR=

and the corresponding column-row expansion is

-

3


(a)

1f A a 3 J( 5 matrix, then the number of leading l 's an a row echelon form of A is at most 3, the number of parameters m a general solution of Ax = 0 is at most 4 (assummg A¥= 0 ), I he rank of A is at most 3, the rank of AT is at most 3, and the nullity of AT is at most 2 (assuming tl -# 0)

(b )

If A as a fl"' 3 rnatnx then the number o f leading l 's in a row echelon form of A is at most 3, the num ber of pararnrters in a. general solution of Ax= 0 is at most 2 {a.ssumang A ¥= 0 ), lhe r1.nk o f A is at most 3. the r ank of AT is at most 3, and the nullity of AT is. al most 4 (nssummg A "# 0)

(c)

If A iR a 4 x 4 matrix, then the number o f leading 1 's in a row echelon form of A' is at most I, tbP number of parameters in a general solution of Ax = 0 is at most 3 (assuming A i 0), l he rnnk o f A is at most 4, the rank of A T is at most. ·1 , and the nullity of A T is at most. 3 (assummg A i 0). If A is an m x n matrix, then the number of leading l's in a ro w echelon form of A is at mos t rn, the n .. mber of parameters in a general solution o f Ax : 0 is at most n - 1 (assuming A # 0), the rank of A is at most min{m, n}, the rank o f A T is at most min{m , n} , a nd the n ullity of AT is at most m- 1 (assuming A# 0).

(d )

D2. The pivot columns of a matrix A are those columns

correspond tv tht: columns of a row ec..,Plon form R o f A which contain a leading 1. For example,

-2 - 10 4 -5

0 4

0

7

6 6

12

-5

12]

28 --. R - 1

=

[1

2

0

3 0

0 0 1 0 0 0 0 0

and :-:o the 1st, 3rd, nnd 5th rolumns of A are tbe pivot columns

0

229

EXERCISE SET 7.7

D3. The vectors v 1 = (4, 0, 1, -4, 5) Mid v 2 = (0, 1,

0, 1) form a bnsis for null( AT).

D4. (a) fa 1, az , a4} is a. basis for col( A). (b) aJ = 4a 1 - 3a2 and a.o = 6a1 + 7a2 + 2a4.

EXERCISE SET 7.7

l

i ).

1. Using Formula (5) we have projaX = 1;:.:1{2 a= ( 1 )(1, 2) = (151 , 2 On the other hand, since sin(} = and cos B = the standard matrix for the projection is Po [

ii !.

.

and we have

Pex =

[li

fs

i_ [- 61} =

2. Using Formula (5) we have proj 8 x On the other hand, since

Pe =

and we have Pex

[_ 29

3.

[.!l] ·¥·

=

=

= [_

29

=

=

and

the st andard mat rix for the projection is 19

:.!!)

·

A vector parallel to l is a = (1, 2). Thus, using Formula 5, the projection of x on lis given by

.

proJaX

a·x a ·""' ( 3)( ) (3 6) = llall2 5 1, 2 = 5• 5

On the other hand, since sin 0 = .1,5 and cos (J

[I 1 2] [i ; and we have Pex = i ! 5

4.

7&,

=

[3] .

5

=

1

15 , the standard rnatrix for the projection is Pe =

v

.')

A vector pnr11.llel to l is a = (3, ·-1). Thus, using Form ula. 5, the projection of x on l is g.iven by

. a ·x -a )( ) ( 33 11 proJ.,X = ll all2a :.::(IO . •3,··· 1. = - ro,w ) On the other hand, since sin O =-

Po =

[_1 . .1] 10

vko

and cosB =

and we have Pex""" [_

10

10

-11

;?to. the standard matrix for =

It)

r-

10

·

!

5. T he vector component of x along a is proj 6 x = 11':"1j2 a = (0, 2, -1) = (0, orth.:.gona l to a is x- proj 8 x = (1, 1, 1) ·- (0, = (1,

-0

6 . prOJ·8 X, -_ iiGlr X·a a = 5 ( l , 2, 3) -.. 5 14• 10 i4 1[)) , an d x - - prOJaX· _ (Z 1 Q, 1) _.. ( Iii• 14 7. The vector component of x along a is projax = {a:lij a = component orthogonal to a is x - proj.x = (2, I, 1, 2) 8. proj ..x =

1,-1, -- 1):::;

x- projax

.

9 . llprOJaxll

=

= (5, 0, -

Ia · xl

=

3, 7} -

e72,

(11,

-¥),

= CZi,

12- 6 + 241 20 20 J4 +9 + 36 = J49 = 7

-

(

-

i , 575 ).

1

), and the component

5 10 15) _ ( 23 14 , i4• 14 - 14•

= fo, - lo) = (

-4 , 2, -2)

the projection is

L{o [o,

10 ·-!4 I ). - 14,

1 10 ),

and the

230

Chapter 7

2 1 J6 v'24=J6-6

Ia · xl

18- 10 + 41

la·xl

1-8+6-2+151

-nail= J4 + 4 +- 16 .

11.

llprOJ nxll =

1 2.

llprOJ aXII

.

Taf = a·

xl

v16

+4+4 + 9

.J33

11

J33 =

=

135-3 + 0- 11

31

= Taf = V49 + I + 0 + 1 = J5I =

-3-

3IJ5i

13. If a = ( - 1, 5, 2) then, from Theorem 7.7.3, the standard matrix for the orthogonal projection of R 3 onto span{a} is given by

P=

;.

aaT

= ..

a a

"

-1] [ 1-1

[ 1-5 -2]

5 2) =

5

2

-5

25

10

-2

10

4

We note from inspection that P is symmetric. lt is also apparent that P has rank I since each of its rows is a scalar multiple of a . Finally, it is easy to check that

p2

= _.!..

30 -2

10

3

4

-2

10

4

= _.!.. 30

- 2

and so P 1s tdempotent.

14. If a = (7, - 2, 2) then the standard matrix for the orthogonal projection of R 3 onto span{ a} is given by

7] lr_

1 T 1 P = - a a = - -2 aTa 57 2

[

1 -2 2] = 57

114 -11] 4

[- 1419 14

-4

4

We note front in!>pection that Pis symmetric. It is also apparent that P has rank 1 since each of its rows is a sc,\lar mult1plc uf a Finally, it is easy to check that P 2 = P and so P is idl.mpot enl.

Hi.

Let M =

3

?.] u

I

:l

- 1

[

.

Then i\f 1 M

16

=[

orthogonal prOJCclton of !1. 3 onto W

P= AI(MTM)-IMT --=

[-!

'l

;] 9 1

and. from Theorem 7 7 5 , the standard matrix for the

= span {a 1, a2}

is given by

[13 -9] [3 -4

1 3 257 - 9

26

2

0

1 ] 3

113 1 -84 257 [ 96

=-

-84

208

56

96] 56

193

We not e from inspection t.hat lhe matrix Pis symmetric. The reduced row echelon form of P is

! and from this we conclude that P has rank 2. Finally, it is easy to check that

P' =

[

and so P ic:: idl'mpotent

,m 2!1 [

,m = 2!1 [

=P 193

231

EXERCISE SET 7.7

16. Let M

[- :

Then MT M

:]and the •tandW"d matrix for the orthogonal

of R 3 onto W = span{ a 1 , a2} is given by

-23] [1 -2 30 -2 4

-102

305

From inspection we see that P is symmetr ic. The reduced row echelon form of P is 0

1 0 and from this we conclude that P has ranK. 2. Finally, it is easy to check that P 2

= P.

17. The standard matrix for the orthogonal projection of R 3 onto the xz-plane is P =

agcees with the following computation rn;ing Po, mula 0

and M(MT M) - MT

[:

[;

:1

:]

(2n

Let M

[: ;

l

This 0 0 1

Then MT M

,

:J.

=

=

18. The standard matrix for t he orthogonal projection of R 3 onto the yz-p]ane is P

This 0

19. We proceeri as iu Example 6. The general ::oolution of the equa tion x

+y +z =

0

I

0 can Le written as

and so the two column vectors on the right form a basis for the plane. If M is the 3 x 2 matrix these vectors as Jts columns, then MT M

=

and the standard matrix of the orthogonal

projection onto the plane is

lJ

[2 -l] [-1 1 0]

2 -1

2

- 1 0

The orthogonal projection of the vector v on the plane is Pv

=

1

=3

-1

-1

2

-1] [ 2] = 1[ 1]

- 2J -1 -1 - 1 2

4 - 1

7 . - 8

232

Chapter 7

20. The general solution of the equation 2x - y + 3z = 0 can be written as

and so the two column vectors on the right form a basis for the plane. 1f M is the 3 x 2 matrix having these vectors as its columns, then NfT M

[!

=

and the standard matnx of the

projection onto the plane is

P = M(MT M)- 1 MT

=

12 30] _!_ [ 10 - 6] [ 1 2 0] = _!_ [ 102 132 -6] 3

[ 0 1 14 -6

5

0 3

1

14

-

6

3

5

2 -6] [ 2] = 141[34]

10 2 14 [ -6

The orthogonal projection of the vector von the plane is Pv = ]_

13

3

1

li

4 - 1

53 •

-5

21. Let A be the matrix having the given vectors as its columns. The reduced row echel<m form of A is

and from this we conclude that t.he a 1 and a 3 form a basts for the II' spanned b}· the given vectors (column space of A). Let M be the 4 x 2 matrix havmg a 1 and a 3 as its columns. Then 6

0

2 -4

-2

4 lj

-6

5]

[

0

72

2 -2

= [- 20

-4

20]

30

5

and the standard matrix for the orthogonal projection of R 4 onto W ts g1ven by

M(MT M)-t MT = [ - :

2 -4

_1_ [ 30 -2 1760 -20 5 89

-105

-105 220 - 3 [ 25

135 - 15 15

= _1_

- 20] [4 72 1

-3 - 15 31 -75

-6

2

0 -2

25] 15 - 75 185

22. Let A be the matrix having the given vectors as its columns. The reduced row echelon form of A is

23:,

EXERCISE SET 7.7

and from this we conclude that the vectors a 1 and

form a basis for the subspace W spanned by the

a2

oiven vectors. Let M be the 4 x 2 matrix having a 1 <md a 2 as its columns. Then M r M = [

o·

·

24 ] 126

matrix for t;he orthogonal projection of R4 onto W is given by

and the

M(MT A-1)-l

5 3] MT = -3 -6 _1_ [ 21 -4] [5 -3 [-11 9 660 - 4 6 3 - 6 0

23.

36 24

The reduced row echelon from of the matrix A is R

_.!._ -89

0

220

9

1

system Ax= 0 can be expressed as

x=

-ts _! + !t1 lt =s

r

2

3 -

l

2

J

s

_.! 2

10

[ 31 -13 - 37 -59

31 -37] -13 -59 7 - 19

-19 193

i1 -i1] ; thus a general solution of the

0

=

153 -89 87

1 -1] =

+t

2

[

-l 2

OJI

where the t,wo column vectors on the nght hand side form a basis for Lhe solution space. Let D be t.hP. matrix having these two vec tors a.S its columns. Then 1

0]1

0

[=! -i] 1 0

0 1

0] = 2 [101OJ

=

and the standar d matr ix for t he orthogonal projection from R 4 onto the solution space is

l-l-! 'l

[-:

2

P

IJ(BT 8) - ' Br

_ .!.

1

?

:;

[0

0 -l 2

1

_!

0

2

0]1 = 3

[-t

- 1

- 1

-1

2

·· 1

0

_:] 0 2

Thus t he orthogonal projectio n of v ·= {5, 6, 7, 2) on the solution space is g iven (in column form ) hy

0 3

24. The ceduced

COW

!com of the matdx A is R

r:

lo

thesyst.m Ax

= 0 can be expcessed as = t X

r- '"'

0

thus the gcnP.ra.l solution of

:

o

-2

(to avoid lcactions)

X

= • [- ] · In ot h& words,

the solution space of Ax= 0 js equal to span{a) where a = ( -3, 0, 1, 2). Thus, from The;).lem 7.7.2, the orthogonal projection of v = (1, 1, 2, 3) on th
.

proJ 8 v =

a· v

11

(

a = i4 ( -3, 0, 1, 2) = -

15

14 , 0,

s

10

14 , 14 )

234

Chapter7

is R =

A=

25. The reduced row echelon form of the

From

3 2 -1 1 0 0 0 0 form a basis for row(A), and the first two columns of

this we conclude that the first two rows of R A form a ba.c:;is for col(A). Orthogonal proJeCt JOn of R4 onto row( A): Let B be the 4 x 2 matrix having the first two rows of R as 1ts columns. ThPn

1 0] [ =

0 1 3 0 1 1 - 2 - 4] 1 -2 3 -4 and the standard matrix fo r the orthogonal p rojection o f

Pr

= B(BTB) - tBT =

1 3

proJCCtJOn of R3

Orthogonal A as its columns. Then

-2 - 11

[21 35 14

1114] [10

11 -14 [-14

21]

R4 onto row(A) is g iven by

01 -21 -43]

=

-1

2_ 35

-7 7

-8 -2

7]

-8

-2

9

11

1l

29

onto col(A): Let C be the 3 x 2 matnx having the first two columns of

cr c =

[t 1 3] 102

= [11 1]

32

75

and the standard matnx for the orthogonal projection of R 3 onto col(A) is given by

P,

= cccrc)-'cr =

26. Tht reduced row echelon form o f

[:

[i IS

R

=

!l

=

H

From this we conclude that the first two

0 0 0 0 0

rows of R form a basis for row(A), a.nd the first two columns o f A form a basis for col(A). Orthogonal projet..tlon onto row(A): Let B be t.he 5 x 2 matrix having the first two rows of R as ItS

column'l Then

G

sT B

and the standard matrix for the orthogonal projection of Rl'> on to

row(A) is given by

Pr

= B(BTB)-I BT = 2_ 24

7

-5

-5

7

2 9

2 -3 9

its columns. Then

ere= [1:

9

-3

2 -3

9

4

6

6

6 5

15 3

3 15 x 2 matrix having t he fi rst two columns of A as

-3 Orthogonal projection onto col(A): Let C be the -1

2

:] and t he stand ard matrix fo r the or t hogonal projection of 2

col(A) is given by 237

P. c

= C(crc)-lcr =

_I_

419

- 73

- 73 369 -13 -95 [ 194 64

29 - 46

1::]

- 46 203

R 4 onto

235

EXERCISE SET 7.7

0

27. The reduced row echelon foxm of the matrix [A I b) is

0

fii:

0I

0

[-,! - ,! •-

its columns, and let C = BTB

tl·

=

Xrow ( A)

1

2

25

25

!R

1

= Pxo =

- 25

-25

'25

25

2

!! 25

·n

7

1

25

b is onnsht
[-i]

11

25

25

28 . The reduced row echelon for m of the matrix [A I b j is

0 0

!RIc] :::

= [-

having the first t wo rows of R as its columns, ami let C

[

1

0

0

1

0

0

n1

=

32 -3I 1I 34] fi fi :- . rrom this we

=

I. .

131

20

299 224 299

13 l 299

32

299 124 299

53 - -m

299

0

is one solution. Let B be the4 x 2 matrix

= BT B = [_

-

matrix for the orthogonA-l projection of R 4 onto row(R) =row( A) is p ::-: of Ax = b which lies in row( A) is given by :!99 20

o,

0

:16

= Pxo

,!']

and the solution of Ax= b which lies in row(A) is

_.!_

25

2!1 11

Xrow (A )

Prom this we

Then the s tandard matrix for the orthogonal projection

= row(A) is P = BC- 1 Br,

r

conclude that

k.

:

0

of R 4 onto row(R) given by

1]

g8 .I - 5

51

1 -

- f.

j

124

l

Then t he s tandard 72

vc-1 BT , and the solutio n

"lr .l I"'l

209

m

90 2!19

-A 299 90

25

- 299

299

- 3 _ 0 -

0

299 48

-m

!M 299

!!l

- 299

29. In Exercise Hl we found t hat the standard ma.trix for the orthogonal projection of R 3 onto the plane W with equation x + y + z = 0 is 1 3

P =-

-1

-

. 1

2

Thus the standard matrix for the orthogonal projection onto

1- p

= 0 0

1

-1

·- 1

2

w.t

(the line perpendicular toW) is 1 1 1

Note that W L is t he !-dimensional space (line) by t he vect or a= (1.1, 1) and so the computation above is consistent with the formula. given in Theorem 7.7.3.

236

Chapter 7

30. In Exercise 20 we found that the standard matrix for the orthogonal prOJection of R 3 onto the plane W with equation 2.c - y + 3z = 0 is

2 -6]

10 2 14 [ -6

13 3

3 5

Thus the standard matrix for the orthogonal projection onto W J. (the line perpendicular to W) !s

1 14 [

I - p =

0 31.

0 1

1

= 14

-6

3

5

6

-3

-:] 9

Let A be the 4 x 2 matnx having the vectors v 1 and v 2 as its columns. Then

[1 -2

!I'TA =

3

0]2

3 -1

4

!]

3

- 1

0

2

14 -8] 30

= [- 8

and the sta ndard matrix for the orthogonal projection of R4 onto the subspace W

=

p -= A(A r A)-1 l r

3 0

- 1 178 2

[15 4] 4

7 l3

Thus the orthogonal projection of the vector v

51 \

Pv

r:]

ll

23 54

28 - 31

89 28 - 31 25 20

59

1

23

[

fl

=

51 -2 4

1:

1]1 =

= 89

23

23 54

[2825 - 3120

28 - 31 59 5

is

25]

20 5 14

w.1. is gtven {in column form) by

(1, 1, 1,1} onto

25] [1]1

1

2]

-1

20

5

0

3

= spa.n{v 1, v 2 }

l

-

66 [127]

89

1

[-38] 23

- 89

DISCUSSION AND DISCOVERY 01. (a) (b)

The rank of the standarrl matrix for the orthogonal projection of Rn onto a line through the on gm 1s 1, and onto its orthogonal complement is n - 1. If n 2, then the rank of the standard matrix for the orthogonal projection of Rn onto a plane through t he ongin is 2, and onto its orthogonal complement is n - 2.

02. A 5 x 5 rnatnx Pis thl' s tandard matrix for an orthogonal projection of R5 onto some 3-dimensional subspace if and only 1f it is symmetric, idempotent, and has rank 3. D3. If x1

= projax and x2 = x- x1 ll x2ll

Thus q

=/

ll x ll 2

-

2

then

llxll 2 ==

llxdl 2 + llx2ll 2 and so

= llxll 2 -llxtll 2 = llxll 2 = llx2ll where x2

2

= llxll 2 -

is the vector component of x orthogonal to a .


237

the standard matrix for the orthogonal projection of Rn on a. subspace W, then P 2 (Pis idempotent) and so p k = P for all k ;:::: 2. In particular, we have pn = P.

D4. If P D5 . (a)

=P

TJ"ue. Since projwu belongs toW and projw.1. u belongs to W .L, the two vectors a.re orthogonal

(b) False. For example, the matrix P

=

satisfies P 2

;::

P but is not symmetric and there-

fore does not correspond to an orthogonal projection. (c) True. See the proof of Theorem 7.7.7. (d) True. Since P 2 = P, we also have (J - P) 2 =I - 2P + P 2 = I- P ; thus I- P is idempotent. (e) False. In fa.ct, since projcoi(Aj b belongs to col(A). the system Ax= projcol(A)b is always consistent. D6. Since (W.L )l. = W (Theorem 7 .7.8), it follows that ((Wl.)..l)l.

D7. The matrix A =

1l 1] [ 1 1 1

= W.1..

is symmetric and has rank 1, but is not the standard matrix of an

1 l 1

orthogonal projection. Note that A 2 = 3A, so A is not idempotent. D8. In this case the row space of A is equal to all of Rn. Thus the orthogonal projection of Rn onto row(A) is the identity transformation, and its matrix is the identity matrix. D9. Suppose that A is ann x n idempotent matrix, and that>. is nn eigenvalue of A with corresponding eigenvector x (x f 0). Then A 2 x = A(Ax) = A(.h) = ..\2 x. On the other hand, since A 2 =A , we have A 2 x = Ax = .>.x. Since x =/= 0, it follows 'that >.? = >. and so >. = 0 or 1. DlO.

Using ca lculus: The reduced row echelon form of lA I bj is

thus the general solution of 0 0 010

Ax = b is x = (7 - 3t, 3 - t , t} where -oo

llxll 2 = {7- 3t) 2 +

< t < oo. We have (3 - t) 2

+ t2

= 58 - 48t + 1lt2

and so the solut.ion vector of smallest length corresponds to

t :.=:

We conclude that

Xrow

= (7=-- ii, 3- H,

ft lllxii 2 J =

-48 + 22t = 0, i.e., to

5 = ( 1p

Using an orthogonal projecti on: The solution Xrow is equal to the orthogonal projection of a.ny solution of Ax = b , e.g., x (7, 3, 0}, onto t he row space of A. From the row reduction alluded to above, we sec that the vectors v 1 = (1, 0, 3) and v 2 == (0, 1, J) form a basis for the row space of

=

.4 . Let. B be the 3 x 2 matrix having these ver,tor as its columns. Then BT B

standard matrix for the orthogonal projection of R 3 onto W

1\

Finally, in agreement with the calculus solution, we have

1

= Px = 11

[;] =

3

1

10

0

and the

= row(A} is given by =

Xrow

=

f [ !] 1

24

3

1

10

238

Chapter 7

Dll. The rows of R Corm a basis for the row space of A, and G = RT has these vectors as its columns. Thus, from Theorem 7.7.5, G(GTG)- 1GT is the standard matrix for the orthogonal projection of Rn onto lV = row{A).

WORKING WITH PROOFS Pl. If x anJ y are vectors in Rn and if a and /3 are scalars, then a· (ax+ /3y) =a( a· x) + /3(a · y). Thus

T(ax + /3y) =

a · (ax + /3y) (a · x) (a · y) llall 2 a = a:lja112a +

= aT(x) + /3T(y)

wh ich shows that T is linear. P 2. If b = ta, then bTb = b · b = (ta) · (ta)

= t 2a · a

= t 2aTa and (similarly) b b T = t2aaT; thus

1 T 1 2T bTb b b = t2aTa t a a P3.

1 T aT a aa

Lf'l P hl' a symmetric n x n matrix that is idempotent. and has rank k. Then W -= col(?) is a k-dimens•onal subspace of R". We will show that P is the standard matrix for the orthogonal projecl ion of Rn onto W, i.e, that Px = projwx for all x in Rn . To this end, we first note that Px belongs to W and that

x

= Px + (x- Px)

= Px t (I- P )x

To show that Px = projwx it suffices (from Theorem 7.7.4) to show that (I- P)x belongs to \1' • .1nd since W - col{ A) = ran(P). this IS equtvalem to showmg that Py · (J - P)x = 0 for all y m R" Finally, since pT = P = P 2 (Pis symmetric and idempotent), we have P(J- P) = P - P 2 - P - P = 0 and so

for evcrv x and y m R" This completes the proof.

EXERCISE SET 7.8 J.

First we note that the columns of A are linearly independent since they are not scalar multiples of each other; thus A has full column rank. It follows from Theorem 7.8.3(b) that the system Ax= b has a unique least squares solution given by

The least squares error vector is

b- A x = 5

4

5

11

8 = 11

15

and it is easy to chP"k that this vector is in fact orthogonal to each of the columns of A . For example, (b- Ax )· Ct(A) = 1\ !(-6){1) + {-27}(2) + (15)(4)] = 0

239

EXERCISE SEl' 7.8

2. The colunms of A are linearly independent and so A has full column rank. Thus the system Ax= b has a unique least squares solution given by

1 2.. 6] -2 I I] 2lj21 [-14] 0

-l [

2 1 3

[

- 1 -

9

The least squares error vector is

and it is easy ,to.check that this vector is orthogonal to each of the columns of A. 3.

From Exercise 1, the least squares solution of Ax= b is x =

-1] 1 [ 2 4

Ax =

3 11 5

f1

= 11

8

[28] 16 40

On the other hand, the standard matrix for the orthogona.J projection of R 3 onto col(A) is

[21 25]-l [ 1 2 4] = _1

p _ A(ATA.)-tAT::::.: 4

5

25

35

-1

3 5

:).nd so we have

projcoi(A)b

= Pb = 20

4.

90

170

From Exercise 2, the least squares solution of Ax = b is x

2 --2].l .!.. [ 9] = [3 1 21 -1 4

Ax = 1

220

[-· f>

= 2\

[

20

90

1 = 11

40

= Ax

-t:]; thus

[ tj6]

..!:_ - 5 21

13

On the other hand, the standard matrix of the orthogonal projection ont.o col(A) is

and so we have 1 17

1 .

= 21

13

=Ax

170

240

Chapter 7

5. The least squares solutions of Ax = b are obtained by solving the associated normal system AT Ax= .4Th which IS

Sinc·p the rnntrix on the left is nonsingular, this system has the unique solution

X= [J:1] = [248 8]-l [12]8 = 2_ [ 6 -8] [12] =[to] 6 80 -8 24 8 l X2

The error vector is

and the least squares error is li b - AxiJ

=

+

+ (0) 2 =

Ji =

6. The least. squares solutions of Ax= b are obtained by solving the normal system AT Ax = ATb which IS

[4214 12642] [Xl] = [124] X2

This (redundant) system has infinJtely many solutions given by

X= [::J = The error ve,.lor

3tl = [!J + t r n

jc:;

and the least squares error is li b - Axil=

+

=

+

J* = :Lfl.

7. The least SCJUf\rf'S solut1ons of Ax= b are obtai;1ed by solving the normal system AT A x = ATb which is

0

The augmented matdx of this system reduces to [: 0

solutions given by

1

'-a7]

' 1 :

i ; thus there are infinitely many

0'

0

EXERCISE SET 7.8

241

T he error vector is

and lhc least squares error is lib - Ax il = 8.

J(!)2

=

+

=

The least squares solutions of Ax= bare obtained by solving AT Ax= ATb which is

1 [ 17 33

[::]

50

=

X3

6

1:-sltf ;

0

Tbe augmented matrix of this system reduces to

1 :

0

0

I

.

thus there are infinitely many

0

:;olutions given by 1

[W,=:]

x = [:;] T he error vt.'C tor is

=

and t he least squares error is llb - Axll

V( !iY + {fr )2 + {- ·fi-)2 =y

9. The hnc"' model fo, the given data b M v

wh"c M

il

,f1j; = f[I .

and y

2

= [;] .

The least squa<"'

M T y which is

solm ion i:> o b tained by sobing the norrnal sys tem

[1:

[i

=

=

Since t he matrix on t he left is nonsingular, this system has a. unique solution given by

1] [ 4 16J-l [IO] [V2 = 16 74 47 = v

3 1 [ 37 -8] [tO] 1 [-3 ] r1 0] 20 ··" 8 2 47 ::::: 10 7 = 7 10

T hus the least squares straight line fit to the given data is y-== y 4 2

12 -I

3

456

7

3 10

+ I0 x.

242

Chapter 7

10. The Hnear model for the given data is Alv solution

[1

and y

= MTy

which is

y where M

obtajned by solving the normal system AfT Mv

r:

[!]·

The leaSt squares

r::J =

Since the matrix on the left is nonsingular, tllls system has a unique solution given by

[Vl] = [48 V1

8] -l [4] 1[22 -8] [4] = 1[16] = 9 = 24 -8

22

4

9

Thus the least squares straight line fit to the given data is y

24

4

!

= j + !x.

y

2

2

-I

3

4

-I

11. The quadratk least squar" model for the gh-en data is Mv- y whe" M The least squares solution

IS

obtained by solving the normal system M 7 Mv

and y

= MT y

which

r .t8 228 6222] [l..'t ] [ 9'] 1Jl

l22 62 178 Since the matrix on the left

JS

nonsingular, this system has a unique solution given by

l [ :] = [-

=[: V3

27

V3

22 62

178

27

J -;] [:] = [-

!

6

-1

Thus the leas1 squares quadratic fit to the given data is y = 1 y

-I

!

3

11 x 6

27

+ jx 2

1 3

1s

[!]·

243

EXERCISE SET 7.8

12. The quadratic least. squares m.odel for the given data is Atv = y where A1

110lj andy= 1 0

=

1 1 1 1 2 4

The le3st squares snl tti.ion is obtained by solving t he normal system Mr llifv =

M'I'y

which is

Since the ma.t.rix on the left is nonsingular, t his system has a unique solution given by

113

Thus the least

[! : [ 6

10

=

t

14

18

-:

_;] [

- 2

1

1

1

normal sysle111 A(r /11 v

r

l

5

15 55

s

4

2 3 4 5

M:::::

r

1

1

27 64 25 125 16

is

225]

55

225

979 225 979 4425 2'25 979 442:> 2051.1

=y 4.9

n a1

a2

a3

-

l 2168 18822.41 916.0

4087.4

and the solution, written in comma delimited form, is (ao, a 1, a2 ,a3) 14. The model for the least squares cubic fit to the given data is Mv

M=

=y

1 1 1 1

0 1

0 1

0

0.9

1

2 3

4

8 27

1

4

3.1 9.4 24 .1 57.3

9 10

64

where

10.8 27.9. y= 60.2 113.0

!)

= lv[T y

[=:]

+

quadratic fit i:.n the give n tlat<'- is y = -1 -

13. The model for t.he lea."3t squares cubic fit to the given data is Mv

T he

== H

(5 .160, -1.864,0.811, {).775). where

Chapter 7

The associated normal system MT M v = MT y is 5 10 30 [ 100

10 30 100 30 100 354 354 1300

aol [ 323.4 94.8] 1300 [ = 1174.4 4890 4396.2 and the solution, written in comma delimited form, is (ao, a1, a2, as) (0.817, 3.586, -2.171, 1.200). 15. If M =

1:tal. and y = [. . 1

:t2

•

•

1 :tn

MTy = [

100] a1 354

a2

a3

312 1 [Yal . , then MT M = [

.

X l

•

1 2:2 n ] [1. :ral = [r: :z:,

1

··

1

%2

•· ·

:J:n

Yn

.• .

1

:

!]' and

L. :t

Xn

1 1··· 1] Y!Ill. = [J-Y• ]. Thus the normal system can be written as 2

XI

:1:2

:Z:n

:

L.x,y,

!in

DISCUSSION AND DISCOVERY Dl. (a)

The distance from the point Po= {1, -2, 1) to the plane tv with equation x + y- z d

= IP)(l) + {1)(-2) ;- (-1)\1)1 /{1)2+(1)2+(-1)2

= _:__ = 2,'3 V3 3

).

and the point in the plane that is closest to Po is Q = ( computing thr

of the vector b

vectors of the matrix A = onto IV

ag

= 0 is

-:---+

= OP0

The latter is found by

onto the plane· ThP column

form a bnsis for tv and so the orthogonal projection of b

given by

(b) The distance from the point P0

= (1, 2, 0, -1) to the hyperplane x 1 - x 2 + 2x 3 - 2x4 = 0 is

d = I( 1)(1) + (-1){2) + (2)(0) + (-2)( -1)1 /(1)2 + (-1)2 + (2)2 + and the point in the plane that is closest to Po is Q == ( 190 , by computing the orthogonal projection of the vector b =

0

P' Jwb =

-!

io [

: -:]

= _ 1_

Jill 2 , 10

= JTij 10

The latter is found onto the hyperplane: ).

H ! Ul

1 1 = 0 [

245


D2.

(a)

= A ( AT A) - 1 ATb .

The vect or in col(A) Lhat is closest to b is projcol( t!)b

(b) The least squares solution of Ax == b is X= (JiT A) - lATh. (c)

The least squares error vector isb -A( AT A)- 1 A'T b .

(d) The least squn.res error is li b-- A(AT A)-1 A1'bll.

(e)

The stan
D3. From Theorem 7.8.4, a vector xis a least squares solution of Ax to coi(A).l . We have A =

= b if and only if b - Ax belongs =

a.nd b - Ax = 4

5

4

thus

= [

s

5

14

s - 14

b - Ax is orthogonal to col( A) if and only if ( 1)(2) + (2)( -7)

+ (4)( s -

14) = 0 ""' ( - 1)(2)

·1s - 68

=0=

+ (3)( - 7) + (5 )( s -

14)

5s - 93

equations are dearly iucornpatible and so we conclude that, for any value of s, the vector x is not a least sq11ates solution of Ax = b.

D4. The given data points nearly fall on a stra ight line; thus it, would be rel\..c;onable to least squares fit and then use the resulting linear formula y = a + bx to

05 . T he m<><;el foe t h;s least "''"'"' fit ;, [:

[

:

'l

Thus

[:]

36

il

+

1:1 : m, •nd the conespond;ng nonnal system ;s

l[

3 11 2 3]1 [ [ LJ= i *

= 251

= 45.

squares solution is

al

rP.'lulting in y

a linear

to x

i

;n

.1)

=

-

l [5]

1 9 ] [1 ¥

-1

...

21

=

as the best least squares fit. by a curve of this type. y

10

6 4

2 I 2

D6. We have

[:]

=

3 4 5 6 7

if and only if Ax + r =b and AT r

= 0.

Note that AT r

= 0 if and only

if r is orthogonal to col(A) . It follows that b - Ax belongs t.o coi{A).L and so, from Theorem 7.8.4, x is a least squares solution of Ax= b and r = b - Ax is the least squares error vec".nr.

246

Chapter 7


If Ax= b is consistent, then b is in the column space of A and any solu t ion of Ax = b is also a least squares solution (since li b - Axil= 0). If, in addition, the columns of A are linearly indep endent lhen there is only one solution of Ax = b and, from Theorem 7 8.3, t he least squares solution is also umque. Thus, in this case, the least squares solution is the same as the exact solution of Ax = b

P 2. If b is orthogonal to the column space of A, then projcoi(A)b = 0 . T h us, since t he columns of A are linearly independent, we have Ax = projcoi(A) b = 0 if and only if x = 0. P 3. The least squares solutions of A x = b are t he solutions of the normal system AT Ax = AT b . From Theorem 3 5 1, the solution space of the latter is the translated subspace X.+ W where :X is any least squares solution and W = null(AT A) = nuii(A).

P 4. If w is in W and w :1: projw b then, as in the proof of Theorem 7.8.1, we have l b lib - projw b ll ; thus projwb is the only best approximation to b from W . P5.

If ao, a 1, a2, ... , am are scalars such t hat aocdM) then ao

+ n1:ri +

+ a 1c2(M)

t a2c3(M )

+ · · · + amxi

wll >

+ .. + amCm+l(M)

= 0,

=0

for each t = 1, 2, ... , n Thus each x, is a root of the polynomial P (x) = a 0 + a 1 x + · · + amxm. But such a polynomial (if not identically zero) can have at most m distinct roots. Thus, if n > m and if at. least. m + 1 of the numbt>rs x 1 , x 2 , . • , Xn are distinct, then ao = a 1 = a2 = · · = am = 0. This shows that the column vectors of M are linearly independent P6.

If at least m + I of the numbers :r1 , :r2 , ...• :rn are distinct then, fr om Exercise P5, the column vect ors of M are linearly independent, thus M has full column rank and M T M is invertible.

EXERCISE SET 7.9 1.

(a) v, · v2 = (2)(3) -t (3)(2) = 12-:/; 0, thus the vectors v 1 , v2 do not form an orthogonal set (b) Vt · v2 = ( 1}(1) t (1}(1) = 0; thus the vectors v 1, v 2 form an orthogonal set. The corresponding Orthonormal SC t IS ql = = = =

u::u

(c)

(-32,

We hnve v1 · vz = V t • v 3 = v2 · v 3 = 0; thus the vectors v 1, v2 , v3 form an orthogonal set. The correspondmgorthonormal set isql = = (7s,O.js), q3 =

(d) Although v1 · v2- Vt · v3 = 0 we have v2 · v3 = (1)(4) vectors v1 , v2, v3 do not form an orthogonal set. 2.

(a)

+ (2)(-3) + (5)(0) = -

2-:/; 0; thus the

.,. · v2 = 0; thus the vectors v 1 , v 2 form an orthogonal set. The corresponding orthonormal set is q l =

?tJ)

(b) v 1 • v2 :1: 0; thus the vectors v 1o v 2 do not form an orthogonal set. (c) V t • v2 -:/; 0, thus the vectors v 1 , v 2 , v 3 do not form an orthogonal set. ( d ) We have v 1 · v 2 = v 1 • v 3 v 2 · v 3 = 0; t hus the vectors v 1 , v 2 , v 3 form an orthogonal seL. The corresponding orthonormal set is q1 = -j, = ! .- i) , q3 = (!,

=

3.

(a) These vectors form an orthonormal set. (b) These vectors do not form an orthogonal set since v2 · v3 (c)

These

form an orthogonal set but not

= - '3; -:/; 0.

orthonormnl set since llv 3ll

= .j3 :1: 1.

247

EXERCISE SET 7.9

4.

(a) Yes.

f. 1 ;,.nd llv zll f. 1. • vz f. 0. vz · vJ f. 0, llv

( b) No; llv1 1l (c)

No;

Vt

6.

(a) projw x = (x · v1)v1 (b) projwx = (x · vt )vl

7.

(n)

l, and

llvJII f.

L

+ (x · v:a)vz = (l )vt + (2)v z = (!. !, 4} + (1, 1, - 1, -1) = + (x · vz)vz + (x · v 3) v3 = (l )vl + (2)v 2 + (O)v3 =

(b) 8.

(a)

(b )

13. Using Formula (6), we have P

=

r·32I

2

1

J

:i

2

-3

3

'2

3 J

14.

Using Formula (6), we h.:we P

=

I

2

?J -76

[

V'6

1 5.

js]- [

72

-

I

7u

76 I

275

2

3

-

I

:.

6

Using t.hf! matrix fouud in Exen;isc 1:3, the or thogona.l projection of w onto W = SJJan{ v 1, v 2 } is

Pw =

r!.. -

s _1 9 9 - 92

On Lhe other hand, using Formula (7) , we have

[ 0] 2

-3

=

[

!§ 9 _L98

Chapter 7

16. Using the matrix found in Exercise 13, the orthogonal projection of w onto W = span{v1 , v 2 } is

On the other hand, u:.mg Formula (7), we have projww = (w · Yt)Vt

!, +

+ (w · v2)v2 =

=

j,

'7s· -76).

(7J, 7J, 7J)

17. We have = and = Thus the standard matrix for the orthogonal projection of R 3 onto W = span{v 1 , v 2 } is given by 1

73 P=

18.

We have 11 :!11 jection of R 3 onto W

1

73

=

1

[

76

73 =

(S,

[! : 0

0

1

Thus the standard matrix for the orthogonal pro-

span{v 1 , v 2} is given by

p

j1] [l

=

1

-il

3 2

3

19. Using the matrix found in Exerctse 17, the orthogonal projection of w onto W =span{ v 1 , v 2} is

Hl [=t]

Pw On t ht> r,t her hand, using Formula (8), we havP

20. Using the matrix found in Exercise 18, the orthogonal projection of w onto W

Pw=

!

!9 ;

-9

[ 2] = [ OJ

On the other hand, using Formula {8), we have

21.

From Exe
[

tt ;

thus tcaoe( P)

j +j +I

2.

= span{v 1 , v 2 }

is

249

EXERCISE SET 7.9

[!

94

'92]

; t hus trace( F )

= .; +

= 2.

22.

From Excrciso !8 we have P ,.

23.

We have pT:;.. I' and i t is easy to check that P 2 = P ; thus Pis the standard matrix. of an orthogonal projection. The dimension of the range of the projection is equal t o tr(P ) = + 251 + = 2.

24.

The dimension of the range is equal to tr(P) =

-

&

ff

19 + ;fg + = 1.

25. We have pT = P and it is ea.'!y to check that P 2 = P ; thus Pis the standard ma.t.rix of an orthogonal projection. T he dimension of t he range of the projection is equal t o tr(P ) = + + = 1.

t

26.

27.

The dimensio n of the range is equal to tr(P)

= + +

= 2.

= wl = (1, - 3) and v 2 = w2 Vt = (2, 2)- ( 1; )(1 , - 3) = (Jj, !> = {:3, I) . Then {v 1, v2} is an orthogona l basis for R 2, and the vectors q 1 = II = ( jfo) and q 2 = = (fro, ftc) forrn an orLet v 1

thonon nal hAS is for R 2 .

28.

Let V J = Wt = (1, 0) and V2 = W 'J Then Q 1 = = (1,0) ami q 2 = for R 2

·-

ft:!n

v 1 = (3, -5) - n)(l,0) = (0, - :>). = {0, - 1} fo rm an orthonormal ba:;is

-s 29. Let

YJ

= w, =(1, 1, 1), voz =

w 2 --

= (-1, 1,0)-

J. , 1) = (--1, 1, 0}, and

form u.n orthon orma l hasls for R 3 .

30. Let. v, ,..

WJ

= (1,

o. 0), Y 2 = w 2 - u;;u! v l =

(3. 7, -2)- (f)(l , O, 0)

= (0 , 7, -2). and

Then {v 1 1 v 2, v3} is an orthogonal basis for R 3 1 and the vectors q,

VJ

= llvdl = (l, O, O),

form an orthonormal basis for

q3

R3 .

VJ

(Q

30

105

= llv3 U= ' ./11925 J uns 1

)

Chapter 7

250

31. Let

Vl

=

Wt

= (0, 2, 1, 0), V2 = w2-

VJ

= (1, -1, 0, 0) -

(

1, 0)

= (1, -i,

and

Then {v 1 , v 2 , v 3 , v 4 } is an orthogonal basis for -

ql -

q 3-

YJ llvdl -VJ

-

11 v3 ll-

{0

2Vs J5 0) '"5• Sl ,

(.{IQ

10 ' 10,-

-

q2 -

and the vectors llv211

v'iO) • s ,--s-

.£@

q4

- ( v'3(i 6

-

I

VJo :.&.Q 0) 30 '

IS

I

'

- (v'iS .ill uTI .iil) 15• 1s , - ts • s

fo rm an orthonormal basis for R4. 32.

Let VJ = WJ

= (1. 2.1. 0). V2 = Wz-

= (1,1, 2, 0)-

2, 1, 0)

=(

0),

and

Then {v 1 , v2, v3 , v 4 } is an orthogonal basis for R 4 , and the vectors

form an orthonormal basis for R4 • 33. The vectors w,

R3.

=(

0), w 2

=(

0), and w 3 = (0, 0, 1) form an "rthonormal basis for

EXERCISE SET 7.9

34.

251

7J,

!) and w2 =

Let A be the 2 x 4 ma.trixhavingthe vectors w 1 =

as

rows.

Then row(A) = s pa.n{w1, w2}, and uull{A) = spa.n{wt, w2}.1. . A basis for null(A) can be found by solving the linPnr system Ax = 0 . The reduced row echelon form of the augmented matrix for this system is

0 _l2

4.!.

:I

l

_!

I

1

by

and oo a general ootut;on ;,

2

4

I

1 ;•]

=

0] 0

[m , [\ - ·[-!]+ •[=;]

the vootorn

w, =

4,

1, 0) and w4 = ( 0, 1) form a basis for span{w1, w2}-'-, and B = {wJ, w2, w3, w 4} is a basis for R 4 . Note also t ha t, in adrlition t o being orthogonal to w1 a.nd W'J, the vectors w 3 and w4 ar<:: orthogonal to each other. Thus B is an orthogonal basis for R4 and application of the Grarn-Schmic.Jt process to t hese vectors amounts to simply n ormnlizing them. This results in the orthon o110td l>asi::; { qt , Q z, Q:$, q 4} where Q1 = Wt (L Q 2 = W z:::: (-7.f, 0), Q3 =

11:; 11 35.

7:3•

= (-js ,- )5, -:76,0), and q4 =

)r8, 7rn,o).

11 :!g = (--dis, -

Note that W 3 = w 1 + Wz . Thus the subspace W spanned by t he given vectors is 2-diinensiooal with basis {w 1, w2} . Let v1 = Wt = (0, 1, 2) and v 2= w 2- w2·V1 llvdl 2 v1 =(-1, 0,l ) - ( 52)(0,1 ,2) ;;- (

2 1)

Then {v 1. ,·l} i::: an Ol lbogonal basis for\\' , anti the vect-or!' Ut

V1 = llvdl

·

l

= tO. -/S'

2 )

,

form an orthonor mal ba.c;is for 36.

NoLe that w.1 = w 1 - W -z t- w 3 • Thus the ::;uhspace \V spanned hy the given vectors { w 1 , W ?, w3}. Let v 1 = w 1 = (-1 , 2, 4, 7) . t\nd let. with v .....

= \1: 2 -

w3 • v 1

= (-3 0 4 -2) ·- (2.)(-l 2 4 ..,, )

Jlv 1 jj 2

llvdl 2 v1- llv

· v2

2

· ;o

' • ' '

jj2

,

(

• • •

9)

v2 = (2, 2, 7, -3)- 70 (- 1, 2,-1,7)9876 3768 5891 = ( 2005 ) 2005 > 2005 I

Then {vt . v2 , VJ} is an orthogonal ha.osis for U2

=

,;!1.,.

Uv211

=

( - 41 - 2 s2 -35 ) JS'iit4 ' •/5614 1 15iii4' J5614 '

v

form an orthonormal basis for W. 37.

that u 1 and

u2

""' (- 1! _J. 11 ' -:,

I

an(

:3-dimensional

2

1 •

31843) ( ( 401

14

41 26 14' 7) 7 '

3032) 200.'5

-

w. and the vectors Uj = n::u = (j.ffl, UJ

=

llvlil =

(

9876 3768 5891 -3032 ) J 155630 105' v'l55630t05 ' ,/155630105 1 J l55630105

projectior. vf w onto the subspace

are orthonormal vectors. Thus the

W spanned by these two vectors is given by WJ

2

= projw w = (w · u t) u 1 + (w · u 2) u 2 =

and the component of w orthogonal to W is Wz = W-Wl =(1,2, 3)

(

0,

+ (2)(0, 1,0) =

2,

252

Chapter 7

38. First we find vD orthonorma.l basis {Q], Q2} for Let = U t = (-1,0 1 11 2) 1 v2 = U 2 -

Vt

w by applying the Gram-Schmidt process to {UtI u 2}·

-!,!},

-1(-1,0,1,2) = (j,l, and let ql = = ( 0, q2 = = ( T hen {qt , Q2} is an orthonormal basis for H' . and so the orthogonal projection of w = (- 1, 2, 6, 0) onto W is given by

--Js, 7s, fs),

=

Jh. Ji2).

and the component of w orthogonal to W is W2

39.

=W -

Wt = ( -1, 2,6, Q) - (

1

=(

1

19 4 1

If w = (a., b, c), then the vector

is an orthonormal for lhe !-dimensional subspace lV spanned by w . Thus, using Formula (6), the sta.ndard matrix for Lhe orthogonal projection of R 3 onto W is

P

= u T u = a2

1-

:2

+c

2

a] Ia [c b

[aab

ab2

a.c

c

2

b c]

.

= a 2 + + c2

b b

acl be .'2

c-


=

=

If a and b" are nonzero, then u 1 (1, 0, a) and u2 (0, 1, b) form a basis for the plane z = ax + by, ami applKHion of tlw Gram-Schm1dt process to these vectors yields '\n orthonormal basis {Q 1, Q2}

where

02. (a) span{vt} =span{ wl}, span{ v 1, v2} = span{w 1, w 2} 1 and spa.n{vJ , v2 , v 3} =span{ Wt, w2 , w 3 (b) v 3 IS orthogonal to span{ wt, w2}.

D3. If the vectors Wt, w2 , . . , are linearly dependent, t hen at least one of the vectors in the list is a linear combination of the previous ones. If w 1 is a linear combination of w 1 , w 2 , ••. , w 1 _ 1 then , when applying the Gram-Schmidt process at the jth step, the vector v i will be 0 . D4. If A has orthonormal columns, then AAT is the standard matrix for the orthogotia.l projection onto the column space of A. D5.

col(M) = col(P ) (b) Find an orthonormal basis for col(P) and use these vectors as the columns of the matrix M . (c) No. Any orthonormal basis for col(P ) can be used to form the columns of M . (a)

253

EX ERCISE SET 7.10

D 6.

(a) True. A ny orthonormal s et of vectors is linearly iudeptmdent. (b ) Pa.lse. An orthogona:l set may contain 0 . However, it is t r ue t hat any orthogonal 'set of no nzero vect.ors is linea.rly independent. (c) False. Strictly speaking, the subspace {0} has no basis, hence no o r thonormal basis. However , it true tha t any nonzero subspace has an orthonormal basiS. (d ) 1'ruc. 'The vect.or q 3 is ort hogonal to the subspace span{w 1, w 2}.

WORKING WITH PROOFS Pl. If {v1, v2, .. . , is an orthogonal basis for W, then {vt/llvl ll , v2/llv2ll. . .. , is a.n orthonor mal basis. Thus , using part (a), the orthogonal projection of a vector x on W can be expressed as 0

proJwX

=

(

x.

Yt

)

V1

llvtll +

(

x.

V2

l!v2ll

)

Y2

llv1ll + ... +

(

x.

)

Vk

PZ. lf A. is symmet.ric and idempot ent, t hen A is the s tandard matrix of an o rthogon al projection operator; namely the ort hogonal projectio n of R" onto W = col(A). Thus A= UlJT where U is any 11 x k matrix whose column vectors form an orthonormal basis for tV . P3 . We rnust. prove that v i E span {w 1, w2 ___ , w,} for each J - 1, 2, . . -- The proof is by induction on J Step 1. Since

v 1 = w 1 , we have v 1

E

span{w 1 } ;

t.hllS

t.he stat ement is true for j

=

L

Step 2 (inductio n step) . Suppose the st atemeot is t.r ne fo r integers k which a re less t han or to j, i.e., fork = l , 2, . . . , j . Then

and since v 1 C ,c;pan{ w d , v'/. E span {Wt, w·2 ), . .. a.nd v; (7 s pan{ Wt , w 2 , . .. , W j } , it follows that v J+ l C span{ Wt , w 2, .. . , wi, wj 1.1} Thus if the s tatement. is true for e ach of the integers k = 1, 2, ... , j then it is also true fork= j + 1. 0

T hesf: two s te ps complete the proof by induction .

EXERCISE SET 7.10 1 . T he column vec.t,o rs of the matrix A are w 1 procE'SS to

= g}

and

Wz

=

_ Applicat ion of t he Gram-Schmidt

vect.or yields

We have W1 := { w, · q i)q 1 = J5q l and w 2 = {w2 · q 1 )q l + (w2 · Q2)Q2 = -/5q l plio.:ation of Form;.:!a (3) yields the following QR·decomposition of A : A===

fll 2

-i] 3

=

v'5

V5

[J5 0

V5j = QR v'5j

+ ../5q 2.

Thus ap-

254

Chapter 7

2. Application of the Gram-Schmidt process to the column vectors

We have Wt

= J2q 1 and

w2

of A yields

w2 = 3J2q 1 + J3q2. This yields the following QR-decomposition of A:

[l [ 1 2

A= 0 1

1 1]

72 -"J!

=

1 4

3.

and

Wt

l

3v'2

J2

13 [ 0

0

1

J3

1

72

73

= QR

l

Application of the Gram-Schmidt process to the column vectors w 1 and

w2

of A yields

aJ26

Q2

We have WJ = (wl · Qt)CJ1 - 3q, and w2 the following QR-decomposilion of A:

A=

[

11]

=[

s0t 8

JJ26

= (w2 · q, )q l + (wz · Q2)Q23726

=

8

-:

3726

l

3

+ " f q z_

This yields

.l

[o

=QR

1\ppliratton nf thP Grn111-Schrnidt process to the column \'ectors w 1 , wz , w3 of A vields

\Ve have w 1 = J2q 1 , w 2 = "lq1 + ing QR-dec01npositinn l.lf A:

A

1 0 2] 011= [1 2 Q

vl1q2, and

0

73'

72

73'

I

l

w3

1q2 +

= .J2q 1 7s

--Ja

¥

Q3·

This yields the follow-

v'2

[v'20

v'3

0

0

5. Application of the Gram-Schmidt process to the column vectors

-

il =QR J2] 3

h1 3

w 11 w 2 , w3

of A yields

Q2 =

719 We have Wt = J2q 1 , w2 = following QR-decomposition of A :

A=

2 1] I

3

1 = l

+

and wa = J2q l

+ '(P q2 + 3

1

738 72 0

1

-73'8 6

73'8

J2] =QR 0

.i!! 19

This yields the

255

EXERCISE SET 7.10

6.

Application of the Gram-Schmidt process to the column vectors w 1, w2, w 3 of 11 yields

We have w 1 = 2ql, ·decomposition of A:

w2

=-

q1

A=

+qz,

and

w3

r :L [ ; oJ

l

- 1 1

7.

From

3, we have .4

-2

= [- 1 2 1 1] =

[-3_:

2 J

+

=

1 2

:Jv"itl

3 2

7

W'16

3

+ ./i-G3·

0] [o2

-1

J]

=QR

1 0

0

1

72

h

f3

I

..r·t u

LO

This yields the following QR-

J=

Q R T hus the nnrmnl

for

3

Ax = b can be expressed as Rx = Qrb, which is:

,J,!l m Lil =

Solving thi.s system hy back substitution yields the least squares solution Xz =

8. From Ex.erci$c 1, we have A =

102]- [-72 [ -jz 0 1 1 -

0

120

-

x1

[v'2o v"2 v'2] ·""' QR. v'3 - :/f.

,16

73 -)6

0

Thus the normal

2..f6

()

:r

system for Ax = b can be e.xprcssed as Rx = Q1'b, which is:

,..

=-

0

v2

1

0

W3

[::] X3

J3

=

2

v·'G

v6

=

Solving this system by back substit ution yields X3 = Xz the syst.etn Ax = b is consistent and this is its exact solution.

9.

F'rom Excrci11e 5, we have A = system for Ax

= b can

12 1] [0 31 =

-·

1 1 1

0

'.;?S8

kl [Vi

o

719

be expressed as Rx = QTb , which is:

w2 .ill 2 0

x,

0

= 0.

41 0

Note that, in this example,

J2]

3'(P

=

QR. Thus the normal

.ill 19

1

v'2 l - v'38 3

'J19

Solving this system by back substitution yields X3 = 16, x2 = -5, x , = -8. Note that, in this example, the system Ax = b is consistent and is its exact solution.

Chapter 7

256

!:] -H [:

10. Prom Exercise 6, we have A=

-1 1 0

I

- I

I

l

2 72

O

ll

= QR. Thus the normal system

for .-\x - b can be expressed as Rx = Qrb, which is

i][=:] [t =

0

0

1

0

X3

72

Solving this system by back substitution yields x 3 = -2, x 2

= 121 , x 1 =

11. The plane 2x - y + 3z = 0 corresponds to a.l. where a= (2, -1, 3). Thus, writing a as a column vector, the standard matrix for the reflection of R3 about the plane is H

=1-

2 T - -aa ara

=

= [ 0

0

6

1

-3

9

-7

and the reflectiOn of I he vee lor b = (1, 2, 2) about that plane is given, in column form, by Hb=

!2.

lhevlane.x 1-y- lz Ocorrespondstoa 1 wherea=(l,l,-4). Thus,writingaMacolumnvector, t.he standard matnx for the reflection of R 3 about the plane tS 18

[ 1 -4] [ I -4

- -l

- 4 16

:;

:1

.:

_l

9 9 o.nd the retlec:ti•>n of the vector b = (1,0,1) about tltat plane is given, in column form, by

H-:

Hb

13.

H = l - -

2

aT a

T

aa -

2 1 14. H =1- - -aa aT a

I

0

Fi

= 1- -aT2 a aaT

1 0 0 0 1 OJ 0 0

=

-1

2 [- :

6

2 11

r 0

I

3 2

3 I

- 1 1

3

2 -2

0 0 15.

3

1

0

u -!l u 2

[-: -:]

1

0

=

Ul

- 1

0

9

4J 0

=

2

3

0

1 - 1 - 1

2

3

0

9

0 2

IT

IT

2

9

1

TI

3 -3

_..§. II

IT

i

II

- ·0] II 6

IT

-n7

EXERCISE SET7.10

257

0 1

2 16. H- l - -aar = aT a

17. (a)

0 0

0

0]

0 0

2

o

10

1

-2

[-21

6 9 3

4 15

4

-3

0 1

T5

6 2

-1

=

']

5

_1

15 4. -15

-ii

--s

-g2

_.i_

:.!

13

15 7

H' 4

2

5 2

2

4

13

-3

5 1

15 -5

15

15

Let a=v-w=-=(3,,1)-(5,0)=(-2,4), Then His the Householder matrix for the reflection about a..l, and Hv = w.

(b) Let a=v-w=(3,4)-(0,5)=(3,-1),

4

•

5

Then H is the Householder matrix for the reflection about a..l, and Hv = w. (c) Let a= v.- w = (3,4)- (¥,- {/) = ( 6 S+l'l). Then the appropriate Householder matrix is:

-;12,

2

T

H = I - --aa aT a

18.

(a)

= [10

0] - ----;:::: 2 1

Let a= v- w,., (1, 1)- (v'2,0)

17-25/2] 2 33+8../2

&0- l7v'2

= (1- v'2, 1).

-2-

a1'a

[(1-

v'2) 1-/2

=

2

(b) Let

v- w

=

2

0

=

1

2

1- v'2] I

2

2

w = ( t, 1)

- (

[ot o]l _[¥v'2

Y1-) ' 1+2v'3) =

(1- /2) 2

4-2J2 1-

2- v'2 -2-

-2

=v-

-1 ../2

[1 0] . . __._2_ [ 1v2 1- v'2]

aTa

Let a

l'n

(1, 1)- {0, /2) = (l, 1- .J2). Then the appropriate Householder matrix is:

H = 1 _ _ aaT =

( <:}

r .1__..

Then the appropriate Householder matrix is:

[1 OJ _ _ _ 2 _ 0 1 ·1-2v'2

H = J .....

=

e-2-n. l-;/3).

=

J:i 2

Then the appropriate

mat.rix is: ( 3-/'3 )( 1-2.,13 )]

e-.i13 ) 2

o]1 _

= (1 0

19. L;;t w

= (llvll, 0, 0) = (3, 0, 0), and a= v- w 2 T lJ =I- --aa aT a

=

-3±2/3] 2

.fl

3-2:/3

2

2

= ( -5, 1, 2).

Then

1 0 0] 2[ 25 -5 -10] [[ 0

1 0 - -

0

0

1

30

-5

1

2 =

-10

2

4

1

3 14

3

15

3

-15

2

is the .standard matrix for the Householder reflection of R 3 about a..l, and H v

-l]

= w.

11

I5

holrlcr

258

Chapter 7

20. Let w

= (llvll, 0, 0) = (3, 0, 0), a= v 2 aar aT a

H =1 - -

w

= (-2, -2, 2).

lq :

0

=

1 0

Then

1

-j

-4 =

4

-4

-4

'2

-4]

4

t

H !l l

j 4

3

is the standard matrix for the Householder reflection of R 3 about a.l, and Hv

= w.

= (1, -1), w = (llvll, 0) = (J2, 0), and a= v- w = (1- -/2, - 1). Then the Householder reflection abouL a..L maps v into w. The standard matrix for this reflection is

21. Let v

H

=1 -

0]

=

[3 -+2J2J2 l'r 4] 2

[_

1

3

0

=R

-

_fl

2

l 2] = [.;2

:D

-1

= [ .;;

=

=[

-1 + J2] 1 J

[ol o] _ 2 + J2 1

We have HA

2 [(1 - J2)2 -1 + v'2

_2_aaT = [1 aTa 0 1 - 4- 2-/2

Y{l

_fl

2

2

and, setting Q

= H - 1 = HT = H,

this

y1elds the following QR-decompositioo of the matrix A:

A=[ - 1 2]=[_fl:f1 3 2

22.

-X}] [h - ]-{/-

-Q

0

R

Let v = (2, 1), w = (llvll, 0) = ( v's, 0), and a = v - w = (2- v's, 1). Then the Ho\lseholder reflection about a.l. nnps v into w The standard matrix for this reflection JS

OJ _ I =

Now let Q,

=

iJ]

= [:

'i

2

10- 4/S

[(2 - ''5) 2 2- v'5l 2 - ,f5

l

OJ 1

5

5

=

[¥

1

s

J :;

and multiply the given matnx on the left by Q 1 This ytelds

_M 5

_w o s

o

1

Js]=R o

and, settmg Q = QT = Q,, we have the following QR-decomposition of the matrix A:

0] [10

y]

-¥

0

=QR

259

EXERCISE SET 7.10

23.

Referring to the construction in Exercise 21, the second entry in the first column of A can be zeroed out by multiplying by the orthogonal matrix Q1 as indicated below: ·

5

-../2]

2¥'2

1]3 = [v-20

0 - 2J2

0

4

5

Although the matrix on the right is not upper triangular, it can be made so by interchanging the 2nd and 3rd rows, a nd this can be achieved by interchanging the corresponding rows of Q 1 • This yields .

QzA

=

fl 2

-f.

0

0

[ _fl 2

and finally, setting Q ==

A=

Q:; 1 "" Qf,

0

4

0 ·0

=

·1

5

0

we obtain the following QR-decomposition of A:

l [1

1 2 -1 -2 [ 0

24.

0

2

-../2] =R

=

v;l [V2

_ 1.1'2 2

1

0

2/2

0

4

0

0

= QR - 2..;'2

Referring to the construction in Exercise 18 (a) , the second enlry in the first column of A can be zeroed out by multiplying hy the orthogonal matrix Q 1 a.<> indicated below!

QI

A

.;_; : 0

:If

0 [l1 23 - 21]

= {/ - Yf : 0 0 -o ---o-: -1---6

o

o: o

1

o

- l

o

0

o: -¥:If- ' -L.}-1 3'{2 .... o r= oi cY o : 1 \ ·- ol o , 1 ) /

From a similar construction , the third entry-in t he second colum'n of Q 1 A multiplyiug by t he orthogomli matrix Q2 as indicated below: 0 I __ ·I

Q·>... Q 1 ; 1 ·-

I

I

r.;

'

I ·

--

-

I

y6 3

I

.

't

r.:.

-V[l =

[v-20 ¥

0

3 f2 :!..lL! 2

,/2 -2

-

'

,J2

-

. ....

.

zeroed out by

w2

0

.i§

0 0

0 0

2

_ill 2

_ _.& 2

"' v'3 1

From a Uurd such construction, the fourt-h entry in the third column of Q2Q1 A can be zeroed out by multiplying by the orthogonal matrix Q3 as indicated below:

- :£1 2

sV2 -2-

v'6 2

2

- ./3

0 0

Finally, setting Q

=

2

0

:&2

0 0

0 0

_ill 2 _:L§ 2

=

R

2 0

= Q[QfQT = Q1Q2Q3, we obtain the following QR-dccomposition of A: :£1 2

A=

\12

::11 2 0

0

.i§

1

-

:il 6

6

2

6

-2

1

:il 6

J

-21

:il 6 fl

0

I

2

2

__ill 0

260

25.

Chapter 7

Since A = QR, the system Ax = b is equivalent to the upper triangular system Rx

[

v'30 - .../2 v'3 0

0

-33]0 4

7s

[:ttl :t2 = X3

= QTb whlcb is:

[730 .:i§ 3

Solving this system by back substitution yields :t3 = 1, :t2 = 1, :t 1 = 1. 26.

(a)

= a (aTx) = (aTx)a, we have Hx =(I- af.aaT)x = /x - af.aaaTx =X- e:1T:)a. Using the formula in pa.rt (a), we have Hx = x- ( 2$:)a = (3,4, 1}- 1: (1, 1, 1) = ( -1, -Jj).

Since aaTx

(b)

On the other ha.nd , we have H = I - ;faaaT

=

j 0 0 l

"" 1 1 1

[-1 -:

a _13

and so _

l3

DISCUSSION AND DISCOVERY Dl. The standard matrix for the reflection of R 3 about

ef is (as should be expected) 2

0

and

0 1

[-

0 0 0

0

0

1

for t.he others.

D2. The standard matnx for the reflection of R 2 about the line y = mx is (taking a = (1, m)) given

by

D3.

D4. D5.

= ±v's3, then llwll = ll vll and tbe Householder reflect1on about. (v- w ).l mnps v into w Since llwll = llvll , the Householder reflection about (v - w).l. maps v into w , We have v - w = (-8, 12), and so (v - w ).l. is the line -8x + 12y = 0, or y = Let a = v - w = ( 1, 2, 2)- (0, 0, 3) = (1, 2, -1). Then the reflection of R 3 about a.l. maps v into w , and the plane a .l corresponds to x + 2y- z = 0 or z = x + 2y. If s

WORKING WITH PROOFS

P2. To show that H = I - af;aaT is orthogonal we must show that HT = H H H T = ( I - -2-aaT) aTa

(I - - 2

aTa

1•

This follows from

' T = I - - 2 aaT - -2-aaT + - -4- a aTaaT aaTJ aTa aTa (aTa )2 2 =I- .2.. aaT - - - aaT a Ta aTa

where we ha,·e used the fact that aaT aaT =a(aT a )a T = (aTa)naT.

4 + -aTa - aaT =I

EXERCISE SET 7.11

261

P 3. One ofthe features of the Gram-Schmidt process is that span{ q 1 , q 2, . for ea.ch j == 1, 2, . .. , k. in the expansion

. .,

q,}

= span {w 1 , w 2 , , . . , Wj}

we must have w 3 • CV -::j:. 0, for otherwise wj would be in span {Q1 , Q2, ... , ClJ -1 } = span{w1 , w2 , ... , w;- l } which would mean that { w 1 , w2 , . . . , w 1} is a linearly dependent set. P4. If A = QR is a QR-decomposition of A, then Q = AR- 1 . From this it follows that the columns of Q belong to the column space of A. In particular, if R- 1 = js,j], then from Q = AR- 1 it follows that c; (Q) = Acj(R- 1 ) = StjCI(A) + s2jc2(A) + · · · + for each j = 1, 2, ... , k. Finally, since dim(col(A)) = k and t.he vectors Ct(Q), c2 (Q), ... ,ck(Q) are linearly independent, it follows that they form a basis for col(A).

EXERCISE SET 7.11 1.

(a)

(b) T h e vect or equation c 1 v 1

(a)

(w )B

=

1 c2 = 8,

= (-2, 5)

=

3 14 .

= {fa, f4)

Thus (w ) B

a nd

lwl o =

(

(b) (w )n =(I , 1}

3. The vector equation c 1 v 1

+ c3 v 3

+

=w

tem by back subi:ititut ion y ir.ldsc.3 = l,c2

4.

and [w]B

+ c2 v 2 = w is equiva.1ent to t 11e Iinear system [ 2 a3) [cc2'] -- [11)' and - 4

the 1:.olut.ion of this system is c 1

2.

= (3, -7)

We have w '"" 3v 1 ... 7v2; thus (w)B

The vector equation c 1v 1

r.2 v 2

is equivalent to

= -2,c 1 = 3.

+ c3 v 3 =

Solving this sys-

=

o

0 3

- :. _; ] 3

3

= (3 ,- 2, 1) and

Thus (w)a

w is equiva.lent to

C3

[win=

=

69C:J

[

Solving this 3

system by row reduction yif!k!s c 1 =' -2, c2 = 0, c3 = 1. Thus (w)B = ( - 2, 0, 1). 5. If (u)n = (7, - 2, 1), then u 6. If (u )B

7.

=

(8, - 5, <1 ), then u = 8v 1

(w }B = (w · v 1 , w · v 2)

8. (w)a = (w · v 1 , w 9.

(w)£J

= 7v 1 -

=

· v 'l , w · v 3 ):::

-

2v 2

+

5v 2

+ 4v3 = 8(1 , 2, 3)- 5{ -1, :>, 6) + 4(7, - 8, 9) =(56, -41 , 30) .

= 7(1, 0, 0) ·- 2(2, 2, 0)

=

..

(0, -2, 1)

=(W • Vt W·v 2 W·v 3 )=(....Q... _..2_ ' ' -/2' ,/3' v6

2 '

_hQ 3

y])

• 6

+ (3, 3, 3) = (6, - 1, 3).

Chapter 7

262

11.

1;). = Using Theoretn 7.11.2: Uu ll = ll(u )all = y'(1) 2 + (1) 2 = /2, llvll = ll(v)sll = J(-1}2 + (4)2 = m. and u. y = (u}s (v)a = (1)(-1) + (1)(4) = 3. Computing directly· llull = 4= = /2, llvll = Je;p + e;p = Jl7. 6 and u · v = (t)( + (-!)( ; ) = = 3.

(a)

We have u

( b)

= v1 +

=

v = - v1 + 4v2

0

12.

!iii=

/¥s

-i,

(a) We have u = -2v1 + v2 + 2v3 = (b) ll u ll = ll(u )sll = j( -2)2 + (1)2 + (2)2

v

= 3vl + Ov2- 2v3 =

.!p, 1).

J(3) 2 + (0) 2 + ( -2)2 =

= 3, llv ll = ll(v )sll = = (u )s · (v )s = {-2}(3) + (1)(0) + (2)(-2) = - 10. ll u ll = + (-D2 + = fj = 3, llv ll = 1 and u · v = -!) + (-i)( 3°) + ( =- 99° = -10. u·v

13.

!lull = ll(u)Dil = J( - 1) 2

1- (2) 2

J¥

=

Jf3, and Jf3,

=

+ (1)2 + (3)2 = Jl5

= li(v)sll = }(0) 2 t (-3)2 + (1)2 + (5) 2 = J35 Uwll - ll( w )sll = ,fi2)2 + (-4)2 + (3)2 + (1)2 = v'30 2 -+----:<"'="'s uv + w 11 = 11 s + <w >s 11 = 11 <- 2, -1, 4, 6) 11 = .;,..,....<-----=2=>2----+--:-< llv- wll = li(v)a- (w )all = 11(2, 1, -2, 4)11 = y'(2)2 + (1)2 + (-2)2 + (4)2 = v · w = (v)a · (w)a = (0)(-2) + (-3)(-4) + (1)(3) + (5)(1) = 20 llvll

=

>""":.!

14.

!lull= ll(u) 8 11 = j(0)2 +

(0)2 + (-1)2 + (-1)2

Hvll

v ·

..l..

w = (viB · (w )a- (5)(3) + (5)(0)

5

= /2 =,/58

= ll(v)o 1- j(5) 2 + (5)2 {-2)2 + (-2)2 llwll = ll(w)all = .j(3)2 + {0) 2 + (-3) 2 + (0)2 = /18 = 3v'2 llv + w U = ll(v)a + (w)all = 11(8,5, -5, -2)11 = llv - w U= il(v)a- (w)all = 11(2, 5, 1, -2)11 = y'(2)'l + (5)2 + (1)2 + 1

.Jf05

+ (-2)(-3) 4- (-2)(0)

= JIT8

-+-;-(

(-2)2 = v'34

= 21

15. Let B - {e 1 ,e2} be the standard basis for R2 , and let B'- {v 1 , v 2 } be the basis corresponding to

the .r'y'-systcm that is described in Ftgure Ex-15. Then Ps· ..... s = [[vi]a Ps-.a·

(PB'-+D)- 1

the equations x' (a) (c) 16.

=x

= - y and y'

= -/2y.

= (1, L), then (x', y') = (0, /2). [f (x,y) = (0,1), thPn (x',y') = (-1, /2).

as described in Figure Ex-16. Then P9•- s -

----=

(b ) If (x,y) (d ) If (x, y)

and so

-

(*f

10 ]

1

= (1,0), then (x',y') = (1,0). = (a, b), then (x', y') = (a- b, v'2b)

= {u 1 , u 2}

= (v'J, 1), then (x',y') = (2,0}. = (1.0), then (x',y') = = (0, L). then (x', y') = (0, I )

be the basis for the x'y'-system

and Pa-s = (Ps .... 8

x'y'-coordinates are related to xy-coordinates by tlre equations x' particu Jar.

If (x,y) lf (x,y)

0 V'1

In particular:

Let B = {i,j} be the standard basis for R 2 , and let B'

(c) (d)

1

=[

It follows that x'y'-coordinates are related to xy-coordinates by

If (x, y)

(a) If (x,y) ( b ) 1f (x,y)

[v2Jsl

=

)

1

=[

and y' =

-73

0 ].

1

Thuc;

+ y. In

263

EXERCISE SET 7.11

(v2)s) (b) The row reduced echelon form of [B 4 3 ] IT Ti [

(c)

l

2

-n H

I

4

rr

li _J.. I

I I

1];

3 - ). -1

thus Ps-+B ::::

ll

.

-J]

= [2

Note that

4

1

We have w

=3e

4 3

.!.. · li [ -1 'J.] -- PS-tB·

-I -

-

(d) We have w = v 1 - v z; thus [w)s

(e)

0)1 IS. [10

:2 - ::J I : [1 4 0

I S'J =

OI

= (2l

=

and

lwls = Ps-ts[w}s =

6e2; thus [w]s

1 -

G-!]

=

=Ps ,s[w]s =

[w]D

[_!]

=

;]

18. (a)

0 8

0

(b) The malrix [B I SJ

01

·40

(J I

13 5

I

:'!:

I

1 0 B10 0 1

l, i

H)

9] as its reduced row echelon

-·.5 ·-·.1 -2 ·-1

Ps_.n .,.,

form;

5 -2 -1

(c)

0.t 0o]: t.hus ( Po -.s)-

lt is easy to chec.k tho.t I 0 8

l 5] I

(d)

-2

-1

[1

0

()

0

0

w = 5et- 3e,

+ e3;

I

thus [wls =

1

[-3l

0

0

I

2.'1!J1

o :.

a - 3 has o 8

(e)

5

I

and lw] s

as its reduceu row echelon form; thus

)6 -391 [- 35] = Ps ·-tH[w)s = [-- 40 13 -5 5 -2 -1

19. (a) The row reduced echelon form of !Bt I B2] -=

(22

4_1

- 1

I I

[11

l

-1] . [)0

3 -l

!S

10

-g2

(b) The row reduced echelon form of [B2I [ 0

_§)

-2 - ;; .

{c)

Note that

=

[

13

·r:

Bd '""

(-5)

= Ps-+B·

1

77 30

I

1

-11 2

-l It 2

-1

i:.

ro Lt) = Ps,-.82 · 10

1

=

lr -239] 77 . 30

264

Chapter 7

= -;i0u 1 +

thus [w]B,

=

[-11

(d )

w

(e)

We ha,·e w - 4v 1 -7v 2i thus (w]s,

21.

Note that (Pe,-.e.)- 1 =

(d )

We have w - 2u1 - u 2; thus [w]B,

(e)

Wf! have

(a)

w

:3v 1

v 2; thus

-

[

01 - 21 - 5]3

1 :•1 2] IS . (1 0 4 2 3

I 1

I

thus Pa,-+B, =

;

= Pe, .... s, .

[_n =

·= Pe, ... a,(w]s, =

(w)e,

[w]B 1 = Pa, ... e ,[wJo,

[w]a 2

The row redured echelon form of [B2l Bt] thus Ps, -BJ =

= [3'

=

(c)

=Pa,.... a, (w)s, =

(w)a,

(b) The row reduced echelon form of (B2I Bt]

-11 [-11 = [=:]. = [-11·

=

and (w]B, =

=

_

-6 -2 -2 -3 -3 1] [Io =[ I

-6 -6 -3: 0

0

6 is

2

7 I -3 - 1 -1

4

0

0

=

0

I I

3

4

3

4

17

1

0

0

q

I

4

0

-u '2

.)

l

i2 -.!1 1 . 1'2 2

3

-H -H

u

(b )

lfw=(-5,8

5).wehave(w )a 1 =(l, l , l )andso 3

4

-1Jm,HJ

17

-12 2

3

(c)

The row reduced e<'helon form of IB2 I w] =

(ti,- g, 22.

(a )

[=:

0

4

[

0 0

0 1 0

(b ) Ifw =

71-5

agrees with the computation in part (b).

The row reduced echelon form of the matdx (B2I B t] I

;

is

0: 3 2 0:-2 - 3 11.'1 I

-4 ; thus Pa,_.a, = 6

[ 3 2 -2 -3 51

=[

-4 . 6

-5), we have (w )a 1 = (9, -9,-5) and so

-S

0

II

!..2]

1

4

o! -H I

I

I -1

1 I

2

2

I

0

I

I

-1

-3

2

1 I

I

l

; tHus (w )s,""

I

265

EXERCISE SET 7.1 1

1

3

(c) The row reduced echeloq form of [B2 I w) = [

(-t, 23.

·-5-3 2 ,},

21 - 5

I

'2

l

0o:•-X ] ¥ ; t hus

0

l :

()

0

!-:] is [o

-

0

=

6), which agrees with the computation in part (b).

The vector equation c 1u 1 + c2u2

+ C3 ll3 =

v1 is equivalent to

[* -t [:]

=

'7J

?s•

solution of this system is c1 = c2 = have (v2)s1 = (v3)s1 =

Thus (vt) s 1

=

-76

= (,h,

the · tTl

Similarly, we is 2

l). Thus the transition matrix PB -4B,

['

1

3

76

Ps1-+B, =

O

C3

_

1

2\/'J

'l

-372

21

0 4.

3¥"2

H is <'MY to check th
2

0

:l,i2

-

0 1 0 0

.!. 6

Thus PFJ 1 4 Bt is an orthogonal matrix, a.nd !lince true of Po, -·•B, .

[l 0 0] 0

= (Pn?-•B,rr ,

=

1

the same is

24.

25.

(a)

We have Vt = (0, 1) = Oe 1 + le2 and vz

( b) 1f P

= Ps-.s =

= (l, 0) = le\

t Oe2 ; thus

then, since P is orthogonal, we have

pT

Geometrically, this corresponds to the fact that reflection about y is an orthogonal transformation. (a)

We have Vt = (cos20,sin28) and Vz

Po .... s

= p- 1

=x

= [o1

1} . 0

= (Ps ..... s)- 1 = Ps-+D ·

preserves length and thus

= (sin20, -cos2B); thus Pa..... s =

sin28] - cos28 ·

(b) II P = PB .... s then, since P is orthogonal, we have pr = p - l = (PB_.s)- 1 = Ps ..... b · Geomctthis corresponds to the fact that reflection about the line preserves length a.nd thus is an orthogonal transformation .

266

27 .

28.

Chapter 7

If

(a)

(x) = (-2], lhen 6 y

sin(34")] -sm( 3; ) cos( 3; } Y

-72

3 : >] [:r:] = r- 72 [x:)= [5]' then [;r] = [cose4") -sin( sm( ft) cos(

(b)

If

(a)

If [:]

(b) If

(:r] =

=[

11

ll

2

3 4

ll

z

r::J =

then

[:J

=

3 ") 4

ll

rJ [:]-= rJ -rJ -rJ =

=

=

* nl l

nl,

29. (a) If [:] = then [::] =[-::(;) (b) rr x'] [ 1] lhen [x] ;: [ =

30

Ul·

(b) If[::]= 31.

We have [::]

=

'

{i]·

(a)

=

=[

1

-

=

[

0

z'

0

1

-3

[-t]. -3

[t :-:][:] -rl nl u:!l [: ] =u: !l Ul = =

r;] m

and

=

0

=

=

:

[! :-!] [: ]

[! r -;)[-t r m

Thus

-4 .:.::] •I


- .J_ [

7

R

l!] IS

2.! •

Let 8 = {v,,v2 , v 3}, where v 1 = (1, 1, 0) , v 2 = (1,0,2),v3 = (0,2, 1) correspond to the column veclors of the matrix P . Then, from Theorem 1.11.8, Pis the transition matrix from 8 to the standard basis S = {e 1 ,e2,e3}. (b) rr pis t he transition matrix from = {e ,,e2,e3} to B = {wl . w 2. W J}, e, = Wt + w2 . e2 = w, + 2w3, a.nd e3 = 2w 2 + W J. Solving these vector equations for w 1 , w 2 , and W J in + = w 2 = %e1 - !e2 + of e1, e2, and e3 results in w 1 = = w 3 = -je, + j e2 + = Note that w w 2, W3 correspond to the column vectors of the matrix p- .

02. (a)

s

i·

\e3

!>·

1,

WORKING WITH PROOFS

f v 1, v, , v 3}

D3 . If 8 v1

==-

267

(1, 1, 1 ) ,

D 4. If lwls B is

=w

V ·l

and if

l
( 4, 3, 0) , and v :l

= 2(1, 1, 0) + ( 1, 0, 0) =

(3, 2, 0).

holds for every w , t hen the transition mat rD.. from the standard basis S to the basis

and soB= S =

: ] is

= 3(1, 1, 0) + (1, 0, 0) =

Ps .... B

D 5. If !x- Yls

[

= ([eds I (ez]s l le3]sl = rel l ez les} =In

= {eb e2, ... ,en}· 0, then {x ]s = IY]B and sox= y .

WORKING WITH PROOFS P l. If c1, Cz, . . . , Ck arc scalars, then (c1v 1 + czv2 + · · · + = c1 (v t ) s + cz(vz )s + · · · + Note also that ( v) 8 = 0 if and only if v = 0. It follows that c1 v 1 + Cz v2 + · · · + ck v 1: = 0 if a nd on ly if c1(v 1)a I· cz(vz)s + · ·· + ck(vk) B = 0. Thus the vectors Y1, v 2, . . , vk arc linearly indc· pendent if and only if (v 1) a, (v;!) B, ... (vk).V a re linearly independe nt. 1

1

P 2. T he vec tors VJ, Vz , . .. ' VI( s pan Rn if and only if every vector v in nn can be expressed as a linear combination of them, i.e., there exist scalars c1 c2, .. . , Ck SllCh t.hat v = r.1 v 1 + c2v 2 + · · · + Since (v)n = c1(v1) B + c2(v2)B + · · · + ck(vk)B and the coordinate mapping v -+ (v)a is onto, it follows tha t the vectors v l, v2,' . . 'Vk span Rn jf and only if (vt)e. (v 2) B,· .. ' (vk)B span 1

nn.

P :l. Since the coorciinate map x -+ !xJn is onto, we have A{xJs = C!xle for every x in R" if and only if Ay =:.: Cy for every y in R 11 • Thus, using Theorem 3.4 .4, we can conclude that. A = C if and only if .4[x ]B -"' C[xJ a for every x in R 11 • P4.. Suppose B ,... {ut , u 2, . . . , u 71 } is a basis for [C' . T hen if v il1 l.I J + l rel="nofollow">z u 2 + · · ·-'- b, u,.. , we have:

V

= a1 u1 + a2 U·l + · · · + a

71

Un

anJ w =

+ a2 u2 + · · ' + O.n Un) -1- (b1U1 + b2U2 + ' ' · + bnUn) =(eLl+ bt) Ut + (a2 + b2)u 2 +··· +(a, + bn )Un ·

+ W=

(a1111

Thus (c:v)s = ca2, .. . , can)= c(a1 , . .. , an.) :::;: c(v)s and (v + w)e =(at + b1, .. . , an + b,.,) (a l,a21. .. , an) + (bt , b2, . .. ,bn) = (v)B + (w)s.

=

CHAPTER 8

Diagonalization EXERCISE SET 8.1

Thus [x]a that

= [.];2

[Tx]a

= [:z\:2:r2 ), and (T]s = I[Tvt]a

(Tx]s

= rxl2l- X2] = [22 l

-ol] [

X2

[Tv z]a] =

we note

X2 X, ] = IT]a[x]a

which is Formula (7).

Thus [x]s

= r i_"':r,' ++

,r,

[Tx]o

=

!% 11

:;:rt-

ITis

and

= IITvda

(Tv2JaJ =

I [

-

".

-"5

Finally, we note that (TxJa

\\"l11ch is Formula (7) 3.

Let P

= Ps-•B where S = {e 1,e2} is the standard basis. Then P = [[vds (v2JsJ = P[TJ 8 p -l

4.

Let P

=

For

[I -1] [2 -1] [ 0 1] = [1 -1] _[TJ 1

0

2

0

-1

1

1

1

o where S- { e1o ez} is the standard basis. Then P = llv!]s

PITJaP 5.

=

P\'ery

1

=

2]1

[-t -..,. :>

-

18 5]

_ .1 5

[

1 2]l [l3

5 -2

=

[vz]s] =

and

= [T]

v€:ctor x m R3 , we have

[::]

(jx,- jx, + jx,)

m H·· +

268

+ jx, +

[!]

+ (jx, + !•· -jx,)

m

Exercise Set 8.1

269

ami

Finally, we note that

[TxJH =

-xz - !x3 ]

[- 4x: + - 2Xt

X2

+

-

-

1 2 X3

which is Formula (7). 6.

.For every vector x in R3 ,

X =

= Hx,

have

- fz, + !x,)

nl

+ tlx, +

+ {,x,)

m

+ (-jx, + lx,-!x,)

and

[T] F = [[Tv,]o [7'v,]o [Tv, ]!,] = [ Finally, we note that

-y [ 12 7

-8

which is Formula (7) .

-J -il

nl

270

Chapter 8

7.

= [lvt]s

P =- Ps .... a where 5' is the standard basis. T hen P 3

1

P[TJaP-

=

I

ilH t] [-l

0

-2

-2

1

I

2

[;

1

l

l

-2

2

l

I

1

2

2

.. 'I nrIy, 1'vI1 Stmt

[T]s =

[()]

1

J =IT

[Tis P

=

l

= [-82 156] =[

Si1mlarly.

_

81i

= Pa ... FJ•

61 ]

=

8

=-

1

4

=

Since

112

3

J

an d

r v2 I

= v; + v2

Vj

1

=[

:

= -

31 2

-1

+

2]

22

4S

-M fl]

1] [_..:!..

11

GJ -

and [Tis· =

[

_;j

26

I

TI

1798[ ") = 15 -1 [

=;

=:) = 9]

31

I

2

-2

_86v -15 1

31 2 v;

S1me

[-1

={Tj

n

[0 -2 0

0

1

0

-2

= [TJ

9

Tv2 =

and

= 11 vl

I] = - rrv l2

4

1

V ?.

= v;- v2,

+

19 2 . + nv

Thus

I

we have p

=

ancl P[T is P -

1

= [1o8

IT

II

L798v 2 and T v 2

v2

= [-3]16 -

=[

and Tv2

v! = 10v1 r 8v2 and v 2 = -

- 21(_§§ -¥J !i: 1

-!] [1:1 2

32] =!Tis

= [A2

2

[Tis where, as before, P

+

7 2s

-

11 . The .-qualion P[T]B p - l = [Tis• is equivalent to [T]a have

-

11

3-AJ

[tSh

11

45

i v; + ;sv;. -

i

1

2•

Thus

¥ v2, we hnve

_..!..] _ = [-¥-7i ¥

= p- 1 [Tis•P. .!.2

6l v 90 1

=

[T]s··

T hus, from Exercise 9, we

[]1 1] = p-• [T]s· P -1

= Pe.-s•.

12. The e>quation P[T]sP

· I=

[Tis· is equivalent to [T]s

= P - 1 [Tl"'•P.

have 86

whf•rt•, ns before.

_;]

1

0

II

'2

lo [8

[TJs

=

- 8 and

v 1I

1

12

8

2 + nv 2

I

..1..

;] and

and

P[l]sP

10. We ha\·e Tvl

-:J

_'J.

II

= [: _:)

8

[2] +IT2 r-3] = nv 3

anti [T]s·

TI

I I

Ps ... a•

=

0

j]

_ !

=-II

=

9. We have T v l

3

=

n

3-1] [-y 2

J] H

-1

0

[i

=

= [lv 1]s Jv,]s lv>lsl =

8. Let P - Ps ..... a where S is the standa rd basis. Then P

3

[v2Js [vJ]sJ

=[

61] 90

45

13 -

90

49 - [ ..!. -;rs 45

P = Po-e··

2

[108

Thus, from Exercise

10, we

Exercise Set 8.1

13. The standard matrix is [T}

= [1

. , o 1 matrices are related by the equation

and, from Exercise 9, we have !T}s

5]

-1

P[T] 8 P - 1 = [ 5 -3

14. The standard matrix is [T]

=

ll l!..l 'l 6

_25

TI

11

22 5 22

22

=

= [-

ft

- rr

.These

u

[10 -2]1 = jTJ =[

_

[-1]

+(4z)

and, from Exercise 10, we have {T)s

These

matrices are related by the equation

where P 1 5.

(a)

= [vl I v:J

For ev€r y x

= (:.c,y) in R 2 , we have

a.nd

16.

(b)

Ju agreement wit b l·hrmu Ia. (7), we hn.vr.

{a)

For every x -" (x.y,z) in R 3 , we ha.ve

and \ .. Tx .,;,

[1]

x+ Y + z1j [

I

( v,

-?

/

!

) J •

'··

[0]

Chapter 8

272

(b) ln agreement with Formula. (7), we have

17. Fo r every vector x in R'l, we have x =

[: 1 =

+ %xz)

+ (- {0x1 + -fo:r2)

=

3

+ ix2)v1 + (- 10xl + -foxz)v2

and

Finally, in agreement with Formula. (26), we h ave

18. For every vector x m R2 , we have

anrl

t hus lxls

=

1 -:tt -

4 Xl

; l· 1 X2

1- ;:t:z

jTxJ s'

=

- 2 :r1 1 [

3

z XI

l

+ 3I :r2 +

, and ITJ s•s

'2 J%2

Finally, in agreement with Formula (26), we have

19.

For every vector x in R 3 , we have

= I!Tv1J B'

Exercise Set 8.1

273

and

+ 2X3] 2x x · = [3Xt

Tx =

1

t hUS

[X IB

=

[-

_

4xt + I l 1

( --·x 5 r 1

-

1 2 xz + '2:2:3

zXl l

2 x1

2

-

- 1.. xz 14

[TX JB' =

,

-

r-3] + (6 2

4 ) ;;XJ ,

- x1 1

[-rrt6 - 1

+ 2 x:z + 2x3

4

1

14X2- 'fX3] 3 2 . 7XJ-f4X2+'f:t3

I

+ 12 - ;[J ) [1]

3 -x2 1<1

1

ann [T]B'B

=

r- -! -!4] I

'f

Ill

.

Iii

Finally, in agreement with Formula (26), we have

20. For every vector x in R 3 , we hnvc

and

-· -!] lf

= 21.

(a) [TvdB

rl -

=

lx1 - .2...t2 7 14'

4 l 7Xl- il.X:.!

I·

ZJ

-

14

-

13

Finally,

GJ· 2GJ =

\.ii

and !Tv:da =

Tvl

(c)


VI -

•

l [-

(b)

=

1

2

2v2

=[

Md T vz

=

3vl +5v2

=

+

=

Thus, using the linearity ofT, it follows t hat

-5) + (-

Tx==(·- .-1 xt+ 52 x2)lf .., 0

7] - [ l 1

2 1 ) [ 5 x1+5x2 11

ix1-

or, in t he comma delimited form , T (xJ> x2) = ( 1

(d) Using the f.--:-mt:la obtained in part (c) , we have T(l , l )

Xt 22x

5

fjx1

= (lri' -

1

+ ll5 x 2

+ Jtx2). ¥ + ¥) = ('5u, 353 ) .

274

Chapter 8

(c)


x -

+

4x2-

+

+ (-

11 2 x4)v1

+

+ + !xJ-

- 2.r2- 4x3

+

!xJ- !x4)v3 +

ofT, it follows

I hu:>, using lht

Tx

- 4x2 - !x3

+

+

+

1; x4)Tv1

+

2x2- !x3

!x3- !x.a)Tv3 +

+

+ !x3-

+

which leads to the following formula forT

(d )

23. 1f T

IS

the identity operator then, since T e 1 = e 1 and Te2

[T]a

On the other hand, Tv1

IT]s• =

and Tv 2 = [

= -{,v}- !:v;; thus ITJs•.B =

24. If Tis the identity operator then IT]

= e2,

=

[

1+

2

=

37 [

36

19 [

?] -36 [g] =

36 - :

II

+ H[_:]

Similarly,

+

=

5I

= IT] B = IT] a• = 37 1 Jij vl

we have IT]=

=

On the other hand, we have Tv1 = 0 0 l

3 ]

= (- 31 1 37, 12).

Using Lhe formula obtained in par t (c), we have T(2, 2, 0, 0)

19 + 36v21

1J 1 36 v 3,

T

_

v'2 -

[Q] -

299 1 144vl

209 + mv2. 1

229 I 144v3,

275

DISCUSSION ANO DISCOVERY

and Tv3 -25.

[9] = 4

173

V.1 + .illv' 48 2 -

115 48 v'· 3•

thus [T}a•' D

=

ill]

299

37

3

m

48

119

'109

.IJ! 36

48 m [ '229 ll5 II - 36 - m -48

.

=

Let B {v 1, v2 , ,.. , v ,t) and B' = {u 1 , u 2, ... , u m} be ba.ses for R" a nd R m respectively. Then, if T is the zero transformation , we have

for each i = 1, 2, ... , n. Thus {TJB',B = [0 I 0 I · · · I OJ is the zero matrix.

26. There is a. scalar k > 0 such that T(x) = kx for all x in R". T hus, if B = {v1 , v 2 , •.• , v,} is any basis for Rn, we have T(v1) = kvi for aJl j = 1, 2, . .. , n and so

1 0 0 1

0 0

[T]a

=k

=

0 0

[H!]

27 .

[T]:

29.

We have Tv1 =

0 0

k

28.

[_

= -4vl

=[

[TI

=

and Tv2 =

= [

= 6vz; thus

{Tla

=

From this we see that the effect of the operator 1' is to stre tch the v 1 component of a

[

vector by a fac tor of 4 and reverse its direction, and to stretch the v 2 component by a factor of 6. If the xy-coordinat e axeg are rotated 45 degrees clockwise to produce an x'y'-coordinate system who::;e axes are aligned with the directions of the vectors v 1 ttnd v 2 , then the effe-.t is st.retch by a fact or of 4 in t.he x'-dircction, reflect about the y'-axis, and stret ch by a factor of 6 in the y'-direction.

30.

Wr.

have

T v3 =

ami

0] [- 3; = - v'3v2 -

v3. Thus [1']a

=

[2o -10 -vf:J0] = 2 [10 -!0 - {/0]. From lhis we sec that 0

J3

- 1

0

:,(f

t he effect of the operator 1' is to rotate vectors counterclockwise by an angle of 120 degrees about the v, axis (looking toward the origin from the tip of v!), l-beu stretch by a factor of 2.

DISCUSSION AND DISCOVERY Dl. Since Tv1

= v2

and Tv2 = vl> the matrix ofT with respect to the basis B = {v 1 ,v2 } is [T]B

On t he other hand, since e1

Te2

=

=

=

and e2 = *v2, we have Te1

thus the standard matrix forT is [Tj ==

= iTv 1 =

t].

=

= 2e2 and

276

Chapter 8

D2. The !lppropriate diagram is

c-•

c (T)Bl

03. The approp1inte diagram is

c-•

D4

D 5.

D

I'luc We ha\·e IT1 (x ))e = !Tds s{x)s = {T2Js• e[x)s = !T2 (x))s·: thus Tt(x) = T2(x). ( b) l· abe. l·ot example, the zero operator has same rn:ttrJX (t hP ZNO nMtnx) with respect to any basis for R 2

(a)

(c)

True If 8 = { v1, v2 , , vn} and [T)s =I, then T(v,.J = v1c for each k = 1, 2, follows from this that T(x ) = x for all x .

(d )

False For example, h:t B = {e1, e2}. 8' = {ez,ct}, and T(x, y) but T not the 1dent1t.y operator.

Ouc 1cason 1s that lhf tresPntation 0f the operator in t. he

. , n and it

= (y , x). Them IT]s •.B =

/2

orht'r b:L.c;is may more clearly reAect

rir effect. of the operator

WORKING WITH PROOFS Pl . If x andy are vectors and cis a scala r then , since T is linear, we have c[x)e [x)o

= [cx]s -7 {T(cx )Ja = [cT (x )Je = c{T(x )]a

+ [Y)s = jx + Y)s

-t[T(x + y )]s

Thts shows that th- mapping {x]s

P2.

Ifx is Thu:-

tn

-7

= jT(x) + T(y)Js -

{Tfx )]e

+ {T(y )]s

[T(x )Js is linear.

R" and y ts m Rk, then we ha"e [T1 (x)J a• = [Tds•.s lx)s and IT2(y)J a" = !T2)B".B'IY)s•. IT2('i JIX))]s"

and rrolll tIll-: It follows thal IT2

0

= IT2] B",B' IT1 (x )Js• = IT2)B".B'!Tds·.alx]s Tds•·,s = (:t;JB" ,B'!Td s·.B·

277

EXERCISE SET 8.2

P3. If x is a vector in Rn, then [T]s[x]D = [Tx] B = 0 if a.nd only if Tx :::: 0 . Thus, if T is one-toone, it follows that [TJB[X]B = 0 if and only if [x] a = J, i.e., that: [T]a is an invertible matrix. Furthermore, since [T'" 1]a[T]B = [T- 1 oT} 8 = [J]B = /,we have [T- 1]s = [TaJ- 1.

P4. [TJs

= [T(vi) a I T(v2)al ··· I T(vn}s] = [Tle. B

EXERCISE SET 8.2 1. We have tr{A) = 3 and tr(B)

= -1;

thus A and Bare not similar.

= 18 and det(B) = 14; thus A and B are not similar.

2.

We have det(A)

3.

We have rank( A) = 3 and rank(B)

4.

We have rank(A)

5.

(a) The size of the matrix corresponds to the deg ree of it.s characteristic polynomial; so in this case we have a 5 x 5 mat rix . The eigenvalues of the matrix wiLh t heir algebraic mu!Liplicities arc A = 0 (wult.iplicity 1), A=- -1 (multiplicity 2), and A= 1 (mult.iplicity 2). The e igenspace corresponding to A = 0 has dimension 1, and the eigenspaces corresponding to A= - 1 or..\= 1 have dimension 1 or 2. (b) The matrix is 11 x 11 with eigenvalues A = - 3 (multiplicity 1), A = - 1 (multiplicity 3), and >. = 8 (multiplicity 7). The eigenspace corresponding to >. ""' -3 has dimension 1; the eigenspace corresponding to ). = - J has dimension 1, 2, or 3; and the eigenspace corresponding to A = 8

= 2; thus A and B

are not similar.

= 1 and rank(B) = 2: thus A and B

are tJOt similar.

have dimension I, 2, 3, 4, 5, 6, 7, or 8.

6.

(a) The matrix is 5 x 5 with eigenvalues A= 0 (multiplicity 1), A= 1 (rm1ltiplicity 1), .A= -2 (multiplicity 1), and .A = 3 (multiplicity 2). The eigenspaces corresponding to .A= 0, >. = 1; and >. = - 2 each have dimension J. The eigenspact! corresponding to A = 3 has dimension 1 or 2. (b) The matrix is 6 x 6 with eigenvalues A= 0 (multiplicity 2), A= 6 (multiplici ty 1), and .A= 2 (multiplicity 3). The eigenspace corresp(1nding to >. = 6 has dimensio n 1, the eigP.nspace corrcspondinr; to A = 0 ha.<> dimension 1 or 2; and the eigenspace corresponding to A = 2 has dimension 1, 2, (){ 3.

7. Since A is triangular, its characteristic polynomial is p(.A) = (..\- l)(A- 1)(.\ - 2) =(.A -- 1)2 (A- 2). Thus the eigelll'alues of !I are .A= 1 and>.= 2, with algebraic multiplicities 2 a nd 1 respectively. The eigenspace corresponding to A= 1 is the solution space of the system (I- A)x = 0, which is

The general solution or this system is x =

[i]

= t [:] ; thus the eigenspa«

is

1-dime.'5ional and so

A = 1 has geometric multiplicity 1. The eigenspace corresponding to A = 2 is the solution space of the system (2/- A)x = 0 which is

Chapter 8

278

[':] = s

The solution space of tbis system ;., x

[l

thus the eigenspace is 1-dimensiotlai and so

). = 2 also has geometric multiplicity 1. 8. The eigenvalues of A are ). = l , >. = 3, and >. rcultipliclty 1.

= 5; each

with algebraic multiplicity 1 and geometric

9. The characteristic polynomial of A is p(>.) = det(>J -A) = (.X- 5) 2 (>. - 3). Thus the eigenvalues of A are>. = 5 and >. = 3, with algebraic multiplicities 2 and 1 respectively. The eigenspace corresponding to ). = 5 is the solution space of the system (51- A)x = 0, which is

m

Tbe gene>al solution of this system is x

t

[!l;

thus the eigenspace is 1-
). = 5 has geometric multiplicity 1. The eigenspace corresponding to ). = 3 is the solution space of the system (31 - A)x = 0 which is

The solution space of this system is x

= s[

= [ - 2s

; thus the eigenspace is ]-dimensional and so -2

). = 3 also has geometric multiplicity l .

10. The characleristic poiynomial of A is (). + 1)(.>.- 3) 2 . Thus the eigenvalues of A are>.= - 1 and >. = 3, with algebraic multiplici ties 1 and 2 respectively. The eigenspace corresponding to >. = -1 JS 1-dirnensional, u.nd so >. = -1 has geometnc multiplicity 1. The eigenspace corresponding to >. = 3 is Lhe solution space of the system (31- A)x = 0 which is

[:;] =

The general solution of Lhis system is x

s

-1"' [

+ t [0] ; thus

the eigenspace is 2-dimensional, and so

). = 3 geometric multiplicity 2. ll. The characteristic polynomial vf A is p(>.) = >. 3 + 3>.2 = >. 2 (>. + 3); thus the eigenvalues are >. = 0 and ..\ = - 3, with algebraic multiplicities 2 and 1 respectively. The rank of the matrix

OJ - A = -A

=[

1 - 1

-1

1

279

EXERCISE SET 6.2

is dearly 1 since each of the rows is a scalar multiple o f the lst row. Thus nullity(OJ - A) and this is tbe geometric multiplicity of ..\ = 0. On the other h and, the matrix

-31 --A =-

-2

>. =

3- _1

= 2,

-1]

-1 - 1 -2

1

-2

[- 1

0 1]

has rank 2 since its reduced row echelon form is this is t he geometric multiplicity of

1

:=:

1

1 .

0

0

Thus nullity( -31- A) == 3 - 2 = 1, and

- 3.

12. The characteris tic polynomial of A is (>.- l)(A 2

-

2A + 2); thus >. = 1 is the only real eigenvalue of

A. The reduced row echelon form of the matrix li - A=

is - l

4

- 6

of 11 ·- A is 2 and the geomet ric multiplicity of ).. = 1 is nullity( ll - A)

0

=

(I

thus the rank 0

3 - 2 "" 1.

13. The characteristic polynomia.l of A is p(>-.) = .>.3 - 11>.2 + 39A - 45 = (A - 5)(>.- 3) 2 ; thus the eigenvalues are >. = 5 and ,\ = 3, with algebraic multipli<.:ities 1 and 2 respectively. The matrix

5/ - A ""' - 1

has
0

l

!

Thus nullity(f rel="nofollow">l --A) - 3 - 2

= l,

and

this is the geometric multiplicity of ,\ = 5. On the o t her ha.nd, the matrix

3T- A = - 1

0

-]

has rank l since each of the rows is a scalar multiple of the 1st row. Thus nuUity(3I - A) = 3 - l = 2, and this is the geometric multiplicity of,\ ""'3. It follows that A hM 1 + 2 '-=-= 3 line<'lrly independent eigenvectors and t hus is diagonali<>able.

1 4 . The characteristic polynomial of A is {A+ 2)(>. - l) 2 ; thus the eigenvalues are,\= - 2 and A= 1, with algeb
:

=

... - 1 - 1

l

=[

has rank 1. Thus rel="nofollow">-

=

=:]

- 2 has

h,. < 2,

1, and

rel="nofollow">. = 1 has geometric multiplicity 2. It follows that A has l + 2 = 3 linearly independent eigenvectors thus is diagonalizable. 15. The characteristic pc,lynomial of A is p(.>-) =.>- 2 - 3>. + 2 = (>.- l) (A- 2); thus A ho.s two distinct eigenvalues,>. = 1 and>.= 2. The eigenspace corresponding to A = 1 is obtained by solving the system (/ - A)x

= 0. which is

=

The general solution of thls system is x

=

Thus,

I 280

Chapter 8

taking t

= 5, we see that ?l

= [:]

is an eigenvector for

for A= 2. Finally, the matrix P = [p 1

p-1 AP = [

p 2]

= [!

>. = 1. Similarly, P 2 =

[!] is an eigenvector

!} has the property that

4 -3]4 [-14 12] [4 3] [l0 0]2 -20 17 5 4 =

-5

16. The characteristic polynomial of A is (.>.- 1)(.>. + 1); thus A ha.s two distinct eigenvalues, .>.1 = 1 and .>.2

=

-1 . Corresponding eigenvectors are P1

=

the property that p -l AP =

and the matrix P = !P 1

and P2 =

P2J has

=

17. The characteristic polynomial of A is p(>.) = .>.(.>.- 1)(,\- 2); thus A has three distinct eigenvalues, >. = 0, >. = 1, and >. = 2. The eigenspace corresponding to >.. = 0 is obtained by solving· the system

(OJ- A)x

0, which is

n-:=!] [::]

Similarly, the genml solution of (I- A)x is x

t [:]. Thus the mat,ix P

p- 1 AP

=

The

[-! :]

-

0 is x =

+l,

solution of this system is x

and the gene
[-

..

002

101

18. The characteristic polynomial of A is (A- 2)( rel="nofollow">. - 3) 2 ; thus tl.e eigenvalues of A are >.

[i] is an eigenvecto. cwesponding to

2, and v 2

linearly independent eigenvectors corresponding to .X = 3. The matrix P property that, p - t AP = 001003001

19. The characteristic polynomiaJ of A is p(>.)

Lhree distinct eigenvalues

v,

m,

and

= >.3 -

;:

=3.

1

-3

Note. The diagonalizing matrix P is not unique· eigenvectors. This is just one possibility.

: 1

it

3.

and v 3

= (v 1

v 3 J has the

= (..>..- 1)(>. - 2)(.>.- 3); thus A has

Corresponding eigenvectors are v 1

v, [:]- Thus A is diagonalizable and P p-IAP =

[!]

= 2 and >.. =

003

6>.2 + 11>.. - 6

>. 1 = 1, .>.2 = 2 and .>.3

0 -1

0

has the P':perty that

011011

The vecto' v 1

+:J

=

[:].

[: : :] has the pmpe
=

3

1 3

4

0 0 3

depends on the choice (and the order) of tht:

EXERCISE SET 8.2

20.

281

The characteristic polynomial of A is p(>.)

= >..3 -

-t >. 2

+ 5>.- 2

= (>.- 2)(>.- 1)2 ;

thus A bas two

:::: -17

9 f

0

7)

il·

which shows I hat the etgenspa corresponding to

solution of (I - A)x

0 is x - s [

>.

=2

has dunension 1

Simi larly, llw general

which shows that the eigenspace conesponding tn A = t also

has dimension 1. It follows that the matrix A is not diagonalizable since it has only two linearly independent eigenvectors.

21. The characteristic polynomial of A is p(>.) = (>. - 5)3 ; thus A has one eigenvalue, >. = 5, which has algebraic multipUcity 3. The eigenspace corresponding to >.. = 5 is obtained by solving the system

(51- A)x = 0, which is

H: [:;]

=

m

The genml solution of this system is

X-

t

which shows that t ht:' eigenspa has dimension 1, i.P , Lhe eigenvalue has geometric mulliplicity 1 lL follows that. A is nnt. since thn sum of t.hr gcomelrif multiplicil.it!S vf its is less than 3 22. The characteristic polynomtaJ nf A is

eigenspace corresponding to A

0

>.2 (>.- 1}; thus the eigenvalues of A are \

hM dimension 2. and the vecto'5 v, = [

hMis lot this space. The "'""' ' '

=

23

=

v,

1 The

[:] fotm a

mfotms a ba.
Thus--A IS diagonalizable and thr matnx P = lv 1

p - t AP

= 0, and >. =

:

v2

=

I.

u! v 3)

has the property that

ThP charactt>risttc polynomtal o f A is p( .\) = (>. + 2) 2 t.\- 3) 2 • thus A has two etgcnwtlues, .-\ = -2 and .>- = 3, each of which has n.lgebraic rnulLiplicity 2. The eigewJpace corresponding L(l ..\ = -2 is

obtained by solving thP. system ( -2/- A)x = 0, which is

The gonctnl solution of this system ts x = r

+s

whteh shows that the eigenspace has dimenSton

2, i.e., that the eigenvalue >. = -2 has geomeLric multiplicity 2. On the other hand, the general

solution of (31 - A )x - 0 is x = I

•nd so

= 3 has geomet d e mu I liphcity I It follows that A is

not d iagonalizable since the sum of the geometric

of its eigenvalues is lt>..ss than 4

282

Chapter 8

The charactenstic polynorrual of A is p(>.) = (>. + 2)2(>.- 3) 2 ; thus A has two eigenvalues A= -2

au.l

3, each of algebrak multiplici
eige "'P"' correspondi og to

v, -

- -2, and
and v 2

rrl

=

and v, - [-

form a

form a basis for

e1genspace corresponding to A= 3. Thus A is diagonalizable and the matrix P = fv 1 has the property that

p-'AP=

1

-1

0 0 0

0 1 0

0 1

0 0

0 0

3 0

-

0 0 30

010 0 01

v2 0

=

for

v3

v 4J

0

-2 •- 0 0 0

0 0

3 0

25.

1f the matrix A is upper triangular with l 's on the main diagonal, then its characteristic polynomial is p(>.) = (>.- 1)" and >. = lss the only eigenvalue. Thus, in order for A to be diagonalizable, the system (1 A)x = 0 m11st haven linearly independent solutions. But, if this is true, then (/- A)x. = 0 for every vector x in R" and so I - A is the zero matrix, i.e., A =I.

26.

If A 1s a 3 x 3 matrix with a three-dimensional eigenspace then A has only one eigenvalue, A= A1 , whsch IS of geometric rnultJphc1ty 3. In other words, the eigenspace corresponding to A1 is all of R3 . It follows that Ax = >. 1 x fos a.! I x in R 3 , and so A = >. 1 I is a diagonal matrix.

27.

If C is similar to A then there is an invertsble matrix P such that C = p-l AP It follows that sf A IS invertible, then C ts invertible since it is a product of invert1ble matnces. Similarly, smce P 1 =-= A, the invertibihty of C implies the invertibility of A.

28.

If P =!P sI P2l · I PnJ. then AP = (Aps I · · · I ApnJ and PD =[A t PI I A2P2I · · · I AnPnJ where \1 -\2. . An at c the diagonal entries of D. Thus A.I: Pk for each k = 1, 2, , n, i.e., is an of an<.l P1r is an eigenvector corresponding to

29. The :>tandard matrix of the linear operator T is A

=

=

and the characteristic polynomial

-1 - 1 -2

+ 9>. =>.(A+ 3) 2 • Thus the eigenvalues ofT are >. = 0 and A= -3, with mulllplicltles 1 and 2 respectively. Since>.= 0 has algebraic multiplicity 1, its -geometric snult tplictty 1S also 1. The eigenspace associated with A = -3 is found by solving ( -31 - A)x = 0 whsth IS

of A ts p(>.) - >.J

+ 6A2

The gen"' ..t svlu
IS

+t [:] ;

diagonali1.able since the sum of the geometric multiplicities of its eigenvalues is 3.

30. Th<' standard rnatnx of the operator Tis A=

and the characteristic polynomial of A l

i-; (>.

- -3 has geome
+ 2)(>.- 1) 2 Thus the eigenvalues ofT are

1

0

A= -2 and

>. = 1, with algebraic

1

283

DISCUSSION ANIJ DISCOVERY

and 2 respedively. Smce >. = -2 has algebraic multiplicity 1, tts geometric multiplicity is also 1. The eigenspace associated with>.= 1 is found by solving the system (J - A)x = 0 Since the matrix

l - A= [

has rnnk 1, the solution space of(!- A)x -1

- 1

=0

is two dimrnsional, i.e.,>.= 1

l

is an eigenvalue of geometnc multiplicity 2. It follows that T is diagonaltzal>le geometric multiplicities of its eigenvalues is 3.

the sum of lhe

31 .

If x is a vector in Rn and >. is scalar, then [Tx)a = [T]a[x]a and [>vc]B = .>.[x]a. lL follows that Tx =Ax if and only if [T}s[x]s = [Tx]B =[Ax:]a = >.[x)a; thus xis an eigenvector ofT corresponding to >. if and only if [x]s is an eigenvector of [T] 8 corresponding to >..

32.

The characteristic polynomial of A is p(>.)

= 1>._:-ca

criminant of thi's quadratic polynomial is (a+

d)2 -

>. -:_b

dl = >.2 -

4(ad- be)

(a+ d)>.+ (ad - be), and the dis-

= (a- d) 2 + 4bc.

(a) lf (a - d) 2 + 4bc > 0, thon p(>.) has two distinct reo.! roots; thus A is (b)

two distinct Pigenvalues. Tf ( o - d) 2 + tbc ·· n, t h"n r(>.) has no nol diagonahzablP.

since it has

roots; t hue: A has no rc:1l t'lgf"nvalues and is the


are not s1mtlar smce rank(A ) 02.

(a)

= 1 and

rank(8)

=2

'ITue. We havP A = p-I AP where P =I

(b) 1Tue. If A 1s s tnHiar to B and B is similar to C. then there arc mvert1ble matrices P 1 nnd P2 surh that A - P1- 1 BP1 and B = F 2- 1 (' P2 . lt follows that A = P1 1 ( P 2 1 C P2 )P1 = (P2P1) 1 C(e'JJ>I ), thus A is sinular to C (c) True. IfA - P 1 BP,thenA- 1 =(P- 1 BP) - 1 = P - 1 B - 1 (P 1 ) 1 p- 1 e- 'P.

( cl) False. This statement does not guarantee that there are enough linearly independent eigenvectors.

has only one (real) eigenvalue, >.

For example, the matrix A= 0 -1

wh1ch has 03 . (a)

=

1,

0

mull1pltcity 1, but A is not dJagonalizable

False. For examplr, 1 -

is diagonalizahle

False. For exc.Lmple, 1f p -l AP is a diagonal matrix then so IS Q 1 AQ where Q = 2P. The diagonalizing matrix (if it exists) is not uruque! (;:) True. Vectors from different eigenspaces correspond to djfferent eigenvalues and are therdore linearly independent. In the situation described {v 1 , v2 , v 3 } is a linearly independent set. (d ) True. If an invert1ble matrix A is similar to a diagtlnal matrix D, then D must also be m1 is the diagoncJ matnx whose diagonal vertible, thus D has nonzero diagonal entries and entries are the reciprocals of the corresponding entries of D. Finally, 1f PIS an mvert1ble ma1 and so A- 1 is sintilar trix such that p -I AP = D, we have p-• A- 1P (P- 1 AP) - 1 = too-l (b)

o-

=

o-

284

Chapter 8

(e) 1'ru_y The vectors in a basis are linearly independent; thus A bas n linear independent \

04.

(a) A is a 6 x 6 matrix. (b) The cigens puce corresponding to >. = 1 has dimension 1 The eigt:nspace corresponding to >. = 3 has dimension l or 2. The eigenspace corresponding to >.= 4 has dtmension 1, 2 or 3. (c) If A is diagonalizable, then the eigenspaces corresponding to >. = 1, >. = 3, and >. = 4 have dimensions 1, 2, and 3 respectively. (d) These vectors must correspond to the eigenvalue >. = 4.

05.

(a) If >. 1 has geometric multiplicity 2 and >.2 has geometric multiplicity 3, then >.3 must have multiplicity l. Thus the sum of the geometric multiplicities is 6 and so A is diagonalizable. (b) In this case the matrix is not diagonalizable since the sum o f the geometric multiplicities of t he e1genvalues is less than 6. (c) The matrix may or may not be diagonalizable. The geomet.ric multiplicity of >.3 .IIWSt be 1 or 2. If the geometric multiplicity of >.3 is 2, then the matrix is diagonalizable. If the geometric mllll.iplicitv of >-1 Is 1, then the matrix is not dia.gonaliza.ble.

WORKING WITH PROOFS Pl. If A and Bare similar, then there is an invertible matnx P such that A = p - t BP. Thus PA = BP and so, usira,v the result of t he ci ted Exercise, we have rank( A) = rank(PA) = rank(BP) = rank(B ) and nullity(A) = nullity(PA ) nullity(BP) = nnllity( B). P2.

If A and 13 are sum lar lhen t here ts an invertible matnx P such that A = p - • BP Thus, using part (e) of Theorem 3 2 12, we have tr(A) = tr(P - 1 BP ) == tr(P 1 ( IJP)) o..: tr(( BP)P - 1 ) = tr(B ).

P3. If

X

= p - l A, we have p - 1 (,\x) = >.P- 1x

f. 0 and Ax = >.x t.hen, since pis invertible and -l C'P - 1 x = p - I Ax =

wit h p -• x -:f 0 T hus p -• x is an eigenvector of C corresponding to >..

P4. If A and B ure :m111lar, lhen there is an invertible matrix P such t ha t A = p - l BP. We will prove, by induction, tha t Air P 1 B" P (thus Ak and BJr are similnr) for every positive integer k . Step l The fact that , \ 1 = .1 = p - 1BP = p-l B 1P is gtven

= p - l BI

Step 2. If Ak

These two s teps compl(·te the proof by induction PS

If A ts d tagonaltzablc, t hen there is an invertible matrix P and a diagonal matrix D such that P - • AP = D . We will prove, by induction, that p - tA'" P = Dk for every posttive integer k . Since D" is diagonal thts s hows that A" IS diagonalizable.

Step 1 The fact that p- 1 A 1 P = P- 1 AP = D

= D1

is given.

= D" , where k is a fixed integer 1, then we have p - lA H 1 p = p - 1 AAkP = (P- 1AP)( P- 1A 4 P ) = DD" = Dlc+ l

Step 2. If p - t Ak P

These two s teps l"omplcte thP proof by induction .

285

EXERCISE SET 8.3

P6. (a )

Let W be the eigenspace corresponding to >-o. Choose a basis {u 1, u z, ... , uk} for W, then extend it to obtain 11. basis B = { U t, u 2, ... , u ,., Uk+l• .. , u n} for R".

(b) If P = [u1 I u2l

I

I · · · I un] =[Btl B2j AP = [-\out I >.o u 2l · · ·

then t.hc product AP has the form

I

I

On the other hand, if Cis ann x n matrix o f the form C

where Z

(c)

=

if Z

=

t hen PC has the form

= p -l AB2 , we have AP =PC.

Since AP =PC, we have p - l AP =C. T hus A is similar to C

=

so A and C

have the·same characteristic polynomjaJ. (d ) Due to the special block structure of C, its characteristic polynomial of has the form

Thus the algebraic multiplicity of >.0 as an eigenvalue of C, and of A, is greater than or equal to k

EXERCISE SET 8.3 1. The charactenslic polynom1'\l of A is p(>.) = ,\2 - 5,\ = >.(>.- 5). Thus the eigenvalues of A are).= 0 and ,\ = 5, and each of the cigcnspaces has climension 1.

+

2 . The characteristic polynomial of A 1s p(,\) == >. 3 - 27...\-54== (,\- 6){>. 3) 2 Thus the eigenvalues of A are,\ = 6 and .>, -3 . Thf> t>igenspace corresponrling to,\ = C hl\.o:; dimension 1, anrllhe e1genspace

=

corresponding to ,\ =- -3 ha.s rlimPnsion 2.

3. T he characteristic polynom1a.l of A is p(>.) = >.3 - 3).2 = ..\2 (.X- 3). Thus t he eigenvalues of A are A= 0 and ,\ 3. The eigensptLCt: corresponding to >. = 0 has dillleu.sion 2, and the eigenspace corresponding to >. - 3 has dimcllSIOn 1 4 . The charactenstic polynomial of A is p(>.) = ,\,3 - 9...\ 2 + 15).- 7- (,\- 7)(>-- 1) 2 Thus the eigenvalues of A are >. = 7 and >. - 1. The eigenspace cor responding to ,\ = 7 has dimension I, and t he eigenspace corrPspondmg to >. = l has dimension 2.

5 . The gen"al solution or the system (0/ - A)x and v 3

[-:]

0 is x

+ s [-:],

r

form a basts for the eigenspace corresponding to

forms a basis for t he eigenspace corresponding to>.= 3. Since follows that the two eigenspnces are

v3

thus the •ectors v 1

= 0. Similarly the vector v, is or thogonal to both

v1

[

=

and

m

v2

it

Chapter 8

286

6. The general wlution or (7f

>i )x

eigenspa<e ro"esponding to >.

0 is

T

[i}

thus the vcctoc

7. Similacly the vecto"' v 2

[-

v, [:]

]•nd v 3

rom•s a basis foe the [-:]

the e tgenspace corres ponding to >. = 1. Since v 1 is orthogonal to both v, and v 3 two etgenspaces a re orthogonal 7.

The characteristic polynomial of A is p(>.) = >.2 are>.= 2 and >.- 4. The vector v 1

=

fvllows that the

6>. + 8 = (>.- 2)(>. - 4); thus the eigenvalues of A

-

forms a basis for the eigenspace corresponding to>.= 2,

forms a basis for the eigenspace corresponding to >. = 4 . These vectors are

and the vector v2 =

orthogonal to each other, and the orthogonal matrix P =

that

PTAP= 72 8.

ll

form a basis for

l]

[31

J2

3

J = [-

[20 0]" = D

=

-12

has the property

\/2

The characteristic polynomial of A is (>.- 2}(>. - 7); thus the eigenvalues of A are >. Correspondmg Pigcnvectors are v 1 =

and v 2 =

[- ..!S

=

ts] [ 6 -2] -2

9. The characteristic polynomial of A is p(>.) A

s

[-l] +

t[

hns the property that

= .>.3 + 6>.2 -

32

I

=D

= (>.- 2j(>. + 4) 2 ; thus the eigenvalues of

A)x

0 is

Thus
corresponding to >. - 2, and the vectors v,

[20

=

75

'75

3

2 and .\ = - 4. The g'ncral solu
of (-If - A)x - 0 is x

7.

respectively. These vectors are odhogonal to

each other, and the orthogonal matrix P = 111 :: 11

pT _.1P =

= 2 and >. =

T

[:]

r:J

[ -:] and v,

m'

and the gcnecal solution

forms a basis foe the eigenspace

form • basis for the eigenspace

corresponding to>.= -4. Application of the Gram-Schmidt process to {vt} and to {v2 , v 3} yield orthonormal bases { u 1} and {u 2 , u 3 } for the eigenspaces, and the orthogonal matnx

P=

lut

u2

u3) =

-73]

?s [

0

...16

73

has the property I

pT AP

=

76

r-12

2]2

0

[7:

=

76 2

76

73

[20

=D

0

Note. The diagonali?.ing matrix I' is not unique. It depends on the choice of bases for lhe eigenspaces. This is j ,ts t onf' possibilit.y.

EXERCISE SET 8.3

287

10. The characteristac polynomial of A is p(>.) = >.3 + 28>. 2 - 1175>.- 3750 = (>. + 3)(>.- 25)(>. +50); thus the eigenvalues of A are >. 1= -3, >.2 = 25, an<.l >.3 = -50. Corresponding eigenvectors are

=

v,

[!],

v,

=[

i],

and v,

=

m.

"!' mutually orthogonal, and the orthogonal

These vc-eto<S

matrix

4

p

[

v2

V1

li::u] =

= llv1l1 llv2ll

0

:J

3

5


0

0

pTAP = :J

-3

-36

0

5 11.

0][ -2 0 -36] [0 i 0

1

0

0 -23

0

3

The charactenstic pnlyn(lmial of A is p(>.) = \ 3 and

> = 2.

( 2/ - A )x

The genoral •oluUon or (OJ - .4)x

= 0 is x

""respondin• to I

>.

I[

l]·

=0

Thus the vedors v,

0. o>nd >he vector v, =

2>.2

-

=

[l]

is

n

-g4

0 25

=

5

=D

0

= ).2 (>,- 2); t.hus the eigf'nvalues of A are>.= 0 X

=T

m

m+

and v,

s [ -:]' and the general soluUon

l

= [-:

or

form a basis £or the eigenspace

£mms a basos £or the eigenspace eonesponding to

= 2 . These VPctors are mutually orthogonal, and t.he orthogonal matrix 1

-72

Vt

F =

[

I

llvdl

72

72

0

0

hn.-, the propt>rtv 1hnl

n [ 12.

1

V2

72 0

72

The <"haracterrsltr. polynomaal of A are

> = 0 and > =3

JS

The ve<:tor v,

i

and the vectors v, = [- ]••d v 3

0 0

0

p(>.) = .>.3

= [:]

= [-

1 -

6>. 2

l[ l

0

=

I)

0

0

0

0

0

0 0

2

+ 9>. = >.(>.- 3) 2 ;

=D

thus the eigenvalues of A

forms a ha.,is £or the eigenspace corresponding to A= 0,

form a basis £or the eigenspace corresponding to A = 3.

Application of Gram-Schmidt to { vt) and to { v 2, v3} yield orthonormal bases { u 1} and {u 2 1 u 3 } for the eigenspaces, and the orthogonal matrix

288

Chapter 8


2

76

2, and A = 4. The

gene,aJ solution of (21 - A)x

Thus the ve
[

-ll

73

2

....!_

0

= >.4 -

6>.3

-1

13. The characteristic polynomial of A is p(.X)

0, A

- 1

-1

-12 - 21

0

I

of A are A

-1]

---

tal [

1

7J 1 "J2

t

0 is x

and v3

lJ,

1 I

l

v3

[

-72

72

[!],

= 0 is ·x

u

[

ll

0

0

0.

fo,ms a basis

0

0

-.J.. v'l

0

72

l

o o lo o o o

1

0

1

0

()

0

1250..\2

[00 00 00 0]0

-

0 -

= >. 4 -

:he

the

0

0 0 2 0 0

0

0

4

+ 390625 = (>.- 25) 2 (..\ + 25) 2 ;

25, and A= -25. The vecto" v 1 =

eigenspace COHespondi og to A

+s

r

2, and v,

0 0 0 0

0 0

14. The characteristic polynomial of A is p(..\)

eigenvalues of A ace

0 is x

rlr

2

00 0 01 00 1] 1 [3 310000] [00 I

3

2)(>. - 4); thus the eigenvalues

and the genecal solution of (4/- A)x

1

1

0 0

7G

fo<m • bO!Sis fo< th<· ci•cnspace coHe;ponding to

:o:::di:;::o >.

T

0 0 0 0 3 0 =D

=

A)x

fo<ms a basi: fo, the eigcn:pace co"esponding to >.

pAP=

2

+ 8.X2 = >.2 (>. -

solution of (01

l[ l

-76

'1.2

v3

25, and the vee to" v 3

thus the

fo<m a basis fo, the

v,

[

-ll

fo<m a basis fo< the

eigenspace corresponding to..\ = -25. These four vectors art> mutually orthogonal , and the orthogcnal matrix P 1: : 11 1has the property that s

PTAP=

0

4

5 0 3

s 0

0 ;!

s 0 4

01

0 s

[-7

24

0

s

3

0

-$

4

5 0

0

3

0

7 0 0 -7

0

0

0

24

24

3

g 4

s

4

5 0 0

25

=

-il

0 0 0

J

EXERCISE SET 8.3

1 5.

289

The cig<>nvalues of the matrix A = eigeuvccturs u 1

l

:l

l] 3

1

= [_

71

= (2) [ -72

16. The eigenvalues of A are and u 2

= [-

=

and u 2

[

>11 = 2

are

and >.. 2

= tl,

with corresponding nOrJ:!lalized

Thus the s pectral Jccompo.'lition of A is 7-i

I

72

>-t = 2 and

.X2

= 1, with corresponding normalized eigenvect ors u 1 = [

Thus the spectral decomposition of A is

1 7.

:::.:·:,:i•

:.-:·

ii

is l

-3

:2]

=2

2

=

[js] [;n

I

76

1.]-• [-72][- ,', ;!; o]-' [-+a]

[! ! l - +t -t

- 73 ;!;] 1

-! =j]

Note fhl' :;pPctral decomposilion Lc; noluni()ue. l t depends on the r hoice of bases for the eigenspaces. Thic; 1s juc;l one possibility.

18.

-2 A= [

0 3G

n

0 -36 -3 0 ....: -3 1 {0 0 -23 0

1

[0

1 OJ + 25 0

0

= -3 0 l 0OJ - 25 [ _g 0 0 0 25 19. T he maln x A has eigenvalues >. Thus the matr ix P

= [- : -

= -1

and ).

o 0

= 2,

n 1

-25

[.

0

9 25

12

0

"o]

25

mll

0 o o lll

25

= lJ = [-

OJ [- 21 -13]

1024

0

25

16

25

with corresponding eigenvectors

has the property that p- t AP

-3]2 [10

lJ _ 50

0

[-11] and [-32].

It follows that

[ 3070 3069] - 2046 -2045

290

Chapter 8

20. The ma•rix A has eigenvalues>.= 2 and>. = -2, with corresponding eigenvectors

;) has the property that p-l AP = D =

the matr;., P =

AlO = PDIOp- 1 = [1

2

- 1, and the vectorn

conesponding

= 1.

Thus the matrix P

[1024 0

and >.

21. The matrix A has elgenv&lues

to >.

-1]

7

= [:

It, follows that

0 ] [-7

1024

Thus

4] -1

2

= [1024 0

0 ] 1024

1. The ve<:to' [:] fonns a basis fo' !he eigenspace

m [!] and

form • basis fo, the eigenspa<e conesponding

=D=

bas the property that p-I AP

4 1 0

and it 0 0 1

follows that

= PDIOOO p - t = [:

AlOOO

4

22.

The mat,ix A has e•genvalu., >. and

[!]

Thu• P

A1ooo

23.

[: ;

!]

1 0

0, >.

0

0

1, and >.

-3

12 -24

-

=

3

0

0 1

!

D

and it follow• that

[ 6:]

(a) The characteristic polynomial of A is p(>.) ..,. [ :

2

- 1, with conesponding eigenvedorn [:]. [:] ,

has the pcope,ty that p-• AP

= potooop-1 =

of A, we have A 2

1

:1

and A 3

= >.3 -

+ 12>.- 8

=

16

-24 - 40 -72

6>. 2

Computing successive powers

thus 36 -72

44

4] [3 -2

6 [ :

12 -24

44

8 + 12 2 -2 16 3 -6

8/

which shows that A satisfies its characteristic equation, i.e., that p(A} = 0.

{b ) Since A 3 =-6A 2 (c)

Since A

3

-

-

12A + 8/, we have A 4

= 6A 3 -l2A 2 + 8A =

6A -1 12A - 81 = 0, we have A(A 2

2

-

6A

24A 2

-

64A ..._ 4 I

+ 12/) = 8 1 and A- 1 =

6A

+ 12/).

EXERCISE SET 8.3

24.

(a)

291

The characteristic polynomial of A is p(>.) =

A, we have A2

=

=I

-

and A 3 =

>.2

-

>. + 1.

Computmg successive powers of

= A, thus

0 0 1

.A2

>.3

-4 0 5

A

-

=

-5 0 6] [1 0 0] [-5 0 6] [ -3 1 3 -4 0 5

0 1 0 0 0 1

-3 -4

3 = 5

l 0

which shows that. A satisfies its characteristic t>quat1on, i.e., that p(A)

(b) Since A =A, we have = 2 1 (c) Since A = I, we have A- =A. 3

A4

= AA =

AA3

_PetD pT = [- -k *) [e 0] [-h -?z] 2

"2

27

'

0.: 11

.,!}

Prom Exercise 9 we have

pT AP =

[

I

=f 76

l

-72 I

vJ

v'3

Thus A= PDP'l' nnd

e'A

Pe'o J>r

=

"2

=

n

../3

[I

-3 2

4

1 [

t e. '

1

76

..;2

-73

-36

0

.. a

I

j

()

2e2t- 2e-4t

+

":!'+elt'

I

n

-/2

-l''lt

Thus A = PDPT and

-72 2

76

= D.

=

2 -e:!-Tt' 11

"'2

[ e"e2t +_ se-" e-·H 28 .

[

1

J;;

I

.J,.

= 0.

= I.

[t

25 . From Exercise 7 we hnve pT AP = -L

A2

l

['' 0

0

e2' _

1

76

72

'l

=

../3 1

0

-rr.. \ (>

0

7J

0 e .,,

0

()

][I-?z

e- 4t

7ti

0

0 - I

=D - I

I

76 I

J2 I

-73

2e2e2' - 2e-"] 2e-4t 21

e-4t

e21 + 5e- 41 2e2t- 2e-"'

-

4e2t

+ 2e- 4t

Prom Exercise 10 we havP

PTAP =

0 0 0] [ -20

_!

[i

Thus A= PDPT and

1

0

i

-36

4

-g

=D

0 3

s

0

-50

292

Chapter 8 4

-5 0 3

5

[ 25 " e'"

+ -2-e-'" 25

= _ 12 e25t 25

29.

Note that

s in('rrA)

l

sln(21f)

o

0 sin( - 4rr)

0

0

0

sin( - 47<)

[

0

= Psin (7rD)Pr =

0

0

I

ts

I

72 -73

-5 0

=

1

0

73

3 4

-g 0 3

5

31.

rr

A

[!

then A'

;J

and A

..!!. e25 t 25

.4 3 -0

cos(1r.A )

=J

[00 0][--:;,to 0 -+a

1

76 1 l

00 ][- 10 0 cos( 507r)

0

1

-1

0 0

0

3

!0] = [-f.0 4

5

2

00 00]

11 0

-73

0

cos(257r) 0

= !,

72

i

0 0

0

7

0

25

il ll 25

-1

24

0

B

Thus A is nilpotent and

210

[00 00 0]0 = [11 01 0]0 !flO

have sin(1rA) = sin(O)/ +1rcos(O}A- t,1r 2 sin(O)A 2 = 'Tf'A

WP

; A

50t

1

0 0 0 0 0

n H !

Smc('

+ !! e 25

proceeding as in Exercise 27:

0]0 + I [01 00 0]0 32.

+ 25 " e-'" ] 0

0

;][oos(t)

5

- 25

0

76 -72 [ l

4

cos("ITA ) - Pcos(1rD}PT

+ 25 12 e-sot

= [0o o0 OJ o • Thus,

76

30.

0 e-31

0

2

=

0 00 00]

[2tr

rr

rr

and

0

[

-T

0

l

33. If P is symmetric nnd orthogonal, then pT = P and pT P = I ; thus P 2 = pr P = I If >. is an eigenvaluE' of P thPn lhere is a nonzero vector x such thal Px = >.x S10ce F 2 = 1 1t follows lhat 2 ..\ x = = Jx = x , thus >. 2 = 1 and so >. = ±1


(a) (b)

True The malrix AAr is symmetric and hence is orthogonally diagonalizable. False If A •s dtagonahzable but not symmetric (therefme not orthogonally daagonalizable), then there as a basis for R" (but not an orthogonal basis) cons1slmg of eJgenvectors of A.

WORKJNG WITH PROOFS

(c)

293

False. An orthogonal matrix need not be symmetric; for ex\\mple A

=

.

{d ) '!Tue If A is an invertible orthogonally diagonalizablc mntnX 1 then there is an orthogonal matrix P such that pT AP = D where D is a diagonal matnx with nonzero entries {lhe eigenvalues of A) on the main diagonal. It follows that pr A - lp - (PT AP)- 1 = D- 1 and t.hat D - 1 is a dtagonal matnx with nonzero entries (the reciprocals of the on the main diagonal Thus the matrix A - l is orthogonally diagonalizable. (e) 1Tue If A is orthogon ally cliagonalizable 1 t hen A is symmetnc and thus has r eal eigenvalues. t

72 D2. (a)

A = P DPT = [ ;

0

; ] 0

(b)

0

0

7

0

-72] [3 0 0] 0

1

=

I

7z

72

0 0

3 4 4 3

No. 'l'he vectors v 2 and v3 correspond to different eigenvalues, b ut are no t orthogona l. T here fore t hey canno t be eigenvectors of a symmetric mat rix.

03. Yes. Since A is cll!lgonn.lizable and the eigenspaces are mutually orthogonlll 1 there 1s an orthonormal basts for R" cons1stmg ol e1genvectors of A. Thus A ts orthogonally d1agouahzablc and therefo re mu11t be symmetric


\'\'e first show that if A and C are orthogonally smular, then there orthonormal ha.c:es with rc.. J • t lv hkb tht:) rt'pn:scnl the same l1ut:ar '-per ll Jr Fur 1!11:. lJUrpose, let T the v1 eratvr defined by T(x) = Ax fhen A = [T], i.e., A is the matnx of T relative to the standard basis B = { e,, e2 1 1en} Smce A and C are orthogonally similar, there 1s an orthogonal matrL< P such that G=PTAP. Let B' ={v 1 , v 21. . , v n} where v 1 , v 2 , .• , vn are the column vectors of P. Then B' is an orthonormal basis for R" , and P = Pa'-•B Thus jTJa = PjTJs· PT and [T]s· = pT[T]sP = pT AP =C. This shows that there exist orthonormal basrs with respect to which .4 n.nJ C repn:!:>Pnt t.he same linear operator. ConwrsPly, lluppost: lhat A = [1']a and C = [TJa• when• T: R" - > Rn IS a linear operator and B, fJ' n.re h1\SCS for If P = PD'-+8 t.hen Pis an ort,hogonnl matrix and C = [T]a· = pT ITlnP pTA P. Thus A and C are orthogonally si1ni ln.r.

nu.

P2 . Suppose A - c1Ut U{ I c2 u 2uf + · · · + Cn llnu ;,· where {u 1, u2, ... , u n} is an orthonormal basis for Rn. Sincf> (u 1 uJ)T = u J'7 ' u J = u1 u f it follov.;s that A'T =A; thus A is symmetnc. Furthermore,

since u{' u1 = u, · u 1

-

o,,

1

\"e have T\

Au1 ={c1u1uf +c2u 2u f -···

2: c, u,u?' u 1 =c1 u 1 I

fo r each j = 112,

, n Thus c 1 , c 2 • •

.

1Cn are e1genvalues of A

P3. T he s pectr a l decomposition A= >q u 1 uf where P = !u tI u 2l · I u nJ and D =

+ A2 u 2ui + ··· -r

ts equivalent. to A= PDPT

A21 . . , >-n); thus

/(A)= P /(D )PT = Pdiag(f( At). /(>•2), ... , /(An))PT = /(AI) ulUJf

+ j(>.2) u 2u f + · · · +

294

Chapter 8

P4. (a)

Suppose A is a symmetric matnx, and >.o is an eigenvalue of A having geometric multiplicity J.. Let l\' be the eigenspace corresponding to >. 0 . Choose an orthonormal basis { u 1, u 2,. , uk} for W, extend it to an orthonormal basis B = {u 1, u2, ... , Uk+J, .. . , u n} for R" , and let P be the orthogonal matrix having the vectors -:>f B as its columns. Then, as ..;hown in Exercise P6(b) of Section 8.2, the product AP can be written asAP= Since P

IS

orthogonal, we have

and since pT AP is a symmetric matrix, it follows that X (b) Since A is similar to C =

= 0.

has the same characteristic polynomial as C, namely

(>.- >.o)" det(>.In-k- Y) = (>.- >.o)kpy(>.) where py(>.) is the characteristic pofynomial of Y . We will now prove that py (>.o) :f. 0 and thus that the a lgebraic multiplicity of >.0 is exactly k. The proof is by contradiction: Suppose py (Ao) - 0, i.e., that >.o is an eigenvo.lue of the matrix Y. Then there is a nonzero vector y in R"-k such that Yy

= ..\oY·

Let x

= [;]

be Lhe vector in R" whose first

k components are 0 and whose last n- k components are those of y

(c)

and sox IS an eigenvector of C corresponding to >.o Since AP = PC, it follows that Px is an eigenvector of A corresponding to >.o But. note that e 1 , •. , are also eigenvectors of C to >.o, and that {e 1 • . . , ek x} IS a linearly independent set It follows that {Pet, .. , Px} 1s a linearly mdependent. set of eigenvectors of A corresponding to >.0 . But this implies that the geometric muJtiplkity of >.0 IS greater than k, a contradiction! It follows from part (b) that the sum of the d1mensions of the eigenspaces of A is equal ton; thus A is dJagonalizable F\u-thermore, smce A IS symmetric, the e1genspaces corresponrling to different eigenvaluPs are orthogonal. Thus we can form an orthonormal bas1s for R" by choosing an orthonormal basis for each of tht> e1genspaces and ing them together. Since the surn of the dimensions is n, this will be an orthonormal bns1s consisting of ejgenvectors uf A . Thus A IS orthogonally diagonalizable.

EXERCISE SET 8.4 1.

(a)

Then

3x?

+

-!xJ

x2]

X

2

][ 4 -3] [X'] -3

-9

X2

295

EXERCISE SET 8.4

5 . The quadratic form Q

= 2xr

- 2x 1 x 2 can be expressed in matrix notation as

1-

The matrix . \ has ds!•n·:alucs ,\ 1 = 1 and ,\ 2 v2

= [-:]

1 2

)

=x

- Py

wilh corresponding dgt:m'l(tor:> v 1

matrix P =

respective)) Thus

of \'&riable [:

= 3,

[tz 7z

72

[Y1 ] Y2

-

GJ

am.l

orthogonally diagona.IJZcs A, and the change eliminates the cross product in Q·

Note that the in verst: relationship between x aad y is

,. = y =

-J'i 7i

pT x = [

[r

1

]

J:'l

•

- I

G.

rho •• v"' """"'' ,.;, !om• ' "" '"

in Lt ,,, notation

The matrix A hns rigcnvahJ<'S ,\ 1 = 1, ).. 2 = 4, ).. 3

v'

Hl'v, m.v, [I] =

tht' (:hang!' of vn.liablc x

= 6,

Q-

1

A X whccc A

:

octhogonally di ..on alizes A' and

= Py eliminates the cross product in Q·

The matrix A has eigenvalues ).. 1 = l, ).. 2 = 4,

nl' v,

= [:]'

]·

l

[

l

0

0

..t

0 0

The given quadiM>e (o, m can be exp<essed in matdx notation as Q = x' Ax whm A =

v, -

[:

with corresponding (orLhogonnl) eigenvectors

Thos the- ma:ix p = [ -

Y2

,.,

M

v, =

Ul

)..3

= 7,

n!-ll

_: -:]·

with cor responding (orthogonal) eigenvectors

Thus the matdx p =

d•agonaH..s A ,

Chapter 8

296

and the change of variable x

= Py eliminates the cross product in Q :

.

Q = xTAx = yr(pT AP)y =

Y'l

ll

0 OJ

Ya} 0 4 0 0 0 7

lYlj Y'2

+ 4yi +

=

Y3

Note that the diagonalizing matrix P is symmetric and so the inverse relationship between x and y is :2 3 1 = y = pTx = Px = 3 2 Y3 . 3 3

[-!

8. The given quadratic form can be expressed as Q

= xT Ax

where A= [

The matrix -2 - 4

A has eigenvalues

I and

10. The

eigenspace w 'espond ing to

vec:o" v, -

1, and v 3

[ _:]

mand v,

[

5

.

fo<m a basis fo< the

fo,m s a basis fo' the eigenspace co,esponding to

v:·,;,> p::d u;: ::::::::::::::::::: A and the change of variable x

= Py eliminates the cross product in Q:

Q = xT Ax = yT (PT AP)y = !Y1 Y'2 YJ)

= + Yi + 10y5 0

10.

[:J +

(b)

!x

yj

(a)

lx

y]

(b)

!x Yl [;

17

[ 1 -l][x]

-8)

y -t 15

[:J -

0

10

Y3

5= 0

8} [:) -3 0 =

11. (a) Ellipse

(b ) Hyperbola.

(c) Parabola

(d) Circle

(a ) Ellipse

(b) Hyperbola

(c) Parabola

(d )

12.

13. The CGUation can be written in matrix form as x T Ax= -8 where A -

are

>.1

= 3 and >.2 =

Lhe matrix P

=[

- 2, ·.vith corresponding eigenvectors v 1 =

ts]

-Vi 7s

orthogonally diagonA-Iizes A. Note

= [- 22

Circle

2

- ]. The ei"envalues of A

- 1

v2 = det(P)

-

q:;

respectively. Thus

= 1, so

P is a rotation

EXERCISE SET 8.4

297

matrix. The equation of lhe conic m the rotated x'y'-coordinate system is

which can be written ns 2y' 2 :.,;g'2 = 8.:. thus. axes have bePn rotated 1s 9 = tan 1 ( -26 6°.

14. The equation can be written m matrix form as At

= 3 and

is a.· hyperbola. The a11gle through which the

xT Ax=

.X2 = 7, with corresponding eigenvectors v1

matrix P = [

9 where A=

=

:] . The eigenvalues of A are

and v2

= GJ


orthogonally diagonalizes A. Note that det(P) = 1, soP is a rotation matrix.

The equation of the conic in the rotated x'y'-coordinate system is

[::]

[x' y'] whkh to

ran

It<:

and !lin 0

=

f·

iy'2 = !1· thns th,. lhus 8 = -45°.

=- 9 ThP angle of rotation corresponds

.1n

15. The equation can be writ len in matrix form as xT Ax= 15 where A A are At

= 20 and

A2 = -5, with corresponding e1genvectors v 1

-n

Thus the matrix P

=

=

The etgenvalues of and v 2

=[

respectively.

orthogonally diagooallzE>s A Note that det(P)-= I , soP is a rotation

matrix The equation of the come in the rotated .x'y'-coordinnte system is

wh1ch we can wntf' as ·1r'2 - y' 2 == 3; thus the conir is a hyperbola axes havP been rotntcd IS (I = U\0 t ( ::::; 36. !) 0 16. The equation co.n be wrinen in rnalrix form as xr Ax= nrc .A1 =

4 and

>.2 ;;:

where A

wit.h cur responding eigenvectors v 1

P

= [_

The angle through which the

=

:] . The eigenvalues of A

and v2 -

Thus 1he matrix

dmgonalizes .4. Note that dct(P) = l, soP is a rotat1on matnx. The

equat1on of lhc conic in the rotntrrl x'y'-coordinate system is

0] [x'] =2 y'

;! 2

•

which we can write as :r' 2 1- 3y'2 = 1; thus thP conic is an ellipse. The angle of rotation corresponds lo cosO= and sm (} = thus(}= -45°_

7i,

17. (a) The e1genvalucs of A= (b)

negative drfinil('

(d) positive semidPfimtc

and A= 1 and .X= 2; thus A is positive defimtP (c)

indefinite

(e)

negative semidefinite

298

18.

Chapter 8

(a)

The eigenvalues of A =

(b)

negative definite

0 ] and A -5

= 2 and A =

(c) positive definjte

'

(e)

(d ) negative semidefinite

positive semidefinite

> 0 for (x 1 . x2) 'f: (0, 0) ; thus Q is positive definite.

19.

We have Q = xi+

20.

negative definite

21.

We have Q = (x1 - x2) 2 > 0 for x1

22.

negative semidefinite

23.

We have Q

24.

indefinite

25.

(a) The eigenvalues of the matrix A = Since

-5; thus A is indefinite.

= x{ -

> 0 for

lSI = 5 and

I

5 -2

x1

5

x2

and Q = 0 for

of 0, x 2

= 0 and Q < 0

I

x1

for

are >. 0

0

= 0, xz ..f 0; thus Q is indefinite.

x 1

are A= 3 and A = 7; thus A is posi tive definite.

= 21 are positive, we reach

(b ) The eigenvalues of A=

= x2; thus Q is positive semidefinite.

the same conclusion using Theorem 8.4.5.

= 1, >. = 3,

5

The determinants of the principal submatrices are

and ,\ = 5; thus A is positive definite.

12 -}

121 =

= 3, and - l 0 I

thus we reach the same conclusion using T heorem 8.4.5. 26.

( a)

The eigenvalues of the matrix A =

121= '2

2 1 ] ,1 2

a nd 1

are .A

2 0

= 1 and >. = 3; thus A is positive definite. Since

are >. = 1, .-\ = 3, and ,\ = 4; thus A is positive definite.

0 -1

3

3 -1

The determinants of the principal submatrices are

131= 3,

= 5, and

- 1

thus we reach the

= (-

Thus the matrix P

B =

12;

3

conclusion using Theorem 8.4.5.

The matrix A h.as eigenvalues >- 1 v2

0

2 - 1 =

0 -1

(a)

5

= 3 an• positive, we reach t.hc sttme conclusion using Theorem 8.4.5

(b) The eigenvalues of A= [-

27.

0

o = 15;

=(

-

t]

orthogonally diagonalizes A , and the matrix

7;]

72

has the property that 8 2

= 3 and >.2 = 7, with correspo nding eigenvectors v 1 = g)

- !':<

v2

= A.

=

[Y..il! +- 4:11. 2

2

and

EXERCISE SET 8.4

299

(b ) The matrix A has eigenvnlues >. 1

nl,

m

Thus P

nnd v,

' 2

I, >.2

[

I

D=

v'3

[

}nhogonWiy diagonWizea A, and

[-

(a ) The n1atrix A has eigenvalues >.1 = 1 a.nd >.2 nnd

v2

= 0

!'2 + _il '2

0

0

= A.

has I he property that 8 2 28.

.J5

0

m,

5, with corresponding ttigcnvecto<S v t

-

0

72

>.,

3,

[-%

= [:]. Thus lhc matrix P =

=3,

=

with corresponding eigenvectors v 1

orthogonally dia.gonalizt>.s A, and the matrix

B=

72 has the property Lhnl 8 2

= A.

(b ) The "'""" A has •i•cn<•Jtoes \ 1 v2

-

[-1]

B-

,

and v 3 -

I

[ ']

Thus P

:

ton

[7.

- 72 0

- "J:i

7a

..;'2

73

-t. \IG I

t

I

has t.h P property that 8 2

2 9 . The q11ad rnt tc form Q = tion as Q - x T Ax where A

I,

!SJ = 5. 5

21 =

12 1

1, and

= 0

0

v'3

0

0

[:.e

3, >.,

' , wu h co,.spondin< eigen"""'"" v 1

--j; ; -

[j 1

1

orthogonally dia.gonalizes A, and

2

7s 0

..;2

7J

[:] ,

1

-7:1

6

72 = I

7:i

I j

-

+ Ifl2

§6

6- T

'

I

-3

3

fl 2

V>l

I

-3

= A.

+

+

+ 4x 1 x2 -

2x 1x3 - 2x2x3 can be

in matrix nota-

=:] . The determinants of the principal submatrices of A are

=[ - 1 - 1

s

>.,

k

2 -1

:! 1 - 1 -1 -1 Jc

= k- 2. Thus Q IS positi\·c d••fiuit<' tf and only if k

3 0 . T he quadrat 1r form can be expressed in matrix notation as Q = xr Ax where A

=

;>

2.

u

0 1

k

The del<'rrnlnanls of the pnncipal submatrices of A are Thus Q 1s positive definite i! :1::u only if

lkl < VIS

J31=

3,

= 3, and

3 0 - 1

-tl

Jc . 2

0 -1

Jc

Jc = 2

5- 3k2

300

Chapter 8

31. (a)

= 3 and >.2 = 15, with

The matrix A has e:genvalues

corresponding eigenvectors

and vz = [:]. Thus A is positive definite, the matrix P = [

=

vJ

orthogonally diagonalues

.4, 1\n:i the matrix

8has the property that 8 2 = A. (b) The LOU-decomposition (pl59-160) of the matrix A is

A and, since L =

!] [i

=

[;

=

= LDU

ur, this can be written as

which is a factorization of the required type. 32. (a)

T(x + y)

(x + y)T A{x + y) = (xT

+ yT)A(x + y) = xTAx+ xT Ay +

= xT Ax+ 2xT Ay + yT Ay

(b)

T(cx.)

33. We have

= (cx)T A(cx) =

(ctXl

c2 (xT Ax)=

+ czxz + · + CnXn) 2

2T(x)

n

n

t=l

t=l J=t+l

[q

c1c2

c,cn (a)

For each

xf -

t

1, .. . ,n we have (x;-

£: x,x1 +

(

J=l

f:

r=1

+2

f: f

J=l k=1 +1

s; = n-1 - 1 -[(x the coefficient of

+

x?

[1 -

is

= -n(n2_ 1 >.

1-

*+

c2

C2Cn

n

2x,x + x

2

= x?-

f: x

J=l

1

+

(t

2

x1 )

=

x1xk). Thus in the 4uadratic form

x) 2 +

(x2-

=

lt foUows that s;

x) 2

+ · · + (xn

1 -n(n-1)

I - n{n-1)

_ _1_ n(n-1)

.!.

I - n(n-1)

1 -n{n-1)

1 - n(n-1)

l

n

- x) 2 ]

and the coefficient of x,x1 for

= xT Ax where

l

n

A=

c2 c.,

=

i) 2

= xr Ax where

C1Cn

c,:c2

A=

34.

= T(x) + 2xT Ay + T(y)

L; c'fx? + 2: 2: 2c.c1 x,x1

=

yT Ax+ yT Ay

n

t

:f. j is


301

si

(b) We have = ": 1 1(x, - i)2 + (xz - i) 2 -t- · • · + (xn- :t)2 J;::: 0, and 0 if a nd only if x 1 = i, Xz = i, ... ,:en= :i:, i.e., if and only if x 1 = X2 = · ·::::: Xn. Thus s; is posit1ve semJdefinite.

35. (a)

T he quad ratic form Q = jx 2

tation as Q

The vectorn and v 3

+

x' Ax whe<e .4

v,

[:]

+

[! !l

[-!]

and v,

+

+

+

can be expressed in matrix

>.

The mal
l

and

>.

form a bas;. for the eigenspace corresponding

forms a hasis for the eigenspace corresponding to

>.

!

j. j,

Application of the

Gram-Schmidt process to {v 1, v 2 , v 3 } produces orthonormal eigenvec tors {p 1, P2, P J}, and t he matrix

P

=

Pt

P2

P•J =

[-i

'

76 I

-"JS 1 -76

73 ange of variable x = Px' converts Q into a quadratic

orthogonally diagonahz.es A. Thus L form m the variables x'- (x',y',z') w1thout cross product .

From this 2j"i =

wP

cnnrludc I hat thr equation Q

to an ellipsoid with a.xis lengths

1

J6 in thP x' and y' directions, and 2.jf =

the z' direction

{b) The matnx A must bP. positive definite.

DISCUSSION AND DISCOVERY D l . (a) False. For cxmnple the matrix A=

hn.s r.igcnvalues

l and 3; tlmii A 1s mdefi nite.

(b ) False. The term 4x 1.z:'2x3 is not quadratic tn the variables x 1 , x2, XJ. (c) True. When cxpuudcd, each of t.he of lhe res uiLing expression is (of degree 2) in t he variables (d) True. The eigenvalues of a pos1tJVe definite matnx A are strictly positive, m particular, 0 Is not an eigenvalue of A and so A is invert.1ble

(e) False For examplr th!C' matnx A= (f )

True.

rr the eigenvalues of. \ are posttive,

is pos1t1ve sem1-definite tlu>n the eigenvalues of - A arc negat1ve

0 2. (a) True. When wr itten in matrix form , we have x · x = xT Ax where A = I . (b) True. Tf A has pos1t1ve eigenvalues, then so does A- 1. (c) True. See Theorem 8.11 3(a) (d ) True. Both of thE' principal S'lbmatrices of A will have a positive determinant (e)

False. If A

=(

:] , then xT

is assumed to be symmetric.

= x' + y'<

On the other hand, the statement is true if A

302

Chapter 8

False If c > 0 the graph is an ellipse. If c

(f)

< 0 the graph is empty.

D3. The eigenvalues of A must be pos1lhe and equal to each other; in other words, A must have a positive eigenvalue of muiLlplicily 2.

WORKING WITH PROOFS Pl. Rotating the coordinate axes through an angle 8 corresponds to the change of va.ria.ble x = Px! where P = [:: -:::], i.e., x = x' cosO- y' sinO andy= x' sin 8 + y' cosO. Substituting these expressions into the quadratic form ax 2 coefficient of the cross product term is

+ 2bxy + cy2

leads to Ax'2

+ Bx'y' + Cy'2 ,

where the

B = -2acos0sin 8 + 2b(cos2 0- sin2 0) + 2ccos OsinO = (-a+ c) sin 20 + 2bc'os20 Thus the resulting quadratic form in the variables x' and y' has no cross product ter):nP.if and only if (-a+ c) sin 28 + 2b cos 20 = 0, or (equivalently) cot 28 = P2 .

From the Print1paJ Ax1s Theorem (8.4.1), there is an orthogonal change of varial>le x = Py for wh1ch xr Ax= yl'Dy - >-.t!Jf 1where )q and >.2 are the eigenvalues of A. Since ,\ 1 and >.2 are nonnegat1ve, it follows that xT Ax 0 for every vector x in Rn.

EXERCISE SET 8.5 l.

{a)

The first partial derivatives ot fare fz(x , y) = 4y- 4x 3 and f 11 (x, y) = 4x - 4y3 To find the cnt1cal pomts we sl'l l:r and fv equal to zero. This yields the equations y = x3 and x = y3. 1-\om Lh1s we conduJe Lhat y = y9 and so y =0 or y = ±1 Since x = y3 the corresponding values of x are x = 0 and x = ±1 respectively Thus there are three critical points: (0, 0), (1, 1), and ( 1,-1).

(b) The Jlc:-.sian matrix

IS H(x.y) = [ln((x.y)) r y

/xy((z,y)] fvv z.y)

= r-l42:r

2 4 2] - 12y

Evaluating this matrix

at thP. cnticf\1 points off yaelds 0 [4

/l(O,O)

•cl]

The eigenvalues of

H(l,l)

I

=

[-12 4] 4

= H( - 1, -

H( - 1,-1)

= [

- 12 4

are A= ±4; thus the matrix H(O, 0) is indefinite and so

saddle poinl at (0, 0) . The eigenvalues of

H(l,l)

-12 '

are>.= -8 and

1

has a

>. = - 16; thus Lhe matrix

1) is negative definHe and so f has a relative maximum at {1,1) and at

(-1,-1)

2.

(a) The first partial derivatives off are f%(x, y) = 3x2 - 6y and fv(x, y) = -6x- 3y2 . To find the 2 and x = points we set 1% and f 11 equal to zero. This yields the equations y = 4 From Lhis we conclude that y = h and so y = 0 or y = 2. The corresponding values of x are

4x

x = 0 and

4

- 2 respectavdy. Thus there are two critical points: (0, 0) and ( -2, 2).

{b) The Hessian matrix is H (x,y) [

are

:z;,

1zv((:z:,y))] = ( 6x -6 11111 %, Y

6

- ]. The eigenvalues of -611

H (O,O} =

>. = ±6; this matrix is indefinite and so f has a saddle point at (0, 0) . The

eigenvalues of H ( -2, 2) = and so

= [/u((x,y)) fvz Y

r-

12

6 - ]

-6 -12

are

>. = -6 and >. = -18, this matnx is negative definite

f has a relnLive maximum at ( -2, 2)

305

EXERCISE SET 8.5

1 3 . The constraint equation 4x 2

+ 8y2 =

16 can be rewritten as ( )2

+ ( )2 =

l. Thus, with the change

of variable (x, y) = (2x', v'2y'), the problem is to find the exvreme values of z = xy = 2v'2x'y' subject lo xn

+ y'2 = 1

The

l

Note that. z of A are

and v2

=[

>. 1

'=

= xy =

2v'2x'y' can be expressed as z =

v'2 and>.,= -J2, with corresponding (normalized) eigenvectors v 1 =

Thus the constrained maximum is z

(x,y) = (J2, 1). Similarly, the constrained minimum is z or (x,y)= (-J2,1) 14. The constraint :r2

+ 3y2 =

The eigenvalue:. of A are >.1

= J2 occurring at (x', y') =

= -v'2 occurdng at (x',y') = (-72,

2

+ (4y)2 = 1. Thus, setting (x, y) = (4:r', x 2 + xy 1 2y 2 = 16x'2 t+ ¥v'2 subject

16 can be rewritten as ( )

lhe problem is Lu find the extreme values of z = x' 2 + y' 2 = 1. Note that z = l6x' 2 + t[

where A= [

xrT Ax'

¥y'

=8

2

can be expressed as z

= xfT Ax'

where A =

with correspondmg eigenvectors v, = [

u:: 11 =

= = [ constrained m;ucimum is z = 536 occurring at (x',y') = ± {4, !>or (x, y) = .J:(2J3, 1s>· the constrained mmimum is z =a occurring at (x', y') = -4) or (x, y) = ±{2. -2)

an:! v·/= [

Normalized

to

are u 1 =

and

[_

ll ?

Thus the Similarly,

15. The level curve correspondmg to the constramed maximum 1s the hyp.:!rbola 5x 2 - y 2 = 5; it touches the unit circle at (x, y) = (± 1, 0). The curve corre• ponuing to the constrained minimum IS the hyperbola 5z• = -1, it touches the unit Circle al (:r,y) = (0,±1)

16.

!'he lt·v.. l cun·p l'nrrPspondmg to th"' constramed maxirnum 1s the hype1bola xy1t touches lhr> un1t circle at (x, y) == The level curvn corresponding to the constrained minimum is the hyperbola xy it touches the unit circle at. {x, y) = ±(

72).

-!;

-J2)

'Y

"_.!.2

xy=.!. 1

-4 I

.ry: 2

-2

x.y = _l 1

-4

17. The area of the inscribed rectangle is z = 4xy. where (x, y) is the corner pomt that lies in the first quadrant. Our problem is to find the maximum value of z = 4xy subject to the constrakts x;::: 0, y ;::: 0, x 2 + = 25 The constraint equat1on can be rewritten as x'2 + y'2 = 1 where x = Sx' and y = y'. In of the vanables x' and y', our problem IS to find the maximum value of z = 20x'y' subject to x' 2 + y' 2

eigenvalue of A area

IS

IS

>.

= 10

= l , x';::: 0, y';::: 0.

Note that z

= x'T Ax' where A =

The largest

1

with c:orresponding (normalized) eigenvector [

z = 10. and th1s occurs whl'n (z',y')

= (-jz,

(x,y)

=

Thus the maximum

306

Chapter 8

18. Our problem is to find the extreme values of z = 4x2 - 4xy + y2 s ubject to x 2 + y2 = 25. Setting x = 5x' and y = 5y' this is equivalent to finding the extreme values of z = 100x'2 - lOOx'y' + 25y'2 subject to x'2 t- y' 2

=1

Note that z = xff Ax' where .-\ = [

The eigenvalues of A are .-\ 1

125 t'.nd >.2 = 0, wit h corresponding (normalized) eigenvectors v 1 = [-

and v 2 =

maximum temperature encountered by the ant is z = 125, and th1s occurs at (x', y')

Thus the

= ±(--;h.

75) or

= 0, and t his occurs at (x', y') =

(x, y) = ± (-2-.15, VS). The minimum temperature encountered is z

±(VS,2VS).

or {x,y) =

[

=

DISCUSSION AND DISCOVERY Dl. (a)

We have fz(x , y) = 4x3 and /,Ax, y ) = 4y 3 ; thus f has a critical point at (0, 0). Simifarly, 9:z(x, y) = 4x3 and g11 (x, y) = - 4y3 , and so g has a critical point a.t {0, 0). The Hessian matrices for

f

and g are HJ(x,y) = [

Since H,(O,O)- H9 (0.0)

=

12 ;

2

°Y ] 12 2

and H 9 (x,y)

second derh·ntl\·e test

IS

= [12;

2

respectively.

mconclusive in both cases.

(b) It is clear that f has a relative minimum at {0, 0) since /{0, 0) = 0 and f(x , y) = x4 + y 4 is strictly positive at all other points (x,y). ln contrast, we have g(O,O) = 0, g(x,O) = x 4 > 0 for x ::/; 0, and g(O, y) = -y4 < 0 for y f. 0. Thus g has a saddle point at (0, 0). D2. The eigenvalues of H

are A= 6 and.,\= -2. Thus His indefinite and so the crit1cal points

=

off (if any} nrc s'\dcile pmnts. Starting from fr.Ax y\(.r y} 2 and /y:r(:r y) = /-ry(xy) = 4 it follows, using partial mlegrat.ion, that. the quadratic form f 1s f(x, y) = :r2 + 4xy + y 2 . This function has one cril!cal point (a saddle) which is located at the origin

D3. lfx is a uni t f>lgenvector corresponding to..\. then q(x) = x1'Ax

= xT(.-\x)

= >.(x T x ) = >.(1)

= >..

WORKING WITH PROOFS Pl. First. note that, as in 03, we have u r,Au m = m and u r,A u M = M On the otl.er hand, since u m and liM are orthogon al, we have = = M(u;: u M) = M (O) = 0 and ur,Au m = 0. It follows that if X r

x TcAXc=

j .f-_":r., UM, then 1

(M-e) M-m u rmAum+O+O+ (e-m ) J\f-m

1

u MAUt.t=

(M-e M -m )m + (e-m) M- m M=c

EXERCISE SET 8.6

m

1. The ch.,.a
of AT A are At = 5 and A2 = 0, and cr1 2. The eigenvalues of AT A = u2

=

J9 =

3 are s mgular values of A.

(t

2

o]

is >.2 (>.

- 5); thus the eigenvalues

= v'5 is a singular value of A. =

are A1

= liJ and A2 = 9; thus u 1 = v'i6 = 4 and

Exercise Set 8.6

307

3. The eigenvalues of A7'A

[_

= 5 (i.e., >. = 5 is an eigeJJ.value J5 and cr2 .:: JS.

>. 1 = 5 and >. 2

=

of mulliplicaty 2); thus the 's ingular values of A are crt = 4.

= ('7,

The eigemalucs of ATA values of A are cr 1 =

J4 -

= [Jz v'22] are >.1 = 4 and >.2 v'l =1.

2 and cr2 =

=

and v2

VJ =

"\ Av2 ""

= ../2 ar:d u2 = .J2. =[

-

A•

=2

unit cagenvectors v 1

[-J

0

] [-

0 -·1

=

and v 2

[v2

72 3

= [90

0 ]

0 -4

=

0

16

] are )q

resperLively

The

=3

A 7.

eigen\'t.!Liors " o2

= 2.

0

l]

=[

1

We hnvc

I

0 ct

and v:::

= [-

..L;\v 1 = c

l

t1 2

= .,1,A v 2 =

eigenwcLors we have u 1

anrl

Vt -

=;

Av1 I

[

2.J L1

arf' .\ 1 = 6-t und

•

= -1, with correspomling umt

The :;mgular values of A arc cr 1 = 8 and

= [-jg] and +. -!.,. • v$ v5

8 0 tl

= [33

3 3 3 3 ] [ ] 3 3 3 3 "2

3

J

]

= (18 18

= 7i

I,.,

u 2 - ..L.4v 2 (1,

4 = ![ 2 0

are

>. 1 = 36 and >.2

4

_JA'

vr>

=

[+1. 72J

Thf' vector

72

72

..l. .;:;

= 0, with corresponding unit Ut

=6,

and

u2 must be chosen so thaL {u1 , u 2}

This results in the following singular value

decomposition.

[33 33] =

6) [•-};] =

rPspccth·ely. The only smgular value of A is

is an or thonormal bas1s for R 2 , e.g., u 2 = (-

A =

=

= UEVT

This results in the fo llowing singular value decomposition:

8. The ('tgcnvn.lucs of AT A

=

= 16 and >.2 = 9, with corresponding smgular values of A are u 1 = 4 and

=

c

1 0 4 6 16 ] = [ of A r A = [' ] (

The

(;

[l0 OJI _UEVT

0 v2

\Vp IHl,·e Ut :
(multiplicity 2), and the vectors

This res ults in t.he following singular value decomposition

[11 -1]1 =

of AT A

The

>.

We have Ut = ;, Av1 =

of A:

6.

is

form an orthonormal basis for the eigenspace (which is all of R2 ). The

singular values of A are crt

u2 =

---

=

5. The only eigenvalue of AT A=

1; thus the singular

[60 0OJ [- 72

72

=UEVT

308

Chapter 8

9.

= [_: -:] are >. 1

eigenvalues of AT A = 2 -2

spondtl'g uniL etgenvectors v1 =

A

and

u 1 = ,/18 = 3,/2, and we have u 1

v2

=

= 18 and

.A2

= 0,

with corre-

respectively The only singular value of

= ;, Av, =

[

=! _:] [-

=

Ul

Wemus',choose

u 2 and u 3 so that { u 1, u 2, u 3 } is an o•thono•mal has;s fo• R 3 , e.g., u, = [i]•nd

the

u , = [ -:] . This results io the foUowU.g singuiOI" value deoomposWon'

A=

[-2 2] [ - 1

1

t -1]

=

2 -2

_.£

!

2

3

3

3

0

0

Note. The singular value decomposition is not unique. It depends on the choice of the (extended) orthonormal basis for R 3 . This is just one possibility.

10. Theeigenvalu"',"f ATA ve

-il

=[

= [=:

J -: J- U_;

=:]a<e .> 1 = 18 and .>1

is a unit eigenvedo<eonesponding to .1,

a basts for the eigenspace corresponding to ,\

= 0,

to these vector:. yields orthononnc.l eigen,·eci:.ors

value of A is o- 1 =

v'18 = 3v'2,

and we have u,

v2

= 18.

= .>3 = 0

The vecto<S [-:]and

= - jsj"

[

= a\ Av1 =

and

v3

=

The eigenvalues of AT A

1

_

2

t [.ll-VfJ

=

-[

=

-1

unit eigenvectors v 1 =

and v 2 ==

fom

. The only smgular

[-22 -1 2] [ t =

M

ll.

m

and application of the Gram-Schmidt process

_

must choose the vector u 2 so that { u 1 , u 2 } is an orthonormal basis ror R2 , e.g., u 2 = [

in

The

are At = 3 and

· We

1- This

= UEVT

3v:> .A2

= 2, with corresponding

I

•tlspectively. The singular values or A are u 1 =

J3

and

Exercise Set 8.6

309

singular value decomposition:

[ l [' .fi

I 0

A=

1 l

=

-73

-1 1

?

and v 2 = [_

and

k

u,

=

'+s] [J3 76 I

I

0

72 -"76 1 72

:1 [: i] = r;: ::1 are

unit eigenvectors v 1 = [

o2

0

l

=UEVT

·,

I

12. The eigenvalues or AT A =

'-

,..

I

= 2. We have

[!l

UJ

= 4, with conespondlng

and

resp,.ctively. The sinf?:nlar valtlf'.s <>fA

and

4 0

= 64

...!..

vf> --

J

= 4 0

[_

=

choose

:iS oJS

so that {u, , u,, u,} Is an octhonocma\-basis roc R3 . This cesults m the rollowmg singular

value decornposttton

-::-sI

0][80]['2 1] = l/EVT 1 ll 2 ../, _"J.

0 ·I 0

] 3.

0

v5

Using the smgular vrdtH' decomposition A =

0 ll

... $

" 5

J.J [s

6) = [.J.. -

75

0

0 ] [

0 2

75

-3; '

-:;a

in Exercise 7, we have thr following polar de>composition nf A:

14.

Using the singular value decowpo.<:ition A =

G

*

=

]

-

UE vr found

= UEVT found

in Exercise 8, we have the following polar decomposition of A:

72

r60 001[- 72

Ji

r

, ,_,:!

72

[-

72 72

= [33

31 0 011= PQ

3

310

15.

Chapter 8

ln Exercise 11 we found the following s:ngular value decomposition·

A=

[

10] 1 1

=

[i1

-1 1

Since A has rank 2, the corresponding reduced singular

decomposition is

A=

and the reduced singular value expansion is

16.

Ir: Exercise 10 we found the following singular value decomposition:

A -

[-2 -1 2] = [--3-272 7z] [3J2 00] [ 2

1 -2

0

0 0

315

Since A has rank 1, the corresponding reduced singular value decomposition is

and the

17.

singular \'alue expansion is

The characteristic polynomial of A is (.>. + 1)(.>. - 3) 2 ; thus>. = - 1 is an eigenvalue cf multiplicity 1 and

>.

= 3 is an Pigenvalue of multiplicity 2.

The vector v 1

ing to >. - -1' and the vee toes v' =

m

and

v'

the etgenspace <'orresponding to>.= 3. Thus the matrix P

il [il

= [-

forms a basis for the eigenspace

form an (orthogonal) basis for

= [1:: 1

onalizes A and the eigenvalue decomposition of A is

120] [--3i 072] [-1030 00] [-720 7,010] [0 0 3 0 0 003 72720

A= 2 1 0

=

fvil) orthogonally diag-

Exerc.se Set 8.6

311

The corresponding singular value decomposition of A is obtained by sh1ft1ng t he negat1ve sign from the diagonal factor t o t he second orthogonal factor·

0

7i

-72

A= 2 1 0 = 003

18.

01

T2

0:30 003

0

1

0

1

0 0

[ l [ lol l [ ] [ 1 2

I

72

The characteristic polynomial of A is(..\+ 1)(..\ + 2)(..\- 4), t hus t h:

.12

- 2, and .13

The matrix P

4. Corresponding unit eigenvecto" are

= [v1

0

.l,..

0

-2 A= [ -2 0

0 3

v, [_;,]

0 0 - 2 0 0

---v5

a•

[-1

0

0

m

v, [;,],1v,

A o.nd lhe eigenvalue <.lec9 n position of A is

v 2 v 3j orthogonally diagonalizes

0-2] = [-35

ozrA are ..\1 =

-

'115

The corresponding singular \'alue decomposition of A is obtained by shiflmg the llt!gative signs from the diagonal factor to the SC'cond vrthogonal factor.

0-2] = [.

2

-2

0

0 3

'5

0

-!,.. '11'5

0I

0

01

0 --l.

0

(a)

-!

u,

-!

(b) A = [

20. Since A

1

2

fnrm n basis for col(.4), u 3

and

vr-

- !l

1

'i

form • basis for col(A).L

for m a bes" for row(A). and v 3

12

-'75

!

I

19.

=

I

0

:

=:

12

0

[

nuii (AT),

and

forms n bas1s for row(At:_= nuli(A).

l

2

2

1

I

1

I

2 -2

10 6

= UEVT and

2 -2 I

2

I

2

V is orthogonal we have AV

= UE.

Written in t.olumn vl'ctor form th1s 1s

312

Chapter 8

and, since T(x) = Ax, it follows that 0

0

.. '

0 0

0

0

0

and [T(v1 )] 8 , - 0 for j - k + 1, ... ,n. Thus [TJ 8 •. 8 =E. 21.

Since A'J'A is positive semidefinite and symmetric its eigenvalues arc nonnegative, and its singular values c;Lre the same as its nonzero eigenvalues. On the other hand, the singular values of A are the squnre roots of the nonzero eigenvalues of AT A. Thus the singular values of AT A are the squares o ft hP sing\llar valnc>s of A.

22.

lf A - U'i:.VT is a singular value decomposition of A, then

"vhcre D IS the diagonal matrix having the eigenvalues of AAT (squares of the singular values of A) on its main diagonal Thus ur AATU = D, i.e' u orthogonally diagonahzes AAT . 23.

We have Q

=

l !] r fl 2

_! 4

;,

.:c.1 l

=

[cos e sm8

.

-Sin

8 ]

cos8

where

e = 330°'

lhus multiplication by Q corresponds to

rJ

rotat1on nhout thP orsgm through an angle ('If 330° The symmetric matrix P has eigenvalues A.= 3 and ,\ = l. \\'llh rorrP.spondmg unil eigenvectors u1 = [

and U2

=[

Thus

v = [ul

U2] is

a clmgonnJi?,ing mat.nx for P:

vrp-v·=

_ ! :i] 2

2

[v'03 J32]

!

2

-4] fl 2

From this we conclude tha.L multiplication by P stretches R2 by n factor of 3 m the directron of u1 and by a factor of 1 in the direction vf u 2.


H A Ul.v·r IS a s1ngular value dccomposiLton of an m x n matrix of rank k, then U has size m x m, E has size m x n and V bas size n x n. {b) If A = Ur Er Vt IS a reduced smgular value decomposition of an m x n matnx of rank k, then U has size m x k, E 1 has size k x k and V has size n x k .

02.

If A 1s an invertible matrix, then it.s eigenvalues are nonzero Thus if A = UEVT is a singular valur decomposition of A, then E is invertible and A- 1 = (VT)- 1 E- 1U- 1 = vE- 1 UT. Note also Ihal the clta.gonal "!1lril''i of r;-t are the reciprocals of the diagonal entnes of r;, which are the vnlurs of A -t. Thus A - 1 = VE- 1UT is a singular value decomposition of A - l

(a)

Exercls a Set 8.7

313

D 3. If A and B are orthogonally similar matrices, then there is an orthogonal matrix P such that B = PAPT. Thus, 'fA = (JEVT is a singular value decomposition of A. then B = PAPT

= P(UEVT)pT =

(PU)E(PV)r

is a singular value decomposition of B. It follows that Band A have the same smgular values (the nonzero diagonal entries of E).

D4. If P is the matrix for the or thogonal projection of Rn onto a subspace W of d1mensioo k, then P 2 = P and the eigenvalues of P are ). = 1 (with multiplicity k) and >. = 0 (with multiplicity n- k). T hus t he singular values of Pare u1 = 1, u2 = 1, ... , U!t = 1.

EXERCISE SET 8.7 = 25; thus A += ( AT A)- 1AT= .}5 (3 4]

1. We have AT A = [3 4] 1 2 ) 3 1

2 . WP. have AT A -

2 I

= [66

6 ), 11

thus the pseudoinversc of A is

A' =(AT A)-1 AT= ....L [ 30

:J

3. We ha-·e AT A - [;

A

5. (a)

+_

= [;: T

-(A A)

_1

A

Ll -6] [I 12] = -6 6 1 3 1

[i

0

T _

1

26 -321 [7 0 5] _ [ ! 74 1 0 5 -

[

[1]=

= [1] [I& is]= (ts fs)

[ts

(c)

AA+ =

=

{d)

A+

=Ill is svmmetnc; thus (.4+ A)T = A+ A .

(e)

T he e1genvalues of AAT:; [

=

rn

and v2

=A+

is symmetnc; thus (AA+)T

9 12 ]

12 16

-:J

=[

we have u 1 = o'. AT v 1 = !(3 4]

0 0

CJ =A

A+AA+

Vt

I

s

- goo -32

(b )

vectors

_!!.IS. ]

-370

thus U>e p"'udotnwse of A ts

AA1-A =

= (ls /ll]

= [fs 2iJ·

are

>q

= 25 and

= AA+ .

= 0, with corresponding unit eigen-

respectively. The only singular value of AT is

[i] = [1].

,1 T = [3 •J

=

0'!

=5, and

This results in the singular value decomposition

III [5 OJ

[-! ll

= UEVT

314

Chapter 8

ThP corresponding reduced singular value decomposition is AT

= (3

4]

= fll [5] [l

= UlEl

vr

and from this we obtam (AT>+ =

(f)

"•Ei"•ur =

The eigenvalues of (A+)T A+ = eigenvectors v 1 and we have

u 1 -

[i] 1

ct ,

and v2

A+v•

[i] UJ 111 = are At=

=

1 25

= (A+>r

and A2 = 0, with corresponcling unit

respectively. The only singular value of A+ is a 1

= 5[,\

fs)

[!] =

= t•

[1). This results in the singular value decompo-

sition

The corresponding reduced singular value decomposition is 3

A+= [ 2 5

= u.E. vt

[1) [t)

=

and from thts we obtam

6.

(a)

(b)

.4 1 A:\

.

[l

4 : 11- .4 =

2

5 __ ]_ 30

=

1

[l ;]

-30

2

5

-1] [i

=

r

[I I]

11

7 - 30

- Jo

2

2

=A

[!

7

s

5

I

6 '2!1

3o

- ygI

l

-.t 13

15

l

=[5 0

(c)

AA'

= [;

i]

-1

is symmetric, thus (AA +)T -= AA+

z

-1$

1] [; i]

(d )

(e) The eigenvalues of AAT =

unol eog•nvocto.s

v, },;;

E

is 'ymme.,ic; thus

3:] v,

are A1

W

= 15, >.2 =

*' Hl·

A)T

A.

'2, and AJ

and ., =

f,o

= 0,

nl

with corresponding

The singular values of

Exercise Set 8.7

315

AT a" u, = Jj5 and

u, = :/rv, =

= .Ji. Setting u , = :, AT

v, =

:J [-!] = f,; [_:].we have

[:

*'

[:

[':]

kJ-u 4

-

726

=

't"'

*' 1:1

and

vr1

and from this it foUows t.hat

(C)

= -

The eignn-.;tlues of (A")rAi-

[

:15]

i& 1;

40

-m

43

arc

ns

;+.

conespondmg uniteigenw>CIO<S v' = ).;; [':]. v, = gular values of A ... are

Ot

=

and

02

,\1

=

=ft . ..\2

H]·

and

VJ

,\3 = 0, wtth

= ;;;\;

n]·

The sin·

Seltmg

-30 7

•

l [ I] _I_

v'26

-3 = _L_ [ 3] -1

\113

-2

we have

and fro111 this it follows that 5

(I 1• ) •

...lm

= r I EI-1 urI =

11 .)t%

[

7. The only eigenvalue of AT A only smgular value of A ts a 1

=-

7

Ji9s

!?.5J is

=5

-!... v26

l

[J}:; .J2 [vp _ip] = :13 713 0

0

=A 2 -1

>. 1 = 25, with corresponding unit eigenvector v 1 = [lj. The

We have u 1

= ;, Av1 =

[!] [Ij = [i], and we choose u 2 = [-:1

so that {u 1 , u2} is an orthonormal basis for R 2 • This results in the singular value decomposition

A=

[!]

= [:

=UEVT

316

Chapter 8

T he corresponding reduced singular value decomposition is

and from this we obtain

8. The eigenvalues of AT A = v1

= -Jb [!] and

[!

are

= 7iJ [

)q

= 15 a nd A2

2

j 1

)]-

[23] =

v 13

we have 3 1

with corresponding unit eigenvectors

ThE' singular values of A are cr 1

= cr1 _!_Avt = v iS

Ut

= 2,

= VIs and <12 = V2.

vl95

Setting

7

l

-726 [v'IS J20

Ji9s 7 :t'9s

0

4

725

and it follows that

.4• 9.

1

= V1Ej UT =

The eigenvalues of AT A v1 =

7S

have u,

and v2 . [_

=

choose u3 =

3

[

14 = [32

..n;

32 ]

are

26

)q =

90 and .X2 = 10, w1th correspondmg unit eigenvectors

The smgnla.r values of A are

i]

Jr,;

[;.

= [; ].

u,

<1 1

= '1/'90 = 3v'i0 and

= :,Av, = ,)rn

i]

a2 =

JIO.

= [_;,] , and

\\'e

we

so that {u 1 , u 2, UJ} is an orthonormal basis for R 3 . This yields the singula r value

decomposition 1

l

0] [3v'0l0 v'lO0

71 0

1 0

l [l [

0

0

The corresponding red uced singular value decomposition is 7 1

A

-= 0 0 = 5t:

i)

72 I I

0

I

I

3 0 [ ViQ

J2 -71

0

l[

2

O jjO

7s

-jw

Exercise Set 8 7

317

ancl from t h1s we obtain

.,+ - \.

•

10. The ouly only

2 7s1][1 W'iO 0][1 72 [-35-?s 0 Jrn

[ Gl

o - :ro] 0 Jo

= I·H] is A1 = 41, with correspouJiug unit. value of A is cr1 = .J41. We have u 1 = ;, Av 1 = Jh [:] Ill = (

v1 =

-

I

0

v2

-t

=

0

of A I' A

r-7fl

11 }.

The

and we choose

so that { U j, U2} is an orthonormal basis for R 2 • This results in the s ingular value 7oii decomposition 4 A= ( ] = = ur;vr 5 74i 74i 0 u'l =

[vf.

[J4I]IIJ

The cora esponding reduced singular value decomposition is

and fm111 th1s we obtain

At

:4+ = rlZJ

.

3]-l [22 -1]1

3 5

=

_!_ [ ;:; -3] 16 -3

5

[:!2

1

[1

1

.j

1

'0:.! ,2_] has f11ll column rank, we have A+= (AT A)- 1 A:r·: thus [

13. The mamx A =

:] does not have full column rank, so Formu la (3) does nol apply. The

eageuv
;.)

has full column rank, we have .4+ = (ATA )- 1 AT : thus

11. Smcc A =

12. Smcc rl

VtE} 1 U[= [lJ[7,rr][Jrr

are )q = 4 and .X2

The only singular value

we ,.hoose

7i the smgular value decompmntion

[I1 l]I

or .4

0, with correspondtng unit eagenwc tors v 1 is

Ot

= 2 We have

Ut

=

= [: :]

= =

that {u 1,u 2 } is an orthonormal basis for R2 • Th1s results m

Ut

A=

::::

=

72

72

[20 0o] [-72

The corresponding reduced singular value decomposition is

v"l

= ur:vr

318

Chapter 8

and fro·n th1s we obtain A•

= v, Et 'ur = [

1l 1 [7, 7,] = [:

i]

14. The m t have fuU column rank, so Formula (3} d.:>es not apply. By inspection. we

see that a smgular value decompositiOn of A is given by

A=

0 0

=

0 0 1

= UEVT

0 0

The corresponding reduced singular value decomposition is

cllld

fr0111

llt iS

WE' 0 htO.in

15. The standard mfltrix for the orthogonal proJectmn of R 3 onto col{A} is \ + _ -

[160 -

11

1I :\

[2 1

8]

·f 30

15

5

5

_

; _!

-

[! 6

30

!

_J..

!

3

A 13

u

15

16. The standard matrix for the orthogonal projection of R 3 onto col(A) is

A A+

=

55

[_

0 G

Q

jQ

17. The giv
l

=

001

= b whm A = [

b

matrix A has thf' fnllowinp; red uced s ingular value decomposition and pseudoinverse:

A+

1

V1 E) Uf

=

72

(! f j) = 18

Thus the least squares solut1on of minimwn norm for the system Ax

9 9

=b

is:

=. [_ ]· The

Working with Proofs

319

18. The g;ven system can he wdtten as Ax= b whece A= [:

_!_ [

11

IS -9

9 1 2 2 - ] ( ] 9 l 3 l

=

0

-fii

-

and b = [:] . The pseudo;,..,,. of

fi,]

0

Thus the least squares solution of

2

19° Since AT has full column rank, we have (A+)T = (AT)+ = (AAT)- 1 A = (14)- 1 (I 2 3]

and so A+=

[H

20 . The matrix A T has full column rank and. from Ext:n: t::.c 18,

have (AT)+= [

= (f.

--& I

0

1

2 -2 $ .4 T-- -n

2 14

1

Lhus

0]

-:I .

[ !1

-l

'"

l

DISCUSSION AND DISCOVERY cnJ

1<;

an m x n matrix with orthogonal (nonzero) column vecto rs, then I

1-1 AT= Note If the columns of A r\re orthonormal, then A+

02. (a) If A = (b )

03

If c

D5o (a)

a u v1', then A;.

A • A= ( IS

= lv uTo a = v( u: u vr = vvT

a nomero scalar, l hen (cA )"' = 0

= AT

and

Uc:tl 2 0

0

0

0

0 0

I

jjCJj"'

0

=

= uur.

i .4+ .

AA+ and A+ A are the standard matrices o f orthoJ!'onal projection operators.

fi)

320

Chapter 8

(b)

Using parts (a) and (b) of Theorem 8.7.2, we have (AA+)(AA+) = A(A+ AA+) (A+ A)(A+ A)= Jl,+(AA+ A)= A+ A; thus AA,. and A+ A are idempotent.

= AA+ and

WORKING WITH PROOFS PI. If A has rank k, then we have U'{U1 == V{V1 = h; thus

...

P5. Using P4 and Pl, we have .4+ AA+ =A+ A H A+= A+ . P6. Using P4 and P2, we have (AA+)'r =(A++ A+)T = A++ A+= AA+. P7. First note that, as in Exercise P2, we have AA+ = (U1 L: 1 = U1 U'[. Thus, since the columns of U1 form an orthonormal basis for col(A), the matrix AA+ = U1 UT is the standard matrix of the orthogonal projection of R" onto col(A). P8. It follows from Exercise P7 t hat AT(AT)+ is the standard matrix of the orthogonal projection of R" onto coi(AT) =·row(A). Furthermore, using parts (d), (e), and (f) of Theorem 8.7.2, we have

and so .4+ A is the matrix of the orthogonal projection of R" onto row(A) .

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