75686648 Solution Manual

Applied Linear Algebra

Instructor’s Solutions Manual

by Peter J. Olver and Chehrzad Shakiban

Table of Contents

Chapter Page

1. Linear Algebraic Systems . . . . . . . . . . . . . . . . . . . . . . . . . 1

2. Vector Spaces and Bases . . . . . . . . . . . . . . . . . . . . . . . . . 46

3. Inner Products and Norms . . . . . . . . . . . . . . . . . . . . . . . 78

4. Minimization and Least Squares Approximation . . . . . . . 114

5. Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

6. Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

7. Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

8. Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

9. Linear Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . 262

10. Iteration of Linear Systems . . . . . . . . . . . . . . . . . . . . . . 306

11. Boundary Value Problems in One Dimension . . . . . . . . . 346

1

Solutions — Chapter 1

1.1.1.(a) Reduce the system to x − y = 7, 3y = −4; then use Back Substitution to solve for

x = 173 , y = − 4

3 .

(b) Reduce the system to 6u + v = 5, − 52 v = 5

2 ; then use Back Substitution to solve foru = 1, v = −1.

(c) Reduce the system to p + q − r = 0, −3q + 5r = 3, −r = 6; then solve for p = 5, q =−11, r = −6.

(d) Reduce the system to 2u − v + 2w = 2, − 32 v + 4w = 2, −w = 0; then solve for

u = 13 , v = − 4

3 , w = 0.

(e) Reduce the system to 5x1 + 3x2 − x3 = 9, 15 x2 − 2

5 x3 = 25 , 2x3 = −2; then solve for

x1 = 4, x2 = −4, x3 = −1.(f ) Reduce the system to x + z − 2w = −3, −y + 3w = 1, −4z − 16w = −4, 6w = 6; then

solve for x = 2, y = 2, z = −3, w = 1.(g) Reduce the system to 3x1 + x2 = 1, 8

3 x2 + x3 = 23 , 21

8 x3 + x4 = 34 , 55

21 x4 = 57 ; then

solve for x1 = 311 , x2 = 2

11 , x3 = 211 , x4 = 3

11 .

1.1.2. Plugging in the given values of x, y and z gives a+2b− c = 3, a−2− c = 1, 1+2b+c = 2.Solving this system yields a = 4, b = 0, and c = 1.

♥ 1.1.3.(a) With Forward Substitution, we just start with the top equation and work down. Thus

2x = −6 so x = −3. Plugging this into the second equation gives 12 + 3y = 3, and soy = −3. Plugging the values of x and y in the third equation yields −3 + 4(−3)− z = 7,and so z = −22.

(b) We will get a diagonal system with the same solution.(c) Start with the last equation and, assuming the coefficient of the last variable is 6= 0, use

the operation to eliminate the last variable in all the preceding equations. Then, againassuming the coefficient of the next-to-last variable is non-zero, eliminate it from all butthe last two equations, and so on.

(d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ).Solving the reduced system by Forward Substitution reproduces the same solution (asit must):

(a) The system reduces to 32 x = 17

2 , x + 2y = 3.

(b) The reduced system is 152 u = 15

2 , 3u− 2v = 5.(c) The method doesn’t work since r doesn’t appear in the last equation.

(d) Reduce the system to 32 u = 1

2 , 72 u− v = 5

2 , 3u− 2w = −1.

(e) Reduce the system to 23 x1 = 8

3 , 4x1 + 3x2 = 4, x1 + x2 + x3 = −1.(f ) Doesn’t work since, after the first reduction, z doesn’t occur in the next to last

equation.(g) Reduce the system to 55

21 x1 = 57 , x2 + 21

8 x3 = 34 , x3 + 8

3 x4 = 23 , x3 + 3x4 = 1.

1.2.1. (a) 3× 4, (b) 7, (c) 6, (d) (−2 0 1 2 ), (e)

0B@

02−6

1CA.

1

1.2.2. (a)

0B@

1 2 34 5 67 8 9

1CA, (b)

1 2 31 4 5

!, (c)

0B@

1 2 3 44 5 6 77 8 9 3

1CA, (d) ( 1 2 3 4 ),

(e)

0B@

123

1CA, (f ) ( 1 ).

1.2.3. x = − 13 , y = 4

3 , z = − 13 , w = 2

3 .

1.2.4.

(a) A =

1 −11 2

!, x =

xy

!, b =

73

!;

(b) A =

6 13 −2

!, x =

uv

!, b =

55

!;

(c) A =

0B@

1 1 −12 −1 3−1 −1 0

1CA, x =

0B@

pqr

1CA, b =

0B@

036

1CA;

(d) A =

0B@

2 1 2−1 3 3

4 −3 0

1CA, x =

0B@

uvw

1CA, b =

0B@

3−2

7

1CA;

(e) A =

0B@

5 3 −13 2 −11 1 2

1CA, x =

0B@

x1x2x3

1CA, b =

0B@

95−1

1CA;

(f ) A =

0BBB@

1 0 1 −22 −1 2 −10 −6 −4 21 3 2 −1

1CCCA, x =

0BBB@

xyzw

1CCCA, b =

0BBB@

−3321

1CCCA;

(g) A =

0BBB@

3 1 0 01 3 1 00 1 3 10 0 1 3

1CCCA, x =

0BBB@

x1x2x3x4

1CCCA, b =

0BBB@

1111

1CCCA.

1.2.5.(a) x− y = −1, 2x + 3y = −3. The solution is x = − 6

5 , y = − 15 .

(b) u + w = −1, u + v = −1, v + w = 2. The solution is u = −2, v = 1, w = 1.(c) 3x1 − x3 = 1, −2x1 − x2 = 0, x1 + x2 − 3x3 = 1.

The solution is x1 = 15 , x2 = − 2

5 , x3 = − 25 .

(d) x + y − z − w = 0, −x + z + 2w = 4, x− y + z = 1, 2y − z + w = 5.The solution is x = 2, y = 1, z = 0, w = 3.

1.2.6.

(a) I =

0BBB@

1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCA, O =

0BBB@

0 0 0 00 0 0 00 0 0 00 0 0 0

1CCCA.

(b) I + O = I , IO = O I = O. No, it does not.

1.2.7. (a) undefined, (b) undefined, (c)

3 6 0−1 4 2

!, (d) undefined, (e) undefined,

(f )

0B@

1 11 93 −12 −127 8 8

1CA, (g) undefined, (h)

0B@

9 −2 14−8 6 −1712 −3 28

1CA, (i) undefined.

2

1.2.8. Only the third pair commute.

1.2.9. 1, 6, 11, 16.

1.2.10. (a)

0B@

1 0 00 0 00 0 −1

1CA, (b)

0BBB@

2 0 0 00 −2 0 00 0 3 00 0 0 −3

1CCCA.

1.2.11. (a) True, (b) true.

♥ 1.2.12. (a) Let A =

x yz w

!. Then AD =

ax byaz bw

!=

ax aybz bw

!= DA, so if a 6= b these

are equal if and only if y = z = 0. (b) Every 2 × 2 matrix commutes with

a 00 a

!= a I .

(c) Only 3 × 3 diagonal matrices. (d) Any matrix of the form A =

0B@

x 0 00 y z0 u v

1CA. (e) Let

D = diag (d1, . . . , dn). The (i, j) entry of AD is aij dj . The (i, j) entry of DA is di aij . If

di 6= dj , this requires aij = 0, and hence, if all the di’s are different, then A is diagonal.

1.2.13. We need A of size m × n and B of size n ×m for both products to be defined. Further,AB has size m×m while BA has size n× n, so the sizes agree if and only if m = n.

1.2.14. B =

x y0 x

!where x, y are arbitrary.

1.2.15. (a) (A + B)2 = (A + B)(A + B) = AA + AB + BA + BB = A2 + 2AB + B2, since

AB = BA. (b) An example: A =

1 20 1

!, B =

0 01 0

!.

1.2.16. If AB is defined and A is m×n matrix, then B is n×p matrix and AB is m×p matrix;on the other hand if BA is defined we must have p = m and BA is n × n matrix. Now,since AB = BA, we must have p = m = n.

1.2.17. A On×p = Om×p, Ol×m A = Ol×n.

1.2.18. The (i, j) entry of the matrix equation cA = O is caij = 0. If any aij 6= 0 then c = 0, so

the only possible way that c 6= 0 is if all aij = 0 and hence A = O.

1.2.19. False: for example,

1 00 0

! 0 01 0

!=

0 00 0

!.

1.2.20. False — unless they commute: AB = BA.

1.2.21. Let v be the column vector with 1 in its jth position and all other entries 0. Then Avis the same as the jth column of A. Thus, the hypothesis implies all columns of A are 0and hence A = O.

1.2.22. (a) A must be a square matrix. (b) By associativity, AA2 = AAA = A2 A = A3.(c) The naıve answer is n − 1. A more sophisticated answer is to note that you can com-

pute A2 = AA, A4 = A2 A2, A8 = A4 A4, and, by induction, A2r

with only r matrixmultiplications. More generally, if the binary expansion of n has r+1 digits, with s nonzerodigits, then we need r + s − 1 multiplications. For example, A13 = A8A4A since 13 is 1101in binary, for a total of 5 multiplications: 3 to compute A2, A4 and A8, and 2 more to mul-tiply them together to obtain A13.

3

1.2.23. A =

0 10 0

!.

♦ 1.2.24. (a) If the ith row of A has all zero entries, then the (i, j) entry of AB is ai1b1j + · · · +ainbnj = 0b1j + · · ·+ 0bnj = 0, which holds for all j, so the ith row of AB will have all 0’s.

(b) If A =

1 10 0

!, B =

1 23 4

!, then BA =

1 13 3

!.

1.2.25. The same solution X =

−1 1

3 −2

!in both cases.

1.2.26. (a)

4 51 2

!, (b)

5 −1−2 1

!. They are not the same.

1.2.27. (a) X = O. (b) Yes, for instance, A =

1 20 1

!, B =

3 2−2 −1

!, X =

1 01 1

!.

1.2.28. A = (1/c) I when c 6= 0. If c = 0 there is no solution.

♦ 1.2.29.(a) The ith entry of A z is 1 ai1+1 ai2+ · · ·+1 ain = ai1+ · · ·+ain, which is the ith row sum.

(b) Each row of W has n− 1 entries equal to1n and one entry equal to

1− nn and so its row

sums are (n− 1)1n +

1− nn = 0. Therefore, by part (a), W z = 0. Consequently, the

row sums of B = AW are the entries of B z = AW z = A0 = 0, and the result follows.

(c) z =

0B@

111

1CA, and so Az =

0B@

1 2 −12 1 3−4 5 −1

1CA

0B@

111

1CA =

0B@

260

1CA, while B = AW =

0BB@

1 2 −1

2 1 3

− 4 5 −1

1CCA

0BBB@

− 23

13

13

13 − 2

313

13

13 − 2

3

1CCCA =

0BB@

− 13 − 4

353

0 1 −1

4 −5 1

1CCA, and so B z =

0B@

000

1CA.

♦ 1.2.30. Assume A has size m× n, B has size n× p and C has size p× q. The (k, j) entry of BC

ispX

l=1

bklclj , so the (i, j) entry of A(BC) isnX

k=1

aik

0@

pX

l=1

bklclj

1A =

nX

k=1

pX

l=1

aikbklclj .

On the other hand, the (i, l) entry of AB iskX

i=1

aikbkl, so the (i, j) entry of (AB)C ispX

l=1

0@

nX

k=1

aikbkl

1A clj =

nX

k=1

pX

l=1

aikbklclj . The two results agree, and so A(BC) =

(AB)C. Remark : A more sophisticated, simpler proof can be found in Exercise 7.1.44.

♥ 1.2.31.(a) We need AB and BA to have the same size, and so this follows from Exercise 1.2.13.(b) AB −BA = O if and only if AB = BA.

(c) (i)

−1 2

6 1

!, (ii)

0 00 0

!, (iii)

0B@

0 1 11 0 1−1 1 0

1CA;

(d) (i) [ cA + dB, C ] = (cA + dB)C − C(cA + dB)

= c(AC − C A) + d(BC − C B) = c [ A, B ] + d [ B, C ],

[ A, cB + dC ] = A(cB + dC)− (cB + dC)A

= c(AB −BA) + d(AC − C A) = c [ A, B ] + d [ A, C ].(ii) [ A, B ] = AB −BA = − (BA−AB) = − [ B, A ].

4

(iii)h[ A, B ], C

i= (AB −BA)C − C (AB −BA) = ABC −BAC − C AB + C BA,

h[ C, A ], B

i= (C A−AC)B −B (C A−AB) = C AB −AC B −BC A + BAC,

h[ B, C ], A

i= (BC − C B)A−A(BC − C B) = BC A− C BA−ABC + AC B.

Summing the three expressions produces O.

♦ 1.2.32. (a) (i) 4, (ii) 0, (b) tr(A + B) =nX

i=1

(aii + bii) =nX

i=1

aii +nX

i=1

bii = tr A + tr B.

(c) The diagonal entries of AB arenX

j =1

aij bji, so tr(AB) =nX

i=1

nX

j =1

aij bji; the diagonal

entries of BA arenX

i=1

bji aij , so tr(BA) =nX

i=1

nX

j =1

bji aij . These double summations are

clearly equal. (d) tr C = tr(AB −BA) = tr AB − tr BA = 0 by part (a).(e) Yes, by the same proof.

♦ 1.2.33. If b = Ax, then bi = ai1x1 + ai2x2 + · · · + ainxn for each i. On the other hand,

cj = (a1j , a2j , . . . , anj)T , and so the ith entry of the right hand side of (1.13) is

x1ai1 + x2ai2 + · · ·+ xnain, which agrees with the expression for bi.

♥ 1.2.34.(a) This follows by direct computation.(b) (i) −2 1

3 2

! 1 −21 0

!=

−2

3

!( 1 −2 ) +

12

!( 1 0 ) =

−2 4

3 −6

!+

1 02 0

!=

−1 4

5 −6

!.

(ii)

1 −2 0−3 −1 2

!0B@

2 5−3 0

1 −1

1CA =

1−3

!( 2 5 ) +

−2−1

!(−3 0 ) +

02

!( 1 −1 )

=

2 5−6 −15

!+

6 03 0

!+

0 02 −2

!=

8 5−1 −17

!.

(iii)0B@

3 −1 1−1 2 1

1 1 −5

1CA

0B@

2 3 03 −1 40 4 1

1CA =

0B@

3−1

1

1CA( 2 3 0 ) +

0B@−1

21

1CA( 3 −1 4 ) +

0B@

11−5

1CA( 0 4 1 )

=

0B@

6 9 0−2 −3 0

2 3 0

1CA+

0B@−3 1 −4

6 −2 83 −1 4

1CA+

0B@

0 4 10 4 10 −20 −5

1CA =

0B@

3 14 −34 −1 95 −18 −1

1CA.

(c) If we set B = x, where x is an n× 1 matrix, then we obtain (1.14).

(d) The (i, j) entry of AB isnX

k=1

aikbkj . On the other hand, the (i, j) entry of ck rk equals

the product of the ith entry of ck, namely aik, with the jth entry of rk, namely bkj .

Summing these entries, aikbkj , over k yields the usual matrix product formula.

♥ 1.2.35.

(a) p(A) = A3 − 3A + 2 I , q(A) = 2A2 + I . (b) p(A) =

−2 −8

4 6

!, q(A) =

−1 0

0 −1

!.

(c) p(A)q(A) = (A3 − 3A + 2 I )(2A2 + I ) = 2A5 − 5A3 + 4A2 − 3A + 2 I , while

p(x)q(x) = 2x5 − 5x3 + 4x2 − 3x + 2.(d) True, since powers of A mutually commute. For the particular matrix from (b),

p(A) q(A) = q(A) p(A) =

2 8−4 −6

!.

5

♥ 1.2.36.

(a) Check that S2 = A by direct computation. Another example: S =

2 00 2

!. Or, more

generally, 2 times any of the matrices in part (c).

(b) S2 is only defined if S is square.

(c) Any of the matrices

±1 0

0 ±1

!,

a bc −a

!, where a is arbitrary and bc = 1− a2.

(d) Yes: for example

0 −11 0

!.

♥ 1.2.37. (a) M has size (i+j)×(k+l). (b) M =

0BBBBB@

1 1 −13 0 11 1 3−2 2 0

1 1 −1

1CCCCCA

. (c) Since matrix addition is

done entry-wise, adding the entries of each block is the same as adding the blocks. (d) Xhas size k ×m, Y has size k × n, Z has size l ×m, and W has size l × n. Then AX + BZwill have size i ×m. Its (p, q) entry is obtained by multiplying the pth row of M times theqth column of P , which is ap1x1q + · · · + apixiq + bp1z1q + · · · + bplzlq and equals the

sum of the (p, q) entries of AX and BZ. A similar argument works for the remaining three

blocks. (e) For example, if X = (1), Y = ( 2 0 ), Z =

01

!, W =

0 −11 0

!, then

P =

0B@

1 2 00 0 −11 1 0

1CA, and so M P =

0BBBBB@

0 1 −14 7 04 5 −1−2 −4 −2

0 1 −1

1CCCCCA

. The individual block products are

04

!=

13

!(1) +

1 −10 1

! 01

!,

0B@

4−2

0

1CA =

0B@

1−2

1

1CA (1) +

0B@

1 32 01 −1

1CA

01

!,

1 −17 0

!=

13

!( 2 0 ) +

1 −10 1

! 0 −11 0

!,

0B@

5 −1−4 −2

1 −1

1CA =

0B@

1−2

1

1CA ( 2 0 ) +

0B@

1 32 01 −1

1CA

0 −11 0

!.

1.3.1.

(a)

1 7−2 −9

˛˛˛

42

!2R1+R2

−→

1 70 5

˛˛˛

410

!. Back Substitution yields x2 = 2, x1 = −10.

(b)

3 −52 1

˛˛˛−1

8

!−

23R1+R2

−→

3 −50 13

3

˛˛˛−1263

!. Back Substitution yields w = 2, z = 3.

(c)

0B@

1 −2 10 2 −8−4 5 9

˛˛˛˛

08−9

1CA

4R1+R3

−→0B@

1 −2 10 2 −80 −3 13

˛˛˛˛

08−9

1CA

3

2R2+R3

−→0B@

1 −2 10 2 −80 0 1

˛˛˛˛

083

1CA.

Back Substitution yields z = 3, y = 16, x = 29.

(d)

0B@

1 4 −2−2 0 −3

3 −2 2

˛˛˛˛

1−7−1

1CA

2R1+R2

−→0B@

1 4 −20 8 −73 −2 2

˛˛˛˛

1−5−1

1CA

−3R1+R3

−→0B@

1 4 −20 8 −70 −14 8

˛˛˛˛

1−5−4

1CA

74R2+R3

−→0B@

1 4 −20 8 −70 0 − 17

4

˛˛˛˛

1−5− 51

4

1CA. Back Substitution yields r = 3, q = 2, p = −1.

6

(e)

0BBB@

1 0 −2 00 1 0 −10 −3 2 0−4 0 0 7

˛˛˛˛˛

−120−5

1CCCA reduces to

0BBB@

1 0 −2 00 1 0 −10 0 2 −30 0 0 −5

˛˛˛˛˛

−126

15

1CCCA.

Solution: x4 = −3, x3 = − 32 , x2 = −1, x1 = −4.

(f )

0BBB@

−1 3 −1 11 −1 3 −10 1 −1 44 −1 1 0

˛˛˛˛˛

−2075

1CCCA reduces to

0BBB@

−1 3 −1 10 2 2 00 0 −2 40 0 0 −24

˛˛˛˛˛

−2−2

8−48

1CCCA.

Solution: w = 2, z = 0, y = −1, x = 1.

1.3.2.(a) 3x + 2y = 2, −4x− 3y = −1; solution: x = 4, y = −5,(b) x + 2y = −3, −x + 2y + z = −6, −2x− 3z = 1; solution: x = 1, y = −2, z = −1,(c) 3x− y + 2z = −3, −2y − 5z = −1, 6x− 2y + z = −3;

solution: x = 23 , y = 3, z = −1,

(d) 2x− y = 0, −x + 2y − z = 1, −y + 2z − w = 1, −z + 2w = 0;solution: x = 1, y = 2, z = 2, w = 1.

1.3.3. (a) x = 173 , y = − 4

3 ; (b) u = 1, v = −1; (c) u = 32 , v = − 1

3 , w = 16 ; (d) x1 =

113 , x2 = − 10

3 , x3 = − 23 ; (e) p = − 2

3 , q = 196 , r = 5

2 ; (f ) a = 13 , b = 0, c = 4

3 , d = − 23 ;

(g) x = 13 , y = 7

6 , z = − 83 , w = 9

2 .

1.3.4. Solving 6 = a + b + c, 4 = 4a + 2b + c, 0 = 9a + 3b + c, yields a = −1, b = 1, c = 6, soy = −x2 + x + 6.

1.3.5.

(a) Regular:

2 11 4

!−→

2 10 7

2

!.

(b) Not regular.

(c) Regular:

0B@

3 −2 1−1 4 −3

3 −2 5

1CA −→

0B@

3 −2 10 10

3 − 83

0 0 4

1CA.

(d) Not regular:

0B@

1 −2 3−2 4 −1

3 −1 2

1CA −→

0B@

1 −2 30 0 50 5 −7

1CA.

(e) Regular:0BBB@

1 3 −3 0−1 0 −1 2

3 3 −6 12 3 −3 5

1CCCA −→

0BBB@

1 3 −3 00 3 −4 20 −6 3 10 −3 3 5

1CCCA −→

0BBB@

1 3 −3 00 3 −4 20 0 −5 50 0 −1 7

1CCCA −→

0BBB@

1 3 −3 00 3 −4 20 0 −5 50 0 0 6

1CCCA.

1.3.6.

(a)

− i 1 + i

1− i 1

˛˛˛−1−3 i

!−→

− i 1 + i0 1− 2 i

˛˛˛−1

1− 2 i

!;

use Back Substitution to obtain the solution y = 1, x = 1− 2 i .

(b)

0B@

i 0 1− i0 2 i 1 + i−1 2 i i

˛˛˛˛

2 i2

1− 2 i

1CA −→

0B@

i 0 1− i0 2 i 1 + i0 0 −2− i

˛˛˛˛

2 i2

1− 2 i

1CA.

solution: z = i , y = − 12 − 3

2 i , x = 1 + i .

(c)

1− i 2− i 1 + i

˛˛˛

i−1

!−→

1− i 2

0 2 i

˛˛˛

i− 3

2 − 12 i

!;

solution: y = − 14 + 3

4 i , x = 12 .

7

(d)

0B@

1 + i i 2 + 2 i1− i 2 i3− 3 i i 3− 11 i

˛˛˛˛

006

1CA −→

0B@

1 + i i 2 + 2 i0 1 −2 + 3 i0 0 −6 + 6 i

˛˛˛˛

006

1CA;

solution: z = − 12 − 1

2 i , y = − 52 + 1

2 i , x = 52 + 2 i .

1.3.7. (a) 2x = 3, −y = 4, 3z = 1, u = 6, 8v = −24. (b) x = 32 , y = −4, z = 1

3 ,

u = 6, v = −3. (c) You only have to divide by each coefficient to find the solution.

♦ 1.3.8. 0 is the (unique) solution since A0 = 0.

♠ 1.3.9.

Back Substitution

start

set xn = cn/unn

for i = n− 1 to 1 with increment −1

set xi =1

uii

0@ ci −

i+1X

j =1

uijxj

1A

next j

end

1.3.10. Since a11 a120 a22

! b11 b120 b22

!=

a11b11 a11b12 + a12b22

0 a22b22

!,

b11 b120 b22

! a11 a120 a22

!=

a11b11 a22b12 + a12b11

0 a22b22

!,

the matrices commute if and only if

a11b12 + a12b22 = a22b12 + a12b11, or (a11 − a22)b12 = a12(b11 − b22).

1.3.11. Clearly, any diagonal matrix is both lower and upper triangular. Conversely, A beinglower triangular requires that aij = 0 for i < j; A upper triangular requires that aij = 0 for

i > j. If A is both lower and upper triangular, aij = 0 for all i 6= j, which implies A is a

diagonal matrix.

♦ 1.3.12.

(a) Set lij =

(aij , i > j,

0, i ≤ j,, uij =

(aij , i < j,

0, i ≥ j,dij =

(aij , i = j,

0, i 6= j.

(b) L =

0B@

0 0 01 0 0−2 0 0

1CA , D =

0B@

3 0 00 −4 00 0 5

1CA , U =

0B@

0 1 −10 0 20 0 0

1CA .

♦ 1.3.13.

(a) By direct computation, A2 =

0B@

0 0 10 0 00 0 0

1CA, and so A3 = O.

(b) Let A have size n × n. By assumption, aij = 0 whenever i > j − 1. By induction, one

proves that the (i, j) entries of Ak are all zero whenever i > j − k. Indeed, to compute

the (i, j) entry of Ak+1 = A Ak you multiply the ith row of A, whose first i entries are 0,

8

by the jth column of Ak, whose first j − k − 1 entries are non-zero, and all the rest arezero, according to the induction hypothesis; therefore, if i > j − k − 1, every term in thesum producing this entry is 0, and the induction is complete. In particular, for k = n,

every entry of Ak is zero, and so An = O.

(c) The matrix A =

1 1−1 −1

!has A2 = O.

1.3.14.(a) Add −2 times the second row to the first row of a 2× n matrix.(b) Add 7 times the first row to the second row of a 2× n matrix.(c) Add −5 times the third row to the second row of a 3× n matrix.

(d) Add 12 times the first row to the third row of a 3× n matrix.

(e) Add −3 times the fourth row to the second row of a 4× n matrix.

1.3.15. (a)

0BBB@

1 0 0 00 1 0 00 0 1 00 0 1 1

1CCCA, (b)

0BBB@

1 0 0 00 1 0 00 0 1 −10 0 0 1

1CCCA, (c)

0BBB@

1 0 0 30 1 0 00 0 1 00 0 0 1

1CCCA, (d)

0BBB@

1 0 0 00 1 0 00 0 1 00 −2 0 1

1CCCA.

1.3.16. L3 L2 L1 =

0B@

1 0 02 1 00 − 1

2 1

1CA 6= L1L2L3.

1.3.17. E3 E2 E1 =

0B@

1 0 0−2 1 0−2 1

2 1

1CA, E1 E2 E3 =

0B@

1 0 0−2 1 0−1 1

2 1

1CA. The second is easier to predict

since its entries are the same as the corresponding entries of the Ei.

1.3.18.(a) Suppose that E adds c 6= 0 times row i to row j 6= i, while eE adds d 6= 0 times row k to

row l 6= k. If r1, . . . , rn are the rows, then the effect of eE E is to replace(i) rj by rl + cri + drk for j = l;

(ii) rj by rj + cri and rl by rl + (cd)ri + drj for j = k;

(iii) rj by rj + cri and rl by rl + drk otherwise.

On the other hand, the effect of E eE is to replace(i) rj by rl + cri + drk for j = l;

(ii) rj by rj + cri + (cd)rk and rl by rl + drk for i = l;

(iii) rj by rj + cri and rl by rl + drk otherwise.

Comparing results, we see that E eE = eE E whenever i 6= l and j 6= k.(b) E1E2 = E2E1, E1E3 6= E3E1, and E3E2 = E2E3.(c) See the answer to part (a).

1.3.19. (a) Upper triangular; (b) both special upper and special lower triangular; (c) lowertriangular; (d) special lower triangular; (e) none of the above.

1.3.20. (a) aij = 0 for all i 6= j; (b) aij = 0 for all i > j; (c) aij = 0 for all i > j and aii = 1

for all i; (d) aij = 0 for all i < j; (e) aij = 0 for all i < j and aii = 1 for all i.

♦ 1.3.21.(a) Consider the product L M of two lower triangular n× n matrices. The last n− i entries

in the ith row of L are zero, while the first j − 1 entries in the jth column of M are zero.So if i < j each summand in the product of the ith row times the jth column is zero,

9

and so all entries above the diagonal in LM are zero.(b) The ith diagonal entry of LM is the product of the ith diagonal entry of L times the ith

diagonal entry of M .(c) Special matrices have all 1’s on the diagonal, and so, by part (b), does their product.

1.3.22. (a) L =

1 0−1 1

!, U =

1 30 3

!, (b) L =

1 03 1

!, U =

1 30 −8

!,

(c) L =

0B@

1 0 0−1 1 0

1 0 1

1CA, U =

0B@−1 1 −1

0 2 00 0 3

1CA, (d) L =

0B@

1 0 012 1 0

0 13 1

1CA, U =

0B@

2 0 30 3 − 1

2

0 0 76

1CA,

(e) L =

0B@

1 0 0−2 1 0−1 −1 1

1CA, U =

0B@−1 0 0

0 −3 00 0 2

1CA, (f ) L =

0B@

1 0 02 1 0−3 1

3 1

1CA, U =

0B@

1 0 −10 3 40 0 − 13

3

1CA,

(g) L =

0BBBB@

1 0 0 00 1 0 0−1 3

2 1 0

0 − 12 3 1

1CCCCA

, U =

0BBBB@

1 0 −1 00 2 −1 −1

0 0 12

72

0 0 0 −10

1CCCCA

, (h) L =

0BBB@

1 0 0 0−1 1 0 0−2 1 1 0

3 −1 −2 1

1CCCA,

U =

0BBB@

1 1 −2 30 3 1 30 0 −4 10 0 0 1

1CCCA, (i) L =

0BBBBB@

1 0 0 012 1 0 032 − 3

7 1 012

17 − 5

22 1

1CCCCCA

, U =

0BBBBB@

2 1 3 1

0 72 − 3

212

0 0 − 227

57

0 0 0 3522

1CCCCCA

.

1.3.23. (a) Add 3 times first row to second row. (b) Add −2 times first row to third row.(c) Add 4 times second row to third row.

1.3.24.

(a)

0BBB@

1 0 0 02 1 0 03 4 1 05 6 7 1

1CCCA

(b) (1) Add −2 times first row to second row. (2) Add −3 times first row to third row.(3) Add −5 times first row to fourth row. (4) Add −4 times second row to third row.(5) Add −6 times second row to fourth row. (6) Add −7 times third row to fourth row.

(c) Use the order given in part (b).

♦ 1.3.25. See equation (4.51) for the general case.

1 1t1 t2

!=

1 0t1 1

! 1 10 t2 − t1

!

0B@

1 1 1t1 t2 t3t21 t22 t23

1CA =

0B@

1 0 0t1 1 0

t21 t1 + t2 1

1CA

0B@

1 1 10 t2 − t1 t3 − t10 0 (t3 − t1)(t3 − t2)

1CA ,

0BBBBB@

1 1 1 1

t1 t2 t3 t4t21 t22 t23 t24t31 t32 t33 t34

1CCCCCA

=

0BBBBB@

1 0 0 0

t1 1 0 0

t21 t1 + t2 1 0

t31 t21 + t1 t2 + t22 t1 + t2 + t3 1

1CCCCCA

0BBBBB@

1 1 1 1

0 t2 − t1 t3 − t1 t4 − t10 0 (t3 − t1)(t3 − t2) (t4 − t1)(t4 − t2)

0 0 0 (t4 − t1)(t4 − t2)(t4 − t3)

1CCCCCA

.

10

1.3.26. False. For instance

1 11 0

!is regular. Only if the zero appear in the (1, 1) position

does it automatically preclude regularity of the matrix.

1.3.27. (n− 1) + (n− 2) + · · ·+ 1 =n(n− 1)

2.

1.3.28. We solve the equation

1 0l 1

! u1 u20 u3

!=

a bc d

!for u1, u2, u3, l, where a 6= 0 since

A =

a bc d

!is regular. This matrix equation has a unique solution: u1 = a, u2 = b,

u3 = d− bc

a, l =

c

a.

♦ 1.3.29. The matrix factorization A = LU is

0 11 0

!=

1 0a 1

! x y0 z

!=

x yax ay + z

!.

This implies x = 0 and ax = 1, which is impossible.

♦ 1.3.30.(a) Let u11, . . . , unn be the pivots of A, i.e., the diagonal entries of U . Let D be the diago-

nal matrix whose diagonal entries are dii = sign uii. Then B = AD is the matrix ob-tained by multiplying each column of A by the sign of its pivot. Moreover, B = LUD =L eU , where eU = UD, is the LU factorization of B. Each column of eU is obtained bymultiplying it by the sign of its pivot. In particular, the diagonal entries of eU , which arethe pivots of B, are uii sign uii = |uii | > 0.

(b) Using the same notation as in part (a), we note that C = DA is the matrix obtainedby multiplying each row of A by the sign of its pivot. Moreover, C = DLU . How-ever, DL is not special lower triangular, since its diagonal entries are the pivot signs.But bL = DLD is special lower triangular, and so C = DLDDU = bL bU , wherebU = DU , is the LU factorization of B. Each row of bU is obtained by multiplying itby the sign of its pivot. In particular, the diagonal entries of bU , which are the pivots ofC, are uii sign uii = |uii | > 0.

(c)0B@−2 2 1

1 0 14 2 3

1CA =

0B@

1 0 0− 1

2 1 0−2 6 1

1CA

0B@−2 2 1

0 1 32

0 0 −4

1CA,

0B@

2 2 −1−1 0 −1−4 2 −3

1CA =

0B@

1 0 0− 1

2 1 0−2 6 1

1CA

0B@

2 2 −10 1 − 3

20 0 4

1CA,

0B@

2 −2 −11 0 1−4 −2 −3

1CA =

0B@

1 0 012 1 0−2 −6 1

1CA

0B@

2 −2 −10 1 3

20 0 4

1CA.

1.3.31. (a) x =

−123

!, (b) x =

0@

1414

1A, (c) x =

0B@

010

1CA, (d) x =

0BBB@

− 472757

1CCCA, (e) x =

0B@−1−1

52

1CA,

(f ) x =

0B@

01−1

1CA, (g) x =

0BBB@

2110

1CCCA, (h) x =

0BBBBBB@

− 3712

− 1712

14

2

1CCCCCCA

, (i) x =

0BBBBBB@

33563517835

1CCCCCCA

.

11

1.3.32.

(a) L =

1 0−3 1

!, U =

−1 3

0 11

!; x1 =

0@ −

511211

1A, x2 =

11

!, x3 =

0@

911311

1A;

(b) L =

0B@

1 0 0−1 1 0

1 0 1

1CA , U =

0B@−1 1 −1

0 2 00 0 3

1CA; x1 =

0B@−1

00

1CA, x2 =

0BBB@

− 16

− 3253

1CCCA;

(c) L =

0BB@

1 0 0

− 23 1 029

53 1

1CCA , U =

0BB@

9 −2 −1

0 − 13

13

0 0 − 13

1CCA; x1 =

0B@

123

1CA, x2 =

0B@−2−9−1

1CA;

(d) L =

0B@

1 0 0.15 1 0.2 1.2394 1

1CA, U =

0B@

2.0 .3 .40 .355 4.940 0 −.2028

1CA;

x1 =

0B@

.6944−1.3889

.0694

1CA, x2 =

0B@

1.1111−82.2222

6.1111

1CA, x3 =

0B@−9.305668.6111−4.9306

1CA

(e) L =

0BBB@

1 0 0 00 1 0 0−1 3

2 1 0

0 − 12 −1 1

1CCCA, U =

0BBBB@

1 0 −1 00 2 3 −1

0 0 − 72

72

0 0 0 4

1CCCCA

; x1 =

0BBBBB@

54

− 141414

1CCCCCA

, x2 =

0BBBBB@

114

− 51411412

1CCCCCA

;

(f ) L =

0BBB@

1 0 0 04 1 0 0−8 − 17

9 1 0−4 −1 0 1

1CCCA, U =

0BBB@

1 −2 0 20 9 −1 −90 0 1

9 00 0 0 1

1CCCA;

x1 =

0BBB@

1040

1CCCA, x2 =

0BBB@

1132

1CCCA, x3 =

0BBB@

108

414

1CCCA.

1.4.1. The nonsingular matrices are (a), (c), (d), (h).

1.4.2. (a) Regular and nonsingular, (b) singular, (c) nonsingular, (d) regular and nonsingular.

1.4.3. (a) x1 = − 53 , x2 = − 10

3 , x3 = 5; (b) x1 = 0, x2 = −1, x3 = 2;

(c) x1 = −6, x2 = 2, x3 = −2; (d) x = − 132 , y = − 9

2 , z = −1, w = −3;

(e) x1 = −11, x2 = − 103 , x3 = −5, x4 = −7.

1.4.4. Solve the equations −1 = 2b+c, 3 = −2a+4b+c, −3 = 2a−b+c, for a = −4, b = −2,c = 3, giving the plane z = −4x− 2y + 3.

1.4.5.(a) Suppose A is nonsingular. If a 6= 0 and c 6= 0, then we subtract c/a times the first row

from the second, producing the (2, 2) pivot entry (ad − bc)/a 6= 0. If c = 0, then thepivot entry is d and so ad − bc = ad 6= 0. If a = 0, then c 6= 0 as otherwise the firstcolumn would not contain a pivot. Interchanging the two rows gives the pivots c and b,and so ad− bc = bc 6= 0.

(b) Regularity requires a 6= 0. Proceeding as in part (a), we conclude that ad− bc 6= 0 also.

1.4.6. True. All regular matrices are nonsingular.

12

♦ 1.4.7. Since A is nonsingular, we can reduce it to the upper triangular form with nonzero diago-nal entries (by applying the operations # 1 and # 2). The rest of argument is the same asin Exercise 1.3.8.

1.4.8. By applying the operations # 1 and # 2 to the system Ax = b we obtain an equivalentupper triangular system Ux = c. Since A is nonsingular, uii 6= 0 for all i, so by Back Sub-

stitution each solution component, namely xn =cn

unn

and xi =1

uii

0@ ci −

nX

k= i+1

uikxk

1A,

for i = n− 1, n− 2, . . . , 1, is uniquely defined.

1.4.9. (a) P1 =

0BBB@

1 0 0 00 0 0 10 0 1 00 1 0 0

1CCCA, (b) P2 =

0BBB@

0 0 0 10 1 0 00 0 1 01 0 0 0

1CCCA,

(c) No, they do not commute. (d) P1 P2 arranges the rows in the order 4, 1, 3, 2, whileP2 P1 arranges them in the order 2, 4, 3, 1.

1.4.10. (a)

0B@

0 1 00 0 11 0 0

1CA, (b)

0BBB@

0 0 0 10 0 1 01 0 0 00 1 0 0

1CCCA, (c)

0BBB@

0 1 0 01 0 0 00 0 0 10 0 1 0

1CCCA, (d)

0BBBBB@

0 0 0 1 01 0 0 0 00 0 1 0 00 1 0 0 00 0 0 0 1

1CCCCCA

.

1.4.11. The (i, j) entry of the following Multiplication Table indicates the product PiPj , where

P1 =

0B@

1 0 00 1 00 0 1

1CA , P2 =

0B@

0 1 00 0 11 0 0

1CA , P3 =

0B@

0 0 11 0 00 1 0

1CA ,

P4 =

0B@

0 1 01 0 00 0 1

1CA , P5 =

0B@

0 0 10 1 01 0 0

1CA , P6 =

0B@

1 0 00 0 10 1 0

1CA .

The commutative pairs are P1Pi = PiP1, i = 1, . . . , 6, and P2P3 = P3P2.

P1 P2 P3 P4 P5 P6

P1 P1 P2 P3 P4 P5 P6

P2 P2 P3 P1 P6 P4 P5

P3 P3 P1 P2 P5 P6 P4

P4 P4 P5 P6 P1 P2 P3

P5 P5 P6 P4 P3 P1 P2

P6 P6 P4 P5 P2 P3 P1

1.4.12. (a)

0BBB@

1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCA,

0BBB@

0 1 0 00 0 0 10 0 1 01 0 0 0

1CCCA,

0BBB@

0 0 0 11 0 0 00 0 1 00 1 0 0

1CCCA,

0BBB@

0 1 0 01 0 0 00 0 1 00 0 0 1

1CCCA,

0BBB@

0 0 0 10 1 0 00 0 1 01 0 0 0

1CCCA,

13

0BBB@

1 0 0 00 0 0 10 0 1 00 1 0 0

1CCCA; (b)

0BBB@

0 1 0 01 0 0 00 0 0 10 0 1 0

1CCCA,

0BBB@

1 0 0 00 0 0 10 1 0 00 0 1 0

1CCCA,

0BBB@

0 0 0 10 1 0 01 0 0 00 0 1 0

1CCCA,

0BBB@

1 0 0 00 1 0 00 0 0 10 0 1 0

1CCCA,

0BBB@

0 0 0 11 0 0 00 1 0 00 0 1 0

1CCCA,

0BBB@

0 1 0 00 0 0 11 0 0 00 0 1 0

1CCCA; (c)

0BBB@

1 0 0 00 0 1 00 1 0 00 0 0 1

1CCCA,

0BBB@

0 0 0 10 0 1 00 1 0 01 0 0 0

1CCCA.

1.4.13. (a) True, since interchanging the same pair of rows twice brings you back to where

you started. (b) False; an example is the non-elementary permuation matrix

0B@

0 0 11 0 00 1 0

1CA.

(c) False; for example P =

−1 0

0 −1

!is not a permutation matrix. For a complete list of

such matrices, see Exercise 1.2.36.

1.4.14. (a) Only when all the entries of v are different; (b) only when all the rows of A aredifferent.

1.4.15. (a)

0B@

1 0 00 0 10 1 0

1CA. (b) True. (c) False — AP permutes the columns of A according to

the inverse (or transpose) permutation matrix P−1 = PT .

♥ 1.4.16.(a) If P has a 1 in position (π(j), j), then it moves row j of A to row π(j) of P A, which is

enough to establish the correspondence.

(b) (i)

0B@

0 1 01 0 00 0 1

1CA, (ii)

0BBB@

0 0 0 10 1 0 00 0 1 01 0 0 0

1CCCA, (iii)

0BBB@

1 0 0 00 0 1 00 0 0 10 1 0 0

1CCCA, (iv)

0BBBBB@

0 0 0 0 10 0 0 1 00 0 1 0 00 1 0 0 01 0 0 0 0

1CCCCCA

.

Cases (i) and (ii) are elementary matrices.

(c) (i)

1 2 32 3 1

!, (ii)

1 2 3 43 4 1 2

!, (iii)

1 2 3 44 1 2 3

!, (iv)

1 2 3 4 52 5 3 1 4

!.

♦ 1.4.17. The first row of an n×n permutation matrix can have the 1 in any of the n positions, sothere are n possibilities for the first row. Once the first row is set, the second row can haveits 1 anywhere except in the column under the 1 in the first row, and so there are n − 1possibilities. The 1 in the third row can be in any of the n− 2 positions not under either ofthe previous two 1’s. And so on, leading to a total of n(n − 1)(n − 2) · · · 2 · 1 = n ! possiblepermutation matrices.

1.4.18. Let ri, rj denote the rows of the matrix in question. After the first elementary row op-

eration, the rows are ri and rj + ri. After the second, they are ri − (rj + ri) = −rj and

rj + ri. After the third operation, we are left with −rj and rj + ri + (−rj) = ri.

1.4.19. (a)

0 11 0

! 0 12 −1

!=

1 00 1

! 2 −10 1

!, x =

0@

52

3

1A;

14

(b)

0B@

0 1 00 0 11 0 0

1CA

0B@

0 0 −41 2 30 1 7

1CA =

0B@

1 0 00 1 00 0 1

1CA

0B@

1 2 30 1 70 0 −4

1CA, x =

0BBB@

5434

− 14

1CCCA;

(c)

0B@

0 0 11 0 00 1 0

1CA

0B@

0 1 −30 2 31 0 2

1CA =

0B@

1 0 00 1 00 2 1

1CA

0B@

1 0 20 1 −30 0 9

1CA, x =

0B@−1

10

1CA;

(d)

0BBB@

1 0 0 00 0 1 00 1 0 00 0 0 1

1CCCA

0BBB@

1 2 −1 03 6 2 −11 1 −7 21 −1 2 1

1CCCA =

0BBB@

1 0 0 01 1 0 03 0 1 01 3 21

5 1

1CCCA

0BBB@

1 2 −1 00 −1 −6 20 0 5 −10 0 0 − 4

5

1CCCA, x =

0BBB@

22−13−5−22

1CCCA;

(e)

0BBB@

0 0 1 01 0 0 00 1 0 00 0 0 1

1CCCA

0BBB@

0 1 0 02 3 1 01 4 −1 27 −1 2 3

1CCCA =

0BBB@

1 0 0 00 1 0 02 −5 1 07 −29 3 1

1CCCA

0BBB@

1 4 −1 20 1 0 00 0 3 −40 0 0 1

1CCCA, x =

0BBB@

−1−1

13

1CCCA;

(f )

0BBBBB@

0 0 1 0 00 1 0 0 00 0 0 1 01 0 0 0 00 0 0 0 1

1CCCCCA

0BBBBB@

0 0 2 3 40 1 −7 2 31 4 1 1 10 0 1 0 20 0 1 7 3

1CCCCCA

=

0BBBBB@

1 0 0 0 00 1 0 0 00 0 1 0 00 0 2 1 00 0 1 7

3 1

1CCCCCA

0BBBBB@

1 4 1 1 10 1 −7 2 30 0 1 0 20 0 0 3 00 0 0 0 1

1CCCCCA

, x =

0BBBBB@

100−1

0

1CCCCCA

.

1.4.20.

(a)

0B@

1 0 00 0 10 1 0

1CA

0B@

4 −4 2−3 3 1−3 1 −2

1CA =

0BB@

1 0 0

− 34 1 0

− 34 0 1

1CCA

0BB@

4 −4 2

0 −2 − 12

0 0 52

1CCA;

solution: x1 = 54 , x2 = 7

4 , x3 = 32 .

(b)

0BBB@

0 0 1 00 1 0 01 0 0 00 0 0 1

1CCCA

0BBB@

0 1 −1 10 1 1 01 −1 1 −31 2 −1 1

1CCCA =

0BBB@

1 0 0 00 1 0 00 1 1 01 3 5

2 1

1CCCA

0BBB@

1 −1 1 −30 1 1 00 0 −2 10 0 0 3

2

1CCCA;

solution: x = 4, y = 0, z = 1, w = 1.

(c)

0BBB@

1 0 0 00 0 0 10 0 1 00 1 0 0

1CCCA

0BBB@

1 −1 2 1−1 1 −3 0

1 −1 1 −31 2 −1 1

1CCCA =

0BBB@

1 0 0 01 1 0 01 0 1 0−1 0 − 1

2 1

1CCCA

0BBB@

1 −1 2 10 3 −3 00 0 2 −40 0 0 1

1CCCA;

solution: x = 193 , y = − 5

3 , z = −3, w = −2.

♦ 1.4.21.(a) They are all of the form P A = LU , where P is a permutation matrix. In the first case,

we interchange rows 1 and 2, in the second case, we interchange rows 1 and 3, in thethird case, we interchange rows 1 and 3 first and then interchange rows 2 and 3.

(b) Same solution x = 1, y = 1, z = −2 in all cases. Each is done by a sequence of elemen-tary row operations, which do not change the solution.

1.4.22. There are four in all:0B@

0 1 01 0 00 0 1

1CA

0B@

0 1 21 0 −11 1 3

1CA =

0B@

1 0 00 1 01 1 1

1CA

0B@

1 0 −10 1 20 0 2

1CA ,

0B@

0 1 00 0 11 0 0

1CA

0B@

0 1 21 0 −11 1 3

1CA =

0B@

1 0 01 1 00 1 1

1CA

0B@

1 0 −10 1 40 0 −2

1CA ,

0B@

0 0 10 1 01 0 0

1CA

0B@

0 1 21 0 −11 1 3

1CA =

0B@

1 0 01 1 00 −1 1

1CA

0B@

1 1 30 −1 −40 0 −2

1CA ,

15

0B@

0 0 11 0 00 1 0

1CA

0B@

0 1 21 0 −11 1 3

1CA =

0B@

1 0 00 1 01 −1 1

1CA

0B@

1 1 30 1 20 0 −2

1CA .

The other two permutation matrices are not regular.

1.4.23. The maximum is 6 since there are 6 different 3× 3 permutation matrices. For example,0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 01 1 0−1 1 1

1CA

0B@

1 0 00 1 00 0 1

1CA ,

0B@

1 0 00 0 10 1 0

1CA

0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 0−1 1 0

1 1 1

1CA

0B@

1 0 00 1 10 0 −1

1CA ,

0B@

0 1 01 0 00 0 1

1CA

0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 01 1 0−1 −2 1

1CA

0B@

1 1 00 −1 00 0 1

1CA ,

0B@

0 1 00 0 11 0 0

1CA

0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 0−1 1 0

1 − 12 1

1CA

0B@

1 1 00 2 10 0 1

2

1CA ,

0B@

0 0 10 1 01 0 0

1CA

0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 0−1 1 0−1 1

2 1

1CA

0B@−1 1 1

0 2 10 0 1

2

1CA ,

0B@

0 0 11 0 00 1 0

1CA

0B@

1 0 01 1 0−1 1 1

1CA =

0B@

1 0 0−1 1 0−1 2 1

1CA

0B@−1 1 1

0 1 10 0 −1

1CA .

1.4.24. False. Changing the permuation matrix typically changes the pivots.

♠ 1.4.25.

Permuted LU factorization

start

set P = I , L = I , U = A

for j = 1 to n

if ukj = 0 for all k ≥ j, stop; print “A is singular”

if ujj = 0 but ukj 6= 0 for some k > j then

interchange rows j and k of U

interchange rows j and k of P

for m = 1 to j − 1 interchange ljm and lkm next m

for i = j + 1 to n

set lij = uij/ujj

add −uij times row j to row i of A

next i

next j

end

16

1.5.1.

(a)

2 3−1 −1

! −1 −3

1 2

!=

1 00 1

!=

−1 −3

1 2

! 2 3−1 −1

!,

(b)

0B@

2 1 13 2 12 1 2

1CA

0B@

3 −1 −1−4 2 1−1 0 1

1CA =

0B@

1 0 00 1 00 0 1

1CA =

0B@

3 −1 −1−4 2 1−1 0 1

1CA

0B@

2 1 13 2 12 1 2

1CA,

(c)

0B@−1 3 2

2 2 −1−2 1 3

1CA

0BB@

− 1 1 147 − 1

7 − 37

− 67

57

87

1CCA =

0B@

1 0 00 1 00 0 1

1CA =

0BB@

− 1 1 147 − 1

7 − 37

− 67

57

87

1CCA

0B@−1 3 2

2 2 −1−2 1 3

1CA.

1.5.2. X =

0B@−5 16 6

3 −8 −3−1 3 1

1CA; X A =

0B@−5 16 6

3 −8 −3−1 3 1

1CA

0B@

1 2 00 1 31 −1 −8

1CA =

0B@

1 0 00 1 00 0 1

1CA.

1.5.3. (a)

0 11 0

!, (b)

1 0−5 1

!, (c)

1 20 1

!,

(d)

0B@

1 0 00 1 30 0 1

1CA, (e)

0BBB@

1 0 0 00 1 0 00 −6 1 00 0 0 1

1CCCA, (f )

0BBB@

0 0 0 10 1 0 00 0 1 01 0 0 0

1CCCA.

1.5.4.

0B@

1 0 0a 1 0b 0 1

1CA

0B@

1 0 0−a 1 0−b 0 1

1CA =

0B@

1 0 00 1 00 0 1

1CA =

0B@

1 0 0−a 1 0−b 0 1

1CA

0B@

1 0 0a 1 0b 0 1

1CA;

M−1 =

0B@

1 0 0−a 1 0

ac− b −c 1

1CA.

1.5.5. The ith row of the matrix multiplied by the ith column of the inverse should be equal 1.This is not possible if all the entries of the ith row are zero; see Exercise 1.2.24.

1.5.6. (a) A−1 =

−1 1

2 −1

!, B−1 =

0@

23

13

− 13

13

1A.

(b) C =

2 13 0

!, C−1 = B−1A−1 =

0@ 0 1

3

1 − 23

1A.

1.5.7. (a) R−1θ =

cos θ sin θ− sin θ cos θ

!. (b)

ab

!= R−1

θ

xy

!=

x cos θ + y sin θ−x sin θ + y cos θ

!.

(c) det(Rθ − a I ) = det

cos θ − a − sin θ

sin θ cos θ − a

!= (cos θ − a)2 + (sin θ)2 > 0

provided sin θ 6= 0, which is valid when 0 < θ < π.

1.5.8.

(a) Setting P1 =

0B@

1 0 00 1 00 0 1

1CA , P2 =

0B@

0 1 00 0 11 0 0

1CA , P3 =

0B@

0 0 11 0 00 1 0

1CA ,

P4 =

0B@

0 1 01 0 00 0 1

1CA , P5 =

0B@

0 0 10 1 01 0 0

1CA , P6 =

0B@

1 0 00 0 10 1 0

1CA ,

we find P−11 = P1, P−1

2 = P3, P−13 = P2, P−1

4 = P4, P−15 = P5, P−1

6 = P6.

(b) P1, P4, P5, P6 are their own inverses.

17

(c) Yes: P =

0BBB@

0 1 0 01 0 0 00 0 0 10 0 1 0

1CCCA interchanges two pairs of rows.

1.5.9. (a)

0BBB@

0 0 0 10 0 1 00 1 0 01 0 0 0

1CCCA, (b)

0BBB@

0 0 0 11 0 0 00 1 0 00 0 1 0

1CCCA, (c)

0BBB@

1 0 0 00 0 1 00 0 0 10 1 0 0

1CCCA, (d)

0BBBBB@

1 0 0 0 00 0 0 1 00 1 0 0 00 0 0 0 10 0 1 0 0

1CCCCCA

.

1.5.10.(a) If i and j = π(i) are the entries in the ith column of the 2 × n matrix corresponding to

the permutation, then the entries in the jth column of the 2× n matrix corresponding tothe permutation are j and i = π−1(j). Equivalently, permute the columns so that thesecond row is in order 1, 2, . . . , n and then switch the two rows.

(b) The permutations correspond to

(i)

1 2 3 44 3 2 1

!, (ii)

1 2 3 44 1 2 3

!, (iii)

1 2 3 41 3 4 2

!, (iv)

1 2 3 4 51 4 2 5 3

!.

The inverse permutations correspond to

(i)

1 2 3 44 3 2 1

!, (ii)

1 2 3 42 3 4 1

!, (iii)

1 2 3 41 4 2 3

!, (iv)

1 2 3 4 51 3 5 2 4

!.

1.5.11. If a = 0 the first row is all zeros, and so A is singular. Otherwise, we make d → 0 byan elementary row operation. If e = 0 then the resulting matrix has a row of all zeros.Otherwise, we make h → 0 by another elementary row operation, and the result is a matrixwith a row of all zeros.

1.5.12. This is true if and only if A2 = I , and so, according to Exercise 1.2.36, A is either of

the form

±1 0

0 ±1

!or

a bc −a

!, where a is arbitrary and bc = 1− a2.

1.5.13. (3 I −A)A = 3A−A2 = I , so 3 I −A is the inverse of A.

1.5.14.

1

cA−1

!(cA) =

1

ccA−1A = I .

1.5.15. Indeed, (An)−1 = (A−1)n.

1.5.16. If all the diagonal entries are nonzero, then D−1D = I . On the other hand, if one ofdiagonal entries is zero, then all the entries in that row are zero, and so D is not invertible.

1.5.17. Since U−1 is also upper triangular, the only nonzero summand in the product of the ith

row of U and the ith column of U−1 is the product of their diagonal entries, which mustequal 1 since U U−1 = I .

♦ 1.5.18. (a) A = I−1A I . (b) If B = S−1AS, then A = S BS−1 = T−1BT , where T = S−1.

(c) If B = S−1AS and C = T−1BT , then C = T−1(S−1AS)T = (S T )−1A(S T ).

♥ 1.5.19. (a) Suppose D−1 =

X YZ W

!. Then, in view of Exercise 1.2.37, the equation DD−1 =

I =

I OO I

!requires AX = I , AY = O, BZ = O, BW = I . Thus, X = A−1, W = B−1

and, since they are invertible, Y = A−1O = O, Z = B−1O = O.

18

(b)

0BBB@

− 13

23 0

23 − 1

3 0

0 0 13

1CCCA,

0BBB@

−1 1 0 0−2 1 0 0

0 0 −5 30 0 2 −1

1CCCA.

1.5.20.

(a) BA =

1 1 0−1 −1 1

!0B@

1 −10 11 1

1CA =

1 00 1

!.

(b) AX = I does not have a solution. Indeed, the first column of this matrix equation is

the linear system

0B@

1 −10 11 1

1CA

xy

!=

0B@

100

1CA, which has no solutions since x− y = 1, y = 0,

and x + y = 0 are incompatible.

(c) Yes: for instance, B =

2 3 −1−1 −1 1

!. More generally, BA = I if and only if B =

1− z 1− 2z z−w 1− 2w w

!, where z, w are arbitrary.

1.5.21. The general solution to AX = I is X =

0B@−2y 1− 2v

y v−1 1

1CA, where y, v are arbitrary.

Any of these matrices serves as a right inverse. On the other hand, the linear systemY A = I is incompatible and there is no solution.

1.5.22.

(a) No. The only solutions are complex, with a =„− 1

2 ± iq

23

«b, where b 6= 0 is any

nonzero complex number.

(b) Yes. A simple example is A =

−1 1−1 0

!, B =

1 00 1

!. The general solution to the

2 × 2 matrix equation has the form A = BM , where M =

x yz w

!is any matrix with

tr M = x + w = −1, and det M = xw − y z = 1. To see this, if we set A = BM ,then ( I + M)−1 = I + M−1, which is equivalent to I + M + M−1 = O. Writing thisout using the formula (1.38) for the inverse, we find that if det M = xw − y z = 1 then

tr M = x+w = −1, while if det M 6= 1, then y = z = 0 and x+x−1+1 = 0 = w+w−1+1,in which case, as in part (a), there are no real solutions.

1.5.23. E =

0BBB@

1 0 0 00 1 0 00 0 7 00 0 0 1

1CCCA, E−1 =

0BBB@

1 0 0 00 1 0 00 0 1

7 00 0 0 1

1CCCA.

1.5.24. (a)

0@ − 1 2

3

1 13

1A, (b)

0@ −

18

38

38 − 1

8

1A, (c)

0@

35

45

− 45

35

1A, (d) no inverse,

(e)

0B@

3 −2 −29 −7 −61 −1 −1

1CA, (f )

0BBB@

− 58

18

58

− 12

12 − 1

278 − 3

818

1CCCA, (g)

0BB@− 5

232

12

2 −1 −12 −1 0

1CCA,

19

(h)

0BBB@

0 2 1 11 −6 −2 −30 −5 0 −30 2 0 1

1CCCA, (i)

0BBB@

−51 8 12 3−13 2 3 1

21 −3 −5 −15 −1 −1 0

1CCCA.

1.5.25.

(a)

1 03 1

! 1 00 3

! 1 −20 1

!=

1 −23 −3

!,

(b)

1 03 1

! 1 00 −8

! 1 30 1

!=

1 33 1

!,

(c)

1 043 1

!0@

35 0

0 1

1A

1 0

0 53

!0@ 1 − 4

3

0 1

1A =

0@

35 − 4

545

35

1A,

(d) not possible,

(e)

0B@

1 0 03 1 00 0 1

1CA

0B@

1 0 00 1 0−2 0 1

1CA

0B@

1 0 00 1 00 −1 1

1CA

0B@

1 0 00 −1 00 0 1

1CA

0B@

1 0 00 1 00 0 −1

1CA

0B@

1 0 −20 1 00 0 1

1CA

0B@

1 0 00 1 −60 0 1

1CA =

0B@

1 0 −23 −1 0−2 1 −3

1CA,

(f )

0B@

1 0 03 1 00 0 1

1CA

0B@

1 0 00 1 02 0 1

1CA

0B@

1 0 00 1 00 3 1

1CA

0B@

1 0 00 −1 00 0 1

1CA

0B@

1 0 00 1 00 0 8

1CA

0B@

1 0 30 1 00 0 1

1CA

0B@

1 0 00 1 40 0 1

1CA

0B@

1 2 00 1 00 0 1

1CA =

0B@

1 2 33 5 52 1 2

1CA,

(g)

0B@

1 0 00 0 10 1 0

1CA

0B@

1 0 00 1 02 0 1

1CA

0B@

2 0 00 1 00 0 1

1CA

0B@

1 0 00 −1 00 0 1

1CA

0B@

1 0 00 1 00 0 −1

1CA

0B@

1 0 10 1 00 0 1

1CA

0B@

1 0 00 1 −10 0 1

1CA

0B@

1 12 0

0 1 00 0 1

1CA =

0B@

2 1 24 2 30 −1 1

1CA,

(h)

0BBB@

1 0 0 00 0 1 00 1 0 00 0 0 1

1CCCA

0BBB@

1 0 0 012 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 00 0 −2 1

1CCCA

0BBB@

2 0 0 00 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 − 1

2 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 12

0 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 30 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 30 0 0 1

1CCCA

0BBB@

1 12 0 0

0 1 0 00 0 1 00 0 0 1

1CCCA =

0BBB@

2 1 0 10 0 1 31 0 0 −10 0 −2 −5

1CCCA,

(i)

0BBB@

1 0 0 00 1 0 00 0 0 10 0 1 0

1CCCA

0BBB@

1 0 0 02 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 03 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 2 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 00 −1 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 −1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 00 0 0 −1

1CCCA

0BBB@

1 0 0 10 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 −20 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 1 −50 0 0 1

1CCCA

0BBB@

1 0 1 00 1 0 00 0 1 00 0 0 1

1CCCA

0BBB@

1 0 0 00 1 1 00 0 1 00 0 0 1

1CCCA

0BBB@

1 −2 0 00 1 0 00 0 1 00 0 0 1

1CCCA =

0BBB@

1 −2 1 12 −3 3 03 −7 2 40 2 1 1

1CCCA.

20

1.5.26. Applying Gaussian Elimination:

E1 =

0@ 1 0

− 1√3

1

1A, E1A =

0B@

√3

2 − 12

0 2√3

1CA,

E2 =

0@ 1 0

0√

32

1A, E2E1A =

0@

√3

2 − 12

0 1

1A,

E3 =

0@

2√3

0

0 1

1A, E3E2E1A =

0@ 1 − 1√

3

0 1

1A,

E4 =

0@ 1 1√

3

0 1

1A, E4E3E2E1A = I =

1 00 1

!,

and hence A = E−11 E−1

2 E−13 E−1

4 =

0@ 1 0

1√3

1

1A0@ 1 0

0 2√3

1A0@

√3

2 0

0 1

1A0@ 1 − 1√

3

0 1

1A.

1.5.27. (a)

0@ −

i2

12

12 − i

2

1A, (b)

−1 1− i

1 + i −1

!,

(c)

0B@

i 0 −11− i − i 1−1 −1 − i

1CA, (d)

0B@

3 + i −1− i − i−4 + 4 i 2− i 2 + i−1 + 2 i 1− i 1

1CA.

1.5.28. No. If they have the same solution, then they both reduce to“

I | x”

under elementary

row operations. Thus, by applying the appropriate elementary row operations to reduce the

augmented matrix of the first system to“

I | x”, and then applying the inverse elemen-

tary row operations we arrive at the augmented matrix for second system. Thus, the firstsystem can be changed into the second by the combined sequence of elementary row opera-tions, proving equivalence. (See also Exercise 2.5.44 for the general case.)

♥ 1.5.29.(a) If eA = EN EN−1 · · · E2 E1 A where E1, . . . , EN represent the row operations applied to

A, then eC = eA B = EN EN−1 · · · E2 E1 AB = EN EN−1 · · · E2 E1 C, which representsthe same sequence of row operations applied to C.

(b)

(E A)B =

0B@

1 2 −12 −3 2−2 −3 −2

1CA

0B@

1 −23 0−1 1

1CA =

0B@

8 −3−9 −2−9 2

1CA =

0B@

1 0 00 1 0−2 0 1

1CA

0B@

8 −3−9 −2

7 −4

1CA = E (AB).

1.5.30. (a)

0@

12

12

14 − 1

4

1A

1

− 2

!=

0@ −

1234

1A; (b)

0@

517

217

− 117

317

1A

2

12

!=

2

2

!;

(c)

0BB@

2 − 52

32

1 −1 0

0 12 − 1

2

1CCA

0BB@

3

− 2

2

1CCA =

0BB@

14

5

− 2

1CCA; (d)

0B@

9 −15 −86 −10 −5−1 2 1

1CA

0B@

3−1

5

1CA =

0B@

230

1CA;

(e)

0B@−4 3 1

2 −1 03 −1 1

1CA

0B@

35−7

1CA =

0B@−4

1−3

1CA; (f )

0BBB@

1 0 1 10 0 −1 −12 −1 −1 02 −1 −1 −1

1CCCA

0BBB@

411−7

6

1CCCA =

0BBB@

314−2

1CCCA;

21

(g)

0BBBBB@

1 1 0 1

− 52 −2 3

2 − 32

− 4 −3 2 −3

− 12 −1 1

2 − 12

1CCCCCA

0BBBBB@

− 2

3

3

2

1CCCCCA

=

0BBBBB@

312

− 1

− 32

1CCCCCA

.

1.5.31. (a)

0@ −

13

− 23

1A , (b)

0@

1414

1A, (c)

0@

75

− 15

1A, (d) singular matrix,

(e)

0B@−1−4−1

1CA, (f )

0BBB@

18

− 1258

1CCCA, (g)

0B@− 1

201

1CA, (h)

0BBB@

4−10−8

3

1CCCA, (i)

0BBB@

−28−7123

1CCCA.

1.5.32.

(a)

1 2−3 1

!=

1 0−3 1

! 1 00 7

! 1 20 1

!;

(b)

0 11 0

! 0 4−7 2

!=

1 00 1

! −7 0

0 4

! 1 − 2

70 1

!

(c)

0B@

2 1 22 4 −10 −2 1

1CA =

0B@

1 0 01 1 00 − 2

3 1

1CA

0B@

2 0 00 3 00 0 −1

1CA

0B@

1 12 1

0 1 −10 0 1

1CA;

(d)

0B@

1 0 00 0 10 1 0

1CA

0B@

1 1 51 1 −22 −1 3

1CA =

0B@

1 0 02 1 01 0 1

1CA

0B@

1 0 00 −3 00 0 −7

1CA

0B@

1 1 50 1 7

30 0 1

1CA;

(e)

0B@

2 −3 21 −1 11 −1 2

1CA =

0B@

1 0 012 1 012 1 1

1CA

0B@

2 0 00 1

2 00 0 1

1CA

0B@

1 − 32 1

0 1 00 0 1

1CA;

(f )

0BBB@

1 −1 1 21 −4 1 51 2 −1 −13 1 1 6

1CCCA =

0BBB@

1 0 0 01 1 0 01 −1 1 03 − 4

3 1 1

1CCCA

0BBB@

1 0 0 00 −3 0 00 0 −2 00 0 0 4

1CCCA

0BBB@

1 −1 1 20 1 0 −10 0 1 00 0 0 1

1CCCA;

(g)

0BBB@

1 0 0 00 1 0 00 0 0 10 0 1 0

1CCCA

0BBB@

1 0 2 −32 −2 0 11 −2 −2 −10 1 1 2

1CCCA =

0BBB@

1 0 0 02 1 0 00 − 1

2 1 01 1 0 1

1CCCA

0BBB@

1 0 0 00 −2 0 00 0 −1 00 0 0 −5

1CCCA

0BBB@

1 0 2 −30 1 2 − 7

2

0 0 1 − 112

0 0 0 1

1CCCA.

1.5.33.

(a)

0@ −

3757

1A, (b)

−8

3

!, (c)

0BBB@

16

− 2323

1CCCA, (d)

0B@

1−2

0

1CA, (e)

0B@−12−3

7

1CA, (f )

0BBBB@

73

25

− 53

1CCCCA

, (g)

0BBB@

010−2

1CCCA.

1.6.1. (a) ( 1 5 ), (b)

1 01 2

!, (c)

1 22 1

!, (d)

0B@

1 22 0−1 2

1CA,

22

(e)

0B@

12−3

1CA, (f )

1 3 52 4 6

!, (g)

0B@

1 0 12 3 1−1 2 5

1CA.

1.6.2. AT =

0B@

3 1−1 2−1 1

1CA, BT =

−1 2 −3

2 0 4

!,

(AB)T = BT AT =

−2 0

2 6

!, (BA)T = AT BT =

0B@−1 6 −5

5 −2 113 −2 7

1CA.

1.6.3. If A has size m × n and B has size n × p, then (AB)T has size p ×m. Further, AT has

size n ×m and BT has size p × n, and so unless m = p the product AT BT is not defined.

If m = p, then AT BT has size n × n, and so to equal (AB)T , we must have m = n = p,

so the matrices are square. Finally, taking the transpose of both sides, AB = (AT BT )T =

(BT )T (AT )T = BA, and so they must commute.

♦ 1.6.4. The (i, j) entry of C = (AB)T is the (j, i) entry of AB, so

cij =nX

k=1

ajk bki =nX

k=1

ebikeakj ,

where eaij = aji and ebij = bji are the entries of AT and BT respectively. Thus, cij equals

the (i, j) entry of the product BT AT .

1.6.5. (ABC)T = CT BT AT

1.6.6. False. For example,

1 10 1

!does not commute with its transpose.

♦ 1.6.7. If A =

a bc d

!, then AT A = AAT if and only if b2 = c2 and (a− d)(b− c) = 0.

So either b = c, or c = −b 6= 0 and a = d. Thus all normal 2× 2 matrices are of the form a bb d

!or

a b−b a

!.

1.6.8.(a) (AB)−T = ((AB)T )−1 = (BT AT )−1 = (AT )−1(BT )−1 = A−T B−T .

(b) AB =

1 02 1

!, so (AB)−T =

1 −20 1

!, while A−T =

0 −11 1

!, B−T =

1 −1−1 2

!,

so A−T B−T =

1 −20 1

!.

1.6.9. If A is invertible, then so is AT by Lemma 1.32; then by Lemma 1.21 AAT and AT A areinvertible.

1.6.10. No; for example,

12

!( 3 4 ) =

3 46 8

!while

34

!( 1 2 ) =

3 64 8

!.

1.6.11. No. In general, BT A is the transpose of AT B.

♦ 1.6.12.(a) The ith entry of Aej is the product of the ith row of A with ej . Since all the entries in

ej are zero except the jth entry the product will be equal to aij , i.e., the (i, j) entry of A.

(b) By part (a), beTi Aej is the product of the row matrix beT

i and the jth column of A. Since

23

all the entries in beTi are zero except the ith entry, multiplication by the jth column of A

will produce aij .

♦ 1.6.13.(a) Using Exercise 1.6.12, aij = eT

i Aej = eTi B ej = bij for all i, j.

(b) Two examples: A =

1 20 1

!, B =

1 11 1

!; A =

0 00 0

!, B =

0 −11 0

!.

♦ 1.6.14.(a) If pij = 1, then P A maps the jth row of A to its ith row. Then Q = P T has qji = 1,

and so it does the reverse, mapping the ith row of A to its jth row. Since this holds forall such entries, the result follows.

(b) No. Any rotation matrix

cos θ − sin θsin θ cos θ

!also has this property. See Section 5.3.

♦ 1.6.15.(a) Note that (AP T )T = P AT , which permutes the rows of AT , which are the columns of

A, according to the permutation P .

(b) The effect of multiplying P AP T is equivalent to simultaneously permuting rows andcolumns of A according to the permutation P . Associativity of matrix multiplicationimplies that it doesn’t matter whether the rows or the columns are permuted first.

♥ 1.6.16.(a) Note that wvT is a scalar, and so

AA−1 = ( I − vwT )( I − cvw

T ) = I − (1 + c)vwT + cv (wT

v)wT

= I − (1 + c− cwTv)vw

T = I

provided c = 1/(vT w − 1), which works whenever wT v 6= 1.

(b) A = I − vwT =

2 −23 −5

!and c =

1

vT w − 1= 1

4 , so A−1 = I − 14 vwT =

54 − 1

234 − 1

2

!.

(c) If vT w = 1 then A is singular, since Av = 0 and v 6= 0, and so the homogeneoussystem does not have a unique solution.

1.6.17. (a) a = 1; (b) a = −1, b = 2, c = 3; (c) a = −2, b = −1, c = −5.

1.6.18.

(a)

0B@

1 0 00 1 00 0 1

1CA,

0B@

0 1 01 0 00 0 1

1CA,

0B@

0 0 10 1 01 0 0

1CA,

0B@

1 0 00 0 10 1 0

1CA.

(b)

0BBB@

1 0 0 00 1 0 00 0 1 00 0 0 1

1CCCA,

0BBB@

0 1 0 01 0 0 00 0 1 00 0 0 1

1CCCA,

0BBB@

0 0 1 00 1 0 01 0 0 00 0 0 1

1CCCA,

0BBB@

0 0 0 10 1 0 00 0 1 01 0 0 0

1CCCA,

0BBB@

1 0 0 00 0 1 00 1 0 00 0 0 1

1CCCA,

0BBB@

1 0 0 00 0 0 10 0 1 00 1 0 0

1CCCA,

0BBB@

1 0 0 00 1 0 00 0 0 10 0 1 0

1CCCA,

0BBB@

0 1 0 01 0 0 00 0 0 10 0 1 0

1CCCA,

0BBB@

0 0 1 00 0 0 11 0 0 00 1 0 0

1CCCA,

0BBB@

0 0 0 10 0 1 00 1 0 01 0 0 0

1CCCA.

1.6.19. True, since (A2)T = (AA)T = AT AT = AA = A2.

♦ 1.6.20. True. Invert both sides of the equation AT = A, and use Lemma 1.32.

24

♦ 1.6.21. False. For example

0 11 0

! 2 11 3

!=

1 32 1

!.

1.6.22.(a) If D is a diagonal matrix, then for all i 6= j we have aij = aji = 0, so D is symmetric.

(b) If L is lower triangular then aij = 0 for i < j, if it is symmetric then aji = 0 for i < j,

so L is diagonal. If L is diagonal, then aij = 0 for i < j, so L is lower triangular and itis symmetric.

1.6.23.(a) Since A is symmetric we have (An)T = (AA . . . A)T = AT AT . . . AT = AA . . . A = An

(b) (2A2 − 3A + I )T = 2(A2)T − 3AT + I = 2A2 − 3A + I

(c) If p(A) = cn An + · · ·+ c1 A + c0 I , then p(A)T = cn An + · · · c1 A + c0 I T = cn (AT )n +

· · · c1 AT + c0 I = p(AT ). In particular, if A = AT , then p(A)T = p(AT ) = p(A).

1.6.24. If A has size m × n, then AT has size n × m and so both products are defined. Also,

KT = (AT A)T = AT (AT )T = AT A = K and LT = (AAT )T = (AT )T AT = AAT = L.

1.6.25.

(a)

1 11 4

!=

1 01 1

! 1 00 3

! 1 10 1

!,

(b)

−2 3

3 −1

!=

1 0− 3

2 1

! −2 0

0 72

! 1 − 3

20 1

!,

(c)

0B@

1 −1 −1−1 3 2−1 2 0

1CA =

0B@

1 0 0−1 1 0−1 1

2 1

1CA

0B@

1 0 00 2 00 0 − 3

2

1CA

0B@

1 −1 −10 1 1

20 0 1

1CA,

(d)

0BBB@

1 −1 0 3−1 2 2 0

0 2 −1 03 0 0 1

1CCCA =

0BBB@

1 0 0 0−1 1 0 0

0 2 1 03 3 6

5 1

1CCCA

0BBB@

1 0 0 00 1 0 00 0 −5 00 0 0 − 49

5

1CCCA

0BBB@

1 −1 0 30 1 2 30 0 1 6

50 0 0 1

1CCCA.

1.6.26. M2 =

1 012 1

! 1 0

0 32

!0@ 1 1

2

0 1

1A, M3 =

0BB@

1 0 012 1 0

0 23 1

1CCA

0BB@

2 0 0

0 32 0

0 0 43

1CCA

0BBB@

1 12 0

0 1 23

0 0 1

1CCCA,

M4 =

0BBBBB@

1 0 0 012 1 0 0

0 23 1 0

0 0 34 1

1CCCCCA

0BBBBB@

2 0 0 0

0 32 0 0

0 0 43 0

0 0 0 54

1CCCCCA

0BBBBB@

1 12 0 0

0 1 23 0

0 0 1 34

0 0 0 1

1CCCCCA

.

♦ 1.6.27. The matrix is not regular, since after the first set of row operations the (2, 2) entry is 0.More explicitly, if

L =

0B@

1 0 0a 1 0b c 1

1CA , D =

0B@

p 0 00 q 00 0 r

1CA , then LDLT =

0B@

p ap bpap a2p + q abp + cqbp abp + cq b2p + c2q + r

1CA.

Equating this to A, the (1, 1) entry requires p = 1, and so the (1, 2) entry requires a = 2,but the (2, 2) entry then implies q = 0, which is not an allowed diagonal entry for D.Even if we ignore this, the (1, 3) entry would set b = 1, but then the (2, 3) entry saysabp + cq = 2 6= −1, which is a contradiction.

♦ 1.6.28. Write A = LDV , then AT = V T DUT = eL eU , where eL = V T and eU = D eL. Thus, AT

is regular since the diagonal entries of eU , which are the pivots of AT , are the same as those

25

of D and U , which are the pivots of A.

♥ 1.6.29. (a) The diagonal entries satisfy jii = −jii and so must be 0. (b)

0 1−1 0

!. (c) No,

because the (1, 1) entry is always 0. (d) Invert both sides of the equation JT = −J and

use Lemma 1.32. (e) (JT )T = J = −JT , (J ±K)T = JT ±KT = −J ∓K = −(J ±K).

J K is not, in general, skew-symmetric; for instance

0 1−1 0

! 0 1−1 0

!=

−1 0

0 −1

!.

(f ) Since it is a scalar, vT J v = (vT J v)T = vT JT (vT )T = −vT J v equals its ownnegative, and so is zero.

1.6.30.(a) Let S = 1

2 (A + AT ), J = 12 (A−AT ). Then ST = S, JT = −J , and A = S + J .

(b)

1 23 4

!=

1 5

252 4

!+

0 − 1

212 0

!;

0B@

1 2 34 5 67 8 9

1CA =

0B@

1 3 53 5 75 7 9

1CA+

0B@

0 −1 −21 0 −12 1 0

1CA.

1.7.1.(a) The solution is x = − 10

7 , y = − 197 . Gaussian Elimination and Back Substitution re-

quires 2 multiplications and 3 additions; Gauss–Jordan also uses 2 multiplications and 3

additions; finding A−1 =

0@

17

27

− 37

17

1A by the Gauss–Jordan method requires 2 additions

and 4 multiplications, while computing the solution x =

0@

17

27

− 37

17

1A

4

−7

!=

0@−

107

− 197

1A

takes another 4 multiplications and 2 additions.

(b) The solution is x = −4, y = −5, z = −1. Gaussian Elimination and Back Substitu-tion requires 17 multiplications and 11 additions; Gauss–Jordan uses 20 multiplications

and 11 additions; computing A−1 =

0B@

0 −1 −12 −8 −532 −5 −3

1CA takes 27 multiplications and 12

additions, while multiplying A−1b = x takes another 9 multiplications and 6 additions.(c) The solution is x = 2, y = 1, z = 2

5 . Gaussian Elimination and Back Substitutionrequires 6 multiplications and 5 additions; Gauss–Jordan is the same: 6 multiplications

and 5 additions; computing A−1 =

0BBB@

− 12

32

32

− 12

12

12

− 25 0 − 1

5

1CCCA takes 11 multiplications and 3

additions, while multiplying A−1b = x takes another 8 multiplications and 5 additions.

1.7.2.(a) For a general matrix A, each entry of A2 requires n multiplications and n − 1 additions,

for a total of n3 multiplications and n3 − n2 additions, and so, when compared withthe efficient version of the Gauss–Jordan algorithm, takes exactly the same amount ofcomputation.

(b) A3 = A2A requires a total of 2n3 multiplications and 2n3 − 2n2 additions, and so isabout twice as slow.

(c) You can compute A4 as A2A2, and so only 2 matrix multiplications are required. In

general, if 2r ≤ k < 2r+1 has j ones in its binary representation, then you need r multi-

plications to compute A2, A4, A8, . . . A2r

followed by j − 1 multiplications to form Ak asa product of these particular powers, for a total of r + j − 1 matrix multiplications, andhence a total of (r + j − 1)n3 multiplications and (r + j − 1)n2(n − 1) additions. See

26

Exercise 1.7.8 and [11] for more sophisticated ways to speed up the computation.

1.7.3. Back Substitution requires about one half the number of arithmetic operations as multi-plying a matrix times a vector, and so is twice as fast.

♦ 1.7.4. We begin by proving (1.61). We must show that 1 + 2 + 3 + . . . + (n − 1) = n(n − 1)/2for n = 2, 3, . . .. For n = 2 both sides equal 1. Assume that (1.61) is true for n = k. Then1+2+3+ . . .+(k−1)+k = k(k−1)/2+k = k(k+1)/2, so (1.61) is true for n = k+1. Nowthe first equation in (1.62) follows if we note that 1 + 2 + 3 + . . . + (n− 1) + n = n(n + 1)/2.

Next we prove the first equation in (1.60), namely 2 + 6 + 12 + . . . + (n − 1)n = 13 n3 − 1

3 nfor n = 2, 3, . . .. For n = 2 both sides equal 2. Assume that the formula is true for n = k.Then 2+6+12+ . . .+(k−1)k+k(k+1) = 1

3 k3− 13 k+k2 +k = 1

3 (k+1)3− 13 (k+1), so the

formula is true for n = k + 1, which completes the induction step. The proof of the secondequation is similar, or, alternatively, one can use the first equation and (1.61) to show that

nX

j =1

(n− j)2 =nX

j =1

(n− j)(n− j + 1)−nX

j =1

(n− j) =n3 − n

3− n2 − n

2=

2n3 − 3n2 + n

6.

♥ 1.7.5. We may assume that the matrix is regular, so P = I , since row interchanges have noeffect on the number of arithmetic operations.

(a) First, according to (1.60), it takes 13 n3 − 1

3 n multiplications and 13 n3 − 1

2 n2 + 16 n

additions to factor A = LU . To solve Lcj = ej by Forward Substitution, the first j − 1

entries of c are automatically 0, the jth entry is 1, and then, for k = j + 1, . . . n, we needk− j − 1 multiplications and the same number of additions to compute the kth entry, fora total of 1

2 (n− j)(n− j − 1) multiplications and additions to find cj . Similarly, to solve

U xj = cj for the jth column of A−1 requires 12 n2 + 1

2 n multiplications and, since the

first j − 1 entries of cj are 0, also 12 n2 − 1

2 n − j + 1 additions. The grand total is n3

multiplications and n(n− 1)2 additions.

(b) Starting with the large augmented matrix M =“

A | I”, it takes 1

2 n2(n − 1) multipli-

cations and 12 n(n− 1)2 additions to reduce it to triangular form

“U | C

”with U upper

triangular and C lower triangular, then n2 multiplications to obtain the special upper

triangular form“

V | B”, and then 1

2 n2(n − 1) multiplications and, since B is upper

triangular, 12 n(n − 1)2 additions to produce the final matrix

“I | A−1

”. The grand

total is n3 multiplications and n(n − 1)2 additions. Thus, both methods take the sameamount of work.

1.7.6. Combining (1.60–61), we see that it takes 13 n3 + 1

2 n2− 56 n multiplications and 1

3 n3− 13 n

additions to reduce the augmented matrix to upper triangular form“

U | c”. Dividing the

jth row by its pivot requires n−j+1 multiplications, for a total of 12 n2+ 1

2 n multiplications

to produce the special upper triangular form“

V | e”. To produce the solved form

“I | d

”

requires an additional 12 n2 − 1

2 n multiplications and the same number of additions for a

grand total of 13 n3 + 3

2 n2 − 56 n multiplications and 1

3 n3 + 12 n2 − 5

6 n additions needed tosolve the system.

1.7.7. Less efficient, by, roughly, a factor of 32 . It takes 1

2 n3 + n2 − 12 n multiplications and

12 n3 − 1

2 n additions.

27

♥ 1.7.8.(a) D1 + D3 −D4 −D6 = (A1 + A4)(B1 + B4) + (A2 −A4)(B3 + B4)−

− (A1 + A2)B4 −A4(B1 −B3) = A1B1 + A2B3 = C1,

D4 + D7 = (A1 + A2)B4 + A1(B2 −B4) = A1B2 + A2B4 = C2,

D5 −D6 = (A3 + A4)B1 −A4(B1 −B3) = A3B1 + A4B3 = C3,D1 −D2 −D5 + D7 = (A1 + A4)(B1 + B4)− (A1 −A3)(B1 + B2)−

− (A3 + A4)B1 + A1(B2 −B4) = A3B2 + A4B4 = C4.(b) To compute D1, . . . , D7 requires 7 multiplications and 10 additions; then to compute

C1, C2, C3, C4 requires an additional 8 additions for a total of 7 multiplications and 18additions. The traditional method for computing the product of two 2 × 2 matrices re-quires 8 multiplications and 4 additions.

(c) The method requires 7 multiplications and 18 additions of n × n matrices, for a total of

7n3 and 7n2(n−1)+18n2 ≈ 7n3 additions, versus 8n3 multiplications and 8n2(n−1) ≈8n3 additions for the direct method, so there is a savings by a factor of 7

8 .(d) Let µr denote the number of multiplications and αr the number of additions to compute

the product of 2r × 2r matrices using Strassen’s Algorithm. Then, µr = 7µr−1, while

αr = 7αr−1 + 18 · 22r−2, where the first factor comes from multiplying the blocks,

and the second from adding them. Since µ1 = 1, α1 = 0. Clearly, µr = 7r, while an

induction proves the formula for αr = 6(7r−1 − 4r−1), namely

αr+1 = 7αr−1 + 18 · 4r−1 = 6(7r − 7 · 4r−1) + 18 · 4r−1 = 6(7r − 4r).

Combining the operations, Strassen’s Algorithm is faster by a factor of

2n3

µr + αr

=23r+1

13 · 7r−1 − 6 · 4r−1,

which, for r = 10, equals 4.1059, for r = 25, equals 30.3378, and, for r = 100, equals678, 234, which is a remarkable savings — but bear in mind that the matrices have sizearound 1030, which is astronomical!

(e) One way is to use block matrix multiplication, in the trivial form

A OO I

! B OO I

!=

C OO I

!where C = AB. Thus, choosing I to be an identity matrix of the appropriate

size, the overall size of the block matrices can be arranged to be a power of 2, and thenthe reduction algorithm can proceed on the larger matrices. Another approach, trickierto program, is to break the matrix up into blocks of nearly equal size since the Strassenformulas do not, in fact, require the blocks to have the same size and even apply to rect-angular matrices whose rectangular blocks are of compatible sizes.

1.7.9.

(a)

0B@

1 2 0−1 −1 1

0 −2 3

1CA =

0B@

1 0 0−1 1 0

0 −2 1

1CA

0B@

1 2 00 1 10 0 5

1CA, x =

0B@−2

30

1CA;

(b)

0BBB@

1 −1 0 0−1 2 1 0

0 −1 4 10 0 −1 6

1CCCA =

0BBB@

1 0 0 0−1 1 0 0

0 −1 1 00 0 −1 1

1CCCA

0BBB@

1 −1 0 00 1 1 00 0 5 10 0 0 7

1CCCA, x =

0BBB@

1012

1CCCA;

28

(c)

0BBB@

1 2 0 0−1 −3 0 0

0 −1 4 −10 0 −1 −1

1CCCA =

0BBB@

1 0 0 0−1 1 0 0

0 1 1 00 0 − 1

4 1

1CCCA

0BBB@

1 2 0 00 −1 0 00 0 4 −10 0 0 − 5

4

1CCCA, x =

0BBBB@

−42

− 25

− 35

1CCCCA

.

1.7.10.

(a)

0B@

2 −1 0−1 2 −1

0 −1 2

1CA =

0B@

1 0 0− 1

2 1 0

0 − 23 1

1CA

0B@

2 −1 00 3

2 −1

0 0 43

1CA,

0BBB@

2 −1 0 0−1 2 −1 0

0 −1 2 −10 0 −1 2

1CCCA =

0BBBB@

1 0 0 0− 1

2 1 0 0

0 − 23 1 0

0 0 − 34 1

1CCCCA

0BBBB@

2 −1 0 00 3

2 −1 0

0 0 43 −1

0 0 0 54

1CCCCA

,

0BBBBB@

2 −1 0 0 0−1 2 −1 0 0

0 −1 2 −1 00 0 −1 2 −10 0 0 −1 2

1CCCCCA

=

0BBBBBBB@

1 0 0 0 0− 1

2 1 0 0 0

0 − 23 1 0 0

0 0 − 34 1 0

0 0 0 − 45 1

1CCCCCCCA

0BBBBBBB@

2 −1 0 0 00 3

2 −1 0 0

0 0 43 −1 0

0 0 0 54 −1

0 0 0 0 65

1CCCCCCCA

;

(b)“

32 , 2, 3

2

”T, ( 2, 3, 3, 2 )T ,

“52 , 4, 9

2 , 4, 52

”T.

(c) The subdiagonal entries in L are li+1,i = − i/(i + 1) → −1, while the diagonal entries in

U uii = (i + 1)/i→ 1.

♠ 1.7.11.

(a)

0B@

2 1 0−1 2 1

0 −1 2

1CA =

0B@

1 0 0− 1

2 1 0

0 − 25 1

1CA

0B@

2 1 00 5

2 1

0 0 125

1CA ,

0BBB@

2 1 0 0−1 2 1 0

0 −1 2 10 0 −1 2

1CCCA =

0BBBB@

1 0 0 0− 1

2 1 0 0

0 − 25 1 0

0 0 − 512 1

1CCCCA

0BBBB@

2 1 0 00 5

2 1 0

0 0 125 1

0 0 0 2912

1CCCCA

,

0BBBBB@

2 1 0 0 01 2 1 0 00 1 2 1 00 0 1 2 10 0 0 1 2

1CCCCCA

=

0BBBBBBB@

1 0 0 0 0− 1

2 1 0 0 0

0 − 25 1 0 0

0 0 − 512 1 0

0 0 0 − 1229 1

1CCCCCCCA

0BBBBBBB@

2 1 0 0 00 5

2 1 0 0

0 0 125 1 0

0 0 0 2912 1

0 0 0 0 7029

1CCCCCCCA

;

(b)“

13 , 1

3 , 23

”T,“

829 , 13

29 , 1129 , 20

29

”T,“

310 , 2

5 , 12 , 2

5 , 710

”T.

(c) The subdiagonal entries in L approach 1 −√

2 = −.414214, and the diagonal entries in

U approach 1 +√

2 = 2.414214.

1.7.12. Both false. For example,0BBB@

1 1 0 01 1 1 00 1 1 10 0 1 1

1CCCA

0BBB@

1 1 0 01 1 1 00 1 1 10 0 1 1

1CCCA =

0BBB@

2 2 1 02 3 2 11 2 3 20 1 2 2

1CCCA,

0BBB@

1 1 0 01 1 1 00 1 1 10 0 1 1

1CCCA

−1

=

0BBB@

1 0 −1 10 0 1 −1−1 1 0 0

1 −1 0 1

1CCCA.

♠ 1.7.13.

0BB@

4 1 1

1 4 1

1 1 4

1CCA =

0BB@

1 0 014 1 014

15 1

1CCA

0BB@

4 1 1

0 154

34

0 0 185

1CCA

29

0BBBBB@

4 1 0 1

1 4 1 0

0 1 4 1

1 0 1 4

1CCCCCA

=

0BBBBB@

1 0 0 014 1 0 0

0 415 1 0

14 − 1

1527 1

1CCCCCA

0BBBBB@

4 1 0 1

0 154 1 − 1

4

0 0 5615

1615

0 0 0 247

1CCCCCA

0BBBBBBB@

4 1 0 0 1

1 4 1 0 0

0 1 4 1 0

0 0 1 4 1

1 0 0 1 4

1CCCCCCCA

=

0BBBBBBBB@

1 0 0 0 014 1 0 0 0

0 415 1 0 0

0 0 1556 1 0

14 − 1

15156

519 1

1CCCCCCCCA

0BBBBBBBB@

4 1 0 0 1

0 154 1 0 − 1

4

0 0 5615 1 1

15

0 0 0 20956

5556

0 0 0 0 6619

1CCCCCCCCA

For the 6× 6 version we have0BBBBBBBBBB@

4 1 0 0 0 1

1 4 1 0 0 0

0 1 4 1 0 0

0 0 1 4 1 0

0 0 0 1 4 1

1 0 0 0 1 4

1CCCCCCCCCCA

=

0BBBBBBBBBBB@

1 0 0 0 0 014 1 0 0 0 0

0 415 1 0 0 0

0 0 1556 1 0 0

0 0 0 56209 1 0

14 − 1

15156 − 1

209726 1

1CCCCCCCCCCCA

0BBBBBBBBBBB@

4 1 0 0 0 1

0 154 1 0 0 − 1

4

0 0 5615 1 0 1

15

0 0 0 20956 1 − 1

56

0 0 0 0 780209

210209

0 0 0 0 0 4513

1CCCCCCCCCCCA

The pattern is that the only the entries lying on the diagonal, the subdiagonal or the lastrow of L are nonzero, while the only nonzero entries of U are on its diagonal, superdiagonalor last column.

♥ 1.7.14.(a) Assuming regularity, the only row operations required to reduce A to upper triangular

form U are, for each j = 1, . . . , n−1, to add multiples of the jth row to the (j +1)st andthe nth rows. Thus, the only nonzero entries below the diagonal in L are at positions(j, j + 1) and (j, n). Moreover, these row operations only affect zero entries in the lastcolumn, leading to the final form of U .

(b)

0B@

1 −1 −1−1 2 −1−1 −1 3

1CA =

0B@

1 0 0−1 1 0−1 2 1

1CA

0B@

1 −1 −10 1 −20 0 2

1CA,

0BBBBB@

1 −1 0 0 −1−1 2 −1 0 0

0 −1 3 −1 00 0 −1 4 −1−1 0 0 −1 5

1CCCCCA

=

0BBBBBB@

1 0 0 0 0−1 1 0 0 0

0 −1 1 0 00 0 − 1

2 1 0

−1 −1 − 12 − 3

7 1

1CCCCCCA

0BBBBBB@

1 −1 0 0 −10 1 −1 0 −10 0 2 −1 −10 0 0 7

2 − 32

0 0 0 0 137

1CCCCCCA

,

0BBBBBBB@

1 −1 0 0 0 −1−1 2 −1 0 0 0

0 −1 3 −1 0 00 0 −1 4 −1 00 0 0 −1 5 −1−1 0 0 0 −1 6

1CCCCCCCA

=

0BBBBBBBB@

1 0 0 0 0 0−1 1 0 0 0 0

0 −1 1 0 0 00 0 − 1

2 1 0 0

0 0 0 − 27 1 0

−1 −1 − 12 − 1

7 − 833 1

1CCCCCCCCA

0BBBBBBBB@

1 −1 0 0 0 −10 1 −1 0 0 −10 0 2 −1 0 −10 0 0 7

2 −1 − 12

0 0 0 0 337 − 8

7

0 0 0 0 0 10433

1CCCCCCCCA

.

The 4× 4 case is a singular matrix.

30

♥ 1.7.15.(a) If matrix A is tridiagonal, then the only nonzero elements in ith row are ai,i−1, aii, ai,i+1.

So aij = 0 whenever | i− j | > 1.

(b) For example,

0BBBBBBB@

2 1 1 0 0 01 2 1 1 0 01 1 2 1 1 00 1 1 2 1 10 0 1 1 2 10 0 0 1 1 2

1CCCCCCCA

has band width 2;

0BBBBBBB@

2 1 1 1 0 01 2 1 1 1 01 1 2 1 1 11 1 1 2 1 10 1 1 1 2 10 0 1 1 1 2

1CCCCCCCA

has band

width 3.(c) U is a matrix that result from applying the row operation # 1 to A, so all zero entries

in A will produce corresponding zero entries in U . On the other hand, if A is of bandwidth k, then for each column of A we need to perform no more than k row replace-ments to obtain zero’s below the diagonal. Thus L which reflects these row replace-ments will have at most k nonzero entries below the diagonal.

(d)

0BBBBBBB@

2 1 1 0 0 01 2 1 1 0 01 1 2 1 1 00 1 1 2 1 10 0 1 1 2 10 0 0 1 1 2

1CCCCCCCA

=

0BBBBBBBBBBB@

1 0 0 0 0 012 1 0 0 0 012

13 1 0 0 0

0 23

12 1 0 0

0 0 34

12 1 0

0 0 0 1 12 1

1CCCCCCCCCCCA

0BBBBBBBBBBB@

2 1 1 0 0 0

0 32

12 1 0 0

0 0 43

23 1 0

0 0 0 1 12 1

0 0 0 0 1 12

0 0 0 0 0 34

1CCCCCCCCCCCA

,

0BBBBBBB@

2 1 1 1 0 01 2 1 1 1 01 1 2 1 1 11 1 1 2 1 10 1 1 1 2 10 0 1 1 1 2

1CCCCCCCA

=

0BBBBBBBBBBB@

1 0 0 0 0 012 1 0 0 0 012

13 1 0 0 0

12

13

14 1 0 0

0 23

12

25 1 0

0 0 34

35

14 1

1CCCCCCCCCCCA

0BBBBBBBBBBB@

2 1 1 1 0 0

0 32

12

12 1 0

0 0 43

13

23 1

0 0 0 54

12

34

0 0 0 0 45

15

0 0 0 0 0 34

1CCCCCCCCCCCA

.

(e)“

13 , 1

3 , 0, 0, 13 , 1

3

”T,“

23 , 1

3 ,− 13 ,− 1

3 , 13 , 2

3

”T.

(f ) For A we still need to compute k multipliers at each stage and update at most 2k2 en-

tries, so we have less than (n−1)(k+2k2) multiplications and (n−1)2k2 additions. Forthe right-hand side we have to update at most k entries at each stage, so we have lessthan (n − 1)k multiplications and (n − 1)k additions. So we can get by with less than

total (n− 1)(2k + 2k2) multiplications and (n− 1)(k + 2k2) additions.(g) The inverse of a banded matrix is not necessarily banded. For example, the inverse of

0B@

2 1 01 2 10 1 2

1CA is

0BBB@

34 − 1

214

− 12 1 − 1

214 − 1

234

1CCCA

1.7.16. (a) (−8, 4 )T , (b) (−10,−4.1 )T , (c) (−8.1,−4.1 )T . (d) Partial pivoting reducesthe effect of round off errors and results in a significantly more accurate answer.

1.7.17. (a) x = 117 ≈ 1.57143, y = 1

7 ≈ .142857, z = − 17 ≈ .142857,

(b) x = 3.357, y = .5, z = −.1429, (c) x = 1.572, y = .1429, z = −.1429.

1.7.18. (a) x = −2, y = 2, z = 3, (b) x = −7.3, y = 3.3, z = 2.9, (c) x = −1.9, y = 2.,z = 2.9, (d) partial pivoting works markedly better, especially for the value of x.

31

1.7.19. (a) x = −220., y = 26, z = .91; (b) x = −190., y = 24, z = .84; (c) x = −210,y = 26, z = 1. (d) The exact solution is x = −213.658, y = 25.6537, z = .858586.Full pivoting is the most accurate. Interestingly, partial pivoting fares a little worse thanregular elimination.

1.7.20. (a)

0BBB@

65

− 135

− 95

1CCCA =

0B@

1.2−2.6−1.8

1CA, (b)

0BBBBBB@

− 14

− 54

1814

1CCCCCCA

, (c)

0BBB@

0110

1CCCA, (d)

0BBBBBB@

− 3235

− 1935

− 1235

− 7635

1CCCCCCA

=

0BBB@

−.9143−.5429−.3429−2.1714

1CCCA.

1.7.21. (a)

0@−

113813

1A =

−.0769

.6154

!, (b)

0BBB@

− 45

− 815

− 1915

1CCCA =

0B@−.8000−.5333−1.2667

1CA,

(c)

0BBBBBB@

21213812159242

− 56121

1CCCCCCA

=

0BBB@

.0165

.3141

.2438−.4628

1CCCA, (d)

0B@−.732−.002

.508

1CA.

1.7.22. The results are the same.

♠ 1.7.23.

Gaussian Elimination With Full Pivoting

start

for i = 1 to n

set σ(i) = τ(i) = i

next i

for j = 1 to n

if mσ(i),j = 0 for all i ≥ j, stop; print “A is singular”

choose i ≥ j and k ≥ j such that mσ(i),τ(k) is maximal

interchange σ(i)←→ σ(j)

interchange τ(k)←→ τ(j)

for i = j + 1 to n

set z = mσ(i),τ(j)/mσ(k),τ(j)

set mσ(i),τ(j) = 0

for k = j + 1 to n + 1

set mσ(i),τ(k) = mσ(i),τ(k) − zmσ(i),τ(k)

next k

next i

next j

end

32

♠ 1.7.24. We let x ∈ Rn be generated using a random number generator, compute b = Hnx and

then solve Hny = b for y. The error is e = x− y and we use e? = max | ei | as a measure ofthe overall error. Using Matlab, running Gaussian Elimination with pivoting:

n 10 20 50 100

e? .00097711 35.5111 318.3845 1771.1

Using Mathematica, running regular Gaussian Elimination:

n 10 20 50 100

e? .000309257 19.8964 160.325 404.625

In Mathematica, using the built-in LinearSolve function, which is more accurate sinceit uses a more sophisticated solution method when confronted with an ill-posed linear sys-tem:

n 10 20 50 100

e? .00035996 .620536 .65328 .516865

(Of course, the errors vary a bit each time the program is run due to the randomness of thechoice of x.)

♠ 1.7.25.

(a) H−13 =

0B@

9 −360 30−36 192 −180

30 −180 180

1CA,

H−14 =

0BBB@

16 −120 240 −140−120 1200 −2700 1680

240 −2700 6480 −4200−140 1680 −4200 2800

1CCCA ,

H−15 =

0BBBBB@

25 −300 1050 −1400 630−300 4080 −18900 26880 −126001050 −18900 79380 −117600 56700−1400 26880 −117600 179200 −88200

630 −12600 56700 −88200 44100

1CCCCCA

.

(b) The same results are obtained when using floating point arithmetic in either Mathe-

matica or Matlab.(c) The product fK10H10, where fK10 is the computed inverse, is fairly close to the 10 × 10

identity matrix; the largest error is .0000801892 in Mathematica or .000036472 inMatlab. As for fK20H20, it is nowhere close to the identity matrix: in Mathemat-

ica the diagonal entries range from −1.34937 to 3.03755, while the largest (in absolutevalue) off-diagonal entry is 4.3505; in Matlab the diagonal entries range from −.4918to 3.9942, while the largest (in absolute value) off-diagonal entry is −5.1994.

1.8.1.(a) Unique solution: (− 1

2 ,− 34 )T ;

(b) infinitely many solutions: (1− 2z,−1 + z, z)T , where z is arbitrary;(c) no solutions;

(d) unique solution: (1,−2, 1)T ;

(e) infinitely many solutions: (5− 2z, 1, z, 0)T , where z is arbitrary;

(f ) infinitely many solutions: (1, 0, 1, w)T , where w is arbitrary;

(g) unique solution: (2, 1, 3, 1)T .

33

1.8.2. (a) Incompatible; (b) incompatible; (c) (1, 0)T ; (d) (1+3x2−2x3, x2, x3)T , where x2

and x3 are arbitrary; (e) (− 152 , 23,−10)T ; (f ) (−5− 3x4, 19− 4x4,−6− 2x4, x4)

T , wherex4 is arbitrary; (g) incompatible.

1.8.3. The planes intersect at (1, 0, 0).

1.8.4. (i) a 6= b and b 6= 0; (ii) b = 0, a 6= −2; (iii) a = b 6= 0, or a = −2 and b = 0.

1.8.5. (a) b = 2, c 6= −1 or b = 12 , c 6= 2; (b) b 6= 2, 1

2 ; (c) b = 2, c = −1, or b = 12 , c = 2.

1.8.6.

(a)“

1 + i − 12 (1 + i )y, y,− i

”T, where y is arbitrary;;

(b) ( 4 i z + 3 + i , i z + 2− i , z )T , where z is arbitrary;

(c) ( 3 + 2 i ,−1 + 2 i , 3 i )T ;

(d) (−z − (3 + 4 i )w,−z − (1 + i )w, z, w )T , where z and w are arbitrary.

1.8.7. (a) 2, (b) 1, (c) 2, (d) 3, (e) 1, (f ) 1, (g) 2, (h) 2, (i) 3.

1.8.8.

(a)

1 11 −2

!=

1 01 1

! 1 10 −3

!,

(b)

2 1 3−2 −1 −3

!=

1 0−1 1

! 2 1 30 0 0

!,

(c)

0B@

1 −1 11 −1 2−1 1 0

1CA =

0B@

1 0 01 1 0−1 1 1

1CA

0B@

1 −1 10 0 10 0 0

1CA,

(d)

0B@

1 0 00 0 10 1 0

1CA

0B@

2 −1 01 1 −12 −1 1

1CA =

0B@

1 0 012 1 01 0 1

1CA

0B@

2 −1 00 3

2 −10 0 1

1CA,

(e)

0B@

30−2

1CA =

0B@

1 0 00 1 0− 2

3 0 1

1CA

0B@

300

1CA,

(f ) ( 0 −1 2 5 ) = ( 1 )( 0 −1 2 5 ),

(g)

0BBB@

0 1 0 01 0 0 00 0 1 00 0 0 1

1CCCA

0BBB@

0 −34 −11 2−1 −5

1CCCA =

0BBB@

1 0 0 00 1 0 014 − 3

4 1 0

− 14 − 7

4 0 1

1CCCA

0BBB@

1 20 70 00 0

1CCCA,

(h)

0BBBBB@

1 −1 2 12 1 −1 01 2 −3 −14 −1 3 20 3 −5 −2

1CCCCCA

=

0BBBBB@

1 0 0 0 02 1 0 0 01 1 1 0 04 1 0 1 00 1 0 0 1

1CCCCCA

0BBBBB@

1 −1 2 10 3 −5 −20 0 0 00 0 0 00 0 0 0

1CCCCCA

,

(i)

0B@

0 1 00 0 11 0 0

1CA

0B@

0 0 0 3 11 2 −3 1 −22 4 −2 1 −2

1CA =

0B@

1 0 02 1 00 0 1

1CA

0B@

1 2 −3 1 −20 0 4 −1 20 0 0 3 1

1CA.

1.8.9. (a)

x = 1,

y = 0,

z = 0.

(b)

x + y = 1,

y + z = 0,

x− z = 1.

(c)

x + y = 1,

y + z = 0,

x− z = 0.

1.8.10. (a)

1 0 00 1 0

!, (b)

0B@

1 0 00 1 00 0 0

1CA, (c)

0B@

1 00 10 0

1CA, (d)

0B@

1 0 00 1 00 0 1

1CA.

34

1.8.11.(a) x2 + y2 = 1, x2 − y2 = 2;

(b) y = x2, x− y + 2 = 0; solutions: x = 2, y = 4 and x = −1, y = 1;

(c) y = x3, x− y = 0; solutions: x = y = 0, x = y = −1, x = y = 1;(d) y = sin x, y = 0; solutions: x = kπ, y = 0, for k any integer.

1.8.12. That variable does not appear anywhere in the system, and is automatically free (al-though it doesn’t enter into any of the formulas, and so is, in a sense, irrelevant).

1.8.13. True. For example, take a matrix in row echelon form with r pivots, e.g., the matrix Awith aii = 1 for i = 1, . . . , r, and all other entries equal to 0.

1.8.14. Both false. The zero matrix has no pivots, and hence has rank 0.

♥ 1.8.15.(a) Each row of A = vwT is a scalar multiple, namely vi w, of the vector w. If necessary,

we use a row interchange to ensure that the first row is non-zero. We then subtract theappropriate scalar multiple of the first row from all the others. This makes all rows be-low the first zero, and so the resulting matrix is in row echelon form has a single non-zero row, and hence a single pivot — proving that A has rank 1.

(b) (i)

−1 2−3 6

!, (ii)

0B@−8 4

0 04 −2

1CA, (iii)

2 6 −2−3 −9 3

!.

(c) The row echelon form of A must have a single nonzero row, say wT . Reversing the ele-mentary row operations that led to the row echelon form, at each step we either inter-change rows or add multiples of one row to another. Every row of every matrix obtained

in such a fashion must be some scalar multiple of wT , and hence the original matrix

A = vwT , where the entries vi of the vector v are the indicated scalar multiples.

1.8.16. 1.

1.8.17. 2.

1.8.18. Example: A =

1 00 0

!, B =

0 10 0

!so AB =

0 10 0

!has rank 1, but BA =

0 00 0

!

has rank 0.

♦ 1.8.19.(a) Under elementary row operations, the reduced form of C will be

“U Z

”where U is the

row echelon form of A. Thus, C has at least r pivots, namely the pivots in A. Examples:

rank

1 2 12 4 2

!= 1 = rank

1 22 4

!, while rank

1 2 12 4 3

!= 2 > 1 = rank

1 22 4

!.

(b) Applying elementary row operations, we can reduce E to

UW

!where U is the row ech-

elon form of A. If we can then use elementary row operations of type #1 to eliminateall entries of W , then the row echelon form of E has the same number of pivots as Aand so rank E = rank A. Otherwise, at least one new pivot appears in the rows below U ,and rank E > rank A. Examples:

rank

0B@

1 22 43 6

1CA = 1 = rank

1 22 4

!, while rank

0B@

1 22 43 5

1CA = 2 > 1 = rank

1 22 4

!.

♦ 1.8.20. By Proposition 1.39, A can be reduced to row echelon form U by a sequence of elemen-tary row operations. Therefore, as in the proof of the LU decomposition, A = E1 E2 · · ·EN U

where E−11 , . . . , E−1

N are the elementary matrices representing the row operations. If A issingular, then U = Z must have at least one all zero row.

35

♦ 1.8.21. After row operations, the augmented matrix becomes N =“

U | c”

where the r = rank A

nonzero rows of U contain the pivots of A. If the system is compatible, then the last m − rentries of c are all zero, and hence N is itself a row echelon matrix with r nonzero rows andhence rank M = rank N = r. If the system is incompatible, then one or more of the lastm − r entries of c are nonzero, and hence, by one more set of row operations, N is placedin row echelon form with a final pivot in row r + 1 of the last column. In this case, then,rank M = rank N = r + 1.

1.8.22. (a) x = z, y = z, where z is arbitrary; (b) x = − 23 z, y = 7

9 z, where z is arbitrary;

(c) x = y = z = 0; (d) x = 13 z − 2

3 w, y = 56 z − 1

6 w, where z and w are arbitrary;

(e) x = 13z, y = 5z, w = 0, where z is arbitrary; (f ) x = 32 w, y = 1

2 w, z = 12 w, where

w is arbitrary.

1.8.23. (a)“

13 y, y

”T, where y is arbitrary; (b)

“− 6

5z, 85z, z

”T, where z is arbitrary;

(c)“− 11

5 z + 35w, 2

5z − 65w, z, w

”T, where z and w are arbitrary; (d) ( z,−2z, z )T , where

z is arbitrary; (e) (−4z, 2z, z )T , where z is arbitrary; (f ) ( 0, 0, 0 )T ; (g) ( 3z, 3z, z, 0 )T ,

where z is arbitrary; (h) ( y − 3w, y, w, w )T , where y and w are arbitrary.

1.8.24. If U has only nonzero entries on the diagonal, it must be nonsingular, and so the onlysolution is x = 0. On the other hand, if there is a diagonal zero entry, then U cannot haven pivots, and so must be singular, and the system will admit nontrivial solutions.

1.8.25. For the homogeneous case x1 = x3, x2 = 0, where x3 is arbitrary. For the inhomoge-

neous case x1 = x3 + 14 (a + b), x2 = 1

2 (a − b), where x3 is arbitrary. The solution to thehomogeneous version is a line going through the origin, while the inhomogeneous solution

is a parallel line going through the point“

14 (a + b), 0, 1

2 (a− b)”T

. The dependence on the

free variable x3 is the same as in the homogeneous case.

1.8.26. For the homogeneous case x1 = − 16 x3 − 1

6 x4, x2 = − 23 x3 + 4

3 x4, where x3 and x4

are arbitrary. For the inhomogeneous case x1 = − 16 x3 − 1

6 x4 + 13 a + 1

6 b, x2 = − 23 x3 +

43 x4 + 1

3 a + 16 b, where x3 and x4 are arbitrary. The dependence on the free variable x3 is

the same as in the homogeneous case.

1.8.27. (a) k = 2 or k = −2; (b) k = 0 or k = 12 ; (c) k = 1.

1.9.1.

(a) Regular matrix, reduces to upper triangular form U =

2 −10 1

!, so determinant is 2;

(b) Singular matrix, row echelon form U =

0B@−1 0 3

0 1 −20 0 0

1CA, so determinant is 0;

(c) Regular matrix, reduces to upper triangular form U =

0B@

1 2 30 1 20 0 −3

1CA, so determinant is −3;

(d) Nonsingular matrix, reduces to upper triangular form U =

0B@−2 1 3

0 1 −10 0 3

1CA after one row

interchange, so determinant is 6;

36

(e) Upper triangular matrix, so the determinant is a product of diagonal entries: −180;

(f ) Nonsingular matrix, reduces to upper triangular form U =

0BBB@

1 −2 1 40 2 −1 −70 0 −2 −80 0 0 10

1CCCA after

one row interchange, so determinant is 40;

(g) Nonsingular matrix, reduces to upper triangular form U =

0BBBBB@

1 −2 1 4 −50 3 −3 −1 20 0 4 −12 240 0 0 −5 100 0 0 0 1

1CCCCCA

after one row interchange, so determinant is 60.

1.9.2. det A = −2, det B = −11 and det AB = det

0B@

5 4 41 5 1−2 10 0

1CA = 22.

1.9.3. (a) A =

2 3−1 −2

!; (b) By formula (1.82),

1 = det I = det(A2) = det(AA) = det A det A = (det A)2, so det A = ±1.

1.9.4. det A2 = (det A)2 = det A, and hence det A = 0 or 1

1.9.5.(a) True. By Theorem 1.52, A is nonsingular, so, by Theorem 1.18, A−1 exists

(b) False. For A =

2 3−1 −2

!, we have 2 det A = −2 and det 2A = −4. In general,

det(2A) = 2n det A.

(c) False. For A =

2 3−1 −2

!and B =

0 10 0

!, we have det(A + B) = det

2 4−1 −2

!=

0 6= −1 = det A + det B.

(d) True. det A−T = det(A−1)T = det A−1 = 1/ det A, where the second equality followsfrom Proposition 1.56, and the third equality follows from Proposition 1.55.

(e) True. det(AB−1) = det A det B−1 = det A/ det B, where the first equality follows fromformula (1.82) and the second equality follows from Proposition 1.55.

(f ) False. If A =

2 3−1 −2

!and B =

0 10 0

!, then det(A+B)(A−B) = det

0 −40 2

!=

0 6= det(A2 −B2) = det

1 00 1

!= 1. However, if AB = BA, then det(A + B)(A−B) =

det(A2 −AB + BA−B2) = det(A2 −B2).(g) True. Proposition 1.42 says rank A = n if and only if A is nonsingular, while Theo-

rem 1.52 implies that det A 6= 0.(h) True. Since det A = 1 6= 0, Theorem 1.52 implies that A is nonsingular, and so B =

A−1O = O.

1.9.6. Never — its determinant is always zero.

1.9.7. By (1.82, 83) and commutativity of numeric multiplication,

det B = det(S−1 AS) = det S−1 det A det S =1

det Sdet A det S = det A.

1.9.8. Multiplying one row of A by c multiplies its determinant by c. To obtain c A, we mustmultiply all n rows by c, and hence the determinant is multiplied by c a total of n times.

1.9.9. By Proposition 1.56, det LT = det L. If L is a lower triangular matrix, then LT is an

37

upper triangular matrix. By Theorem 1.50, det LT is the product of its diagonal entrieswhich are the same as the diagonal entries of L.

1.9.10. (a) See Exercise 1.9.8. (b) If n is odd, det(−A) = − det A. On the other hand, if

AT = −A, then det A = det AT = − det A, and hence det A = 0. (c) A =

0 1−1 0

!.

♦ 1.9.11. We have

det

a b

c + ka d + kb

!= ad + akb− bc− bka = ad− bc = det

a bc d

!,

det

c da b

!= cb− ad = −(ad− bc) = − det

a bc d

!,

det

ka kbc d

!= kad− kbc = k (ad− bc) = k det

a bc d

!,

det

a b0 d

!= ad− b0 = ad.

♦ 1.9.12.(a) The product formula holds if A is an elementary matrix; this is a consequence of the

determinant axioms coupled with the fact that elementary matrices are obtained by ap-plying the corresponding row operation to the identity matrix, with det I = 1.

(b) By induction, if A = E1 E2 · · ·EN is a product of elementary matrices, then (1.82) alsoholds. Proposition 1.25 then implies that the product formula is valid whenever A isnonsingular.

(c) The first result is in Exercise 1.2.24(a), and so the formula follows by applying Lemma 1.51to Z and Z B.

(d) According to Exercise 1.8.20, every singular matrix can be written as A = E1 E2 · · ·EN Z,where the Ei are elementary matrices, while Z, its row echelon form, is a matrix with arow of zeros. But then Z B = W also has a row of zeros, and so AB = E1 E2 · · ·ENWis also singular. Thus, both sides of (1.82) are zero in this case.

1.9.13. Indeed, by (1.82), det A det A−1 = det(AA−1) = det I = 1.

♦ 1.9.14. Exercise 1.6.28 implies that, if A is regular, so is AT , and they both have the same piv-ots. Since the determinant of a regular matrix is the product of the pivots, this implies

det A = det AT . If A is nonsingular, then we use the permuted LU decomposition to write

A = PT LU where PT = P−1 by Exercise 1.6.14. Thus, det A = det P T det U = ±det U ,

while det AT = det(UT LT P ) = det U det P = ±det U where det P−1 = det P = ±1.Finally, if A is singular, then the same computation holds, with U denoting the row echelon

form of A, and so det A = det U = 0 = ± det AT .

1.9.15.

det

0BBB@

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34a41 a42 a43 a44

1CCCA =

a11 a22 a33 a44 − a11 a22 a34 a43 − a11 a23 a32 a44 + a11 a23 a34 a42 − a11 a24 a33 a42

+ a11 a24 a32 a43 − a12 a21 a33 a44 + a12 a21 a34 a43 + a12 a23 a31 a44 − a12 a23 a34 a41

+ a12 a24 a33 a41 − a12 a24 a31 a43 + a13 a21 a32 a44 − a13 a21 a34 a42 − a13 a22 a31 a44

+ a13 a22 a34 a41 − a13 a24 a32 a41 + a13 a24 a31 a42 − a14 a21 a32 a43 + a14 a21 a33 a42

+ a14 a22 a31 a43 − a14 a22 a33 a41 + a14 a23 a32 a41 − a14 a23 a31 a42.

38

♦ 1.9.16.(i) Suppose B is obtained from A by adding c times row k to row l, so

bij =

8<:

alj + caij , i = l,

aij , i 6= l.Thus, each summand in the determinantal formula for

det B splits into two terms, and we find that det B = det A + c det C, where C is thematrix obtained from A by replacing row l by row k. But rows k and l of C are identi-cal, and so, by axiom (ii), if we interchange the two rows det C = − det C = 0. Thus,det B = det A.

(ii) Let B be obtained from A by interchanging rows k and l. Then each summand in theformula for det B equals minus the corresponding summand in the formula for det A,since the permutation has changed sign, and so det B = − det A.

(iii) Let B be obtained from A by multiplying rows k by c. Then each summand in the for-mula for det B contains one entry from row k, and so equals c times the correspondingterm in det A, hence det B = c det A.

(iv) The only term in det U that does not contain at least one zero entry lying below thediagonal is for the identity permutation π(i) = i, and so det U is the product of its diag-onal entries.

♦ 1.9.17. If U is nonsingular, then, by Gauss–Jordan elimination, it can be reduced to the iden-tity matrix by elementary row operations of types #1 and #3. Each operation of type #1doesn’t change the determinant, while operations of type #3 multiply the determinant bythe diagonal entry. Thus, det U = u11u22 · · ·unn det I . On the other hand, U is singular ifand only if one or more of its diagonal entries are zero, and so det U = 0 = u11u22 · · ·unn.

♦ 1.9.18. The determinant of an elementary matrix of type #2 is −1, whereas all elementary ma-trices of type #1 have determinant +1, and hence so does any product thereof.

♥ 1.9.19.(a) Since A is regular, a 6= 0 and ad− bc 6= 0. Subtracting c/a times the first from from the

second row reduces A to the upper triangular matrix

a b0 d + b(−c/a)

!, and its pivots

are a and d− bca =

ad− bca =

det Aa .

(b) As in part (a) we reduce A to an upper triangular form. First, we subtract c/a timesthe first row from the second row, and g/a times the first row from third row, resulting

in the matrix

0BBB@

a b e

0ad− bc

aaf − ce

a

0ah− bg

aaj − cg

a

1CCCA. Performing the final row operation reduces

the matrix to the upper triangular form U =

0BBB@

a b e

0ad− bc

a −af − cea

0 0 P

1CCCA, whose pivots

are a,ad− bc

a, and

aj − eg

a− (af − ce)(ah− bg)

a (ad− bc)=

adj + bf g + ech− af h− bcj − edg

ad− bc=

det A

ad− bc.

(c) If A is a regular n × n matrix, then its first pivot is a11, and its kth pivot, for k =2, . . . , n, is det Ak/det Ak−1, where Ak is the k × k upper left submatrix of A with en-tries aij for i, j = 1, . . . , k. A formal proof is done by induction.

♥ 1.9.20. (a–c) Applying an elementary column operation to a matrix A is the same as apply-

ing the elementary row operation to its transpose AT and then taking the transpose of theresult. Moreover, Proposition 1.56 implies that taking the transpose does not affect the de-

39

terminant, and so any elementary column operation has exactly the same effect as the cor-responding elementary row operation.

(d) Apply the transposed version of the elementary row operations required to reduce AT

to upper triangular form. Thus, if the (1, 1) entry is zero, use a column interchange toplace a nonzero pivot in the upper left position. Then apply elementary column operationsof type #1 to make all entries to the right of the pivot zero. Next, make sure a nonzeropivot is in the (2, 2) position by a column interchange if necessary, and then apply elemen-tary column operations of type #1 to make all entries to the right of the pivot zero. Con-tinuing in this fashion, if the matrix is nonsingular, the result is an lower triangular matrix.(e) We first interchange the first and second columns, and then use elementary column op-erations of type #1 to reduce the matrix to lower triangular form:

det

0B@

0 1 2−1 3 5

2 −3 1

1CA = − det

0B@

1 0 23 −1 5−3 2 1

1CA

= − det

0B@

1 0 03 −1 −1−3 2 7

1CA = − det

0B@

1 0 03 −1 0−3 2 5

1CA = 5.

♦ 1.9.21. Using the LU factorizations established in Exercise 1.3.25:

(a) det

1 1t1 t2

!= t2 − t1, (b) det

0B@

1 1 1t1 t2 t3t21 t22 t23

1CA = (t2 − t1)(t3 − t1)(t3 − t2),

(c) det

0BBBB@

1 1 1 1t1 t2 t3 t4t21 t22 t23 t24t31 t32 t33 t34

1CCCCA

= (t2 − t1)(t3 − t1)(t3 − t2)(t4 − t1)(t4 − t2)(t4 − t3).

The general formula is found in Exercise 4.4.29.

♥ 1.9.22.(a) By direct substitution:

ax + by = apd− bq

ad− bc+ b

aq − pc

ad− bc= p, cx + dy = c

pd− bq

ad− bc+ d

aq − pc

ad− bc= q.

(b) (i) x = − 1

10det

13 30 2

!= −2.6, y = − 1

10det

1 134 0

!= 5.2;

(ii) x =1

12det

4 −2−2 6

!=

5

3, y =

1

12det

1 43 −2

!= − 7

6.

(c) Proof by direct substitution, expanding all the determinants.

(d) (i) x =1

9det

0B@

3 4 02 2 10 1 −1

1CA = − 1

9, y =

1

9det

0B@

1 3 04 2 1−1 0 −1

1CA =

7

9,

z =1

9det

0B@

1 4 34 2 2−1 1 0

1CA =

8

9; (ii) x = − 1

2det

0B@

1 2 −12 −3 23 −1 1

1CA = 0,

y = − 1

2det

0B@

3 1 −11 2 22 3 1

1CA = 4, z = − 1

2det

0B@

3 2 11 −3 22 −1 3

1CA = 7.

(e) Assuming A is nonsingular, the solution to Ax = b is xi = det Ai/ det A, where Ai

is obtained by replacing the ith column of A by the right hand side b. See [60] for acomplete justification.

40

♦ 1.9.23.(a) We can individually reduce A and B to upper triangular forms U1 and U2 with the

determinants equal to the products of their respective diagonal entries. Applying theanalogous elementary row operations to D will reduce it to the upper triangular form

U1 O

O U2

!, and its determinant is equal to the product of its diagonal entries, which

are the diagonal entries of both U1 and U2, so det D = det U1 det U2 = det A det B.(b) The same argument as in part (a) proves the result. The row operations applied to A

are also applied to C, but this doesn’t affect the final upper triangular form.

(c) (i) det

0B@

3 2 −20 4 −50 3 7

1CA = det(3) det

4 −53 7

!= 3 · 43 = 129,

(ii) det

0BBB@

1 2 −2 5−3 1 0 −5

0 0 1 30 0 2 −2

1CCCA = det

1 2−3 1

!det

1 32 −2

!= 7 · (−8) = −56,

(iii) det

0BBB@

1 2 0 4−3 1 4 −1

0 3 1 80 0 0 −3

1CCCA = det

0B@

1 2 0−3 1 4

0 3 1

1CA det(−3) = (−5) · (−3) = 15,

(iv) det

0BBB@

5 −1 0 02 5 0 02 4 4 −23 −2 9 −5

1CCCA = det

5 −12 5

!det

4 −29 −5

!= 27 · (−2) = −54.

41

Solutions — Chapter 2

2.1.1. Commutativity of Addition:

(x + i y) + (u + i v) = (x + u) + i (y + v) = (u + i v) + (x + i y).

Associativity of Addition:

(x + i y) +h(u + i v) + (p + i q)

i= (x + i y) +

h(u + p) + i (v + q)

i

= (x + u + p) + i (y + v + q)

=h(x + u) + i (y + v)

i+ (p + i q) =

h(x + i y) + (u + i v)

i+ (p + i q).

Additive Identity : 0 = 0 = 0 + i 0 and

(x + i y) + 0 = x + i y = 0 + (x + i y).

Additive Inverse: −(x + i y) = (−x) + i (−y) and

(x + i y) +h(−x) + i (−y)

i= 0 =

h(−x) + i (−y)

i+ (x + i y).

Distributivity :

(c + d)(x + i y) = (c + d)x + i (c + d)y = (cx + dx) + i (cy + dy) = c(x + i y) + d(x + i y),

c[ (x + i y) + (u + i v) ] = c(x + u) + (y + v) = (cx + cu) + i (cy + cv) = c(x + i y) + c(u + i v).

Associativity of Scalar Multiplication:

c [d(x + i y) ] = c [ (dx) + i (dy) ] = (cdx) + i (cdy) = (cd)(x + i y).

Unit for Scalar Multiplication: 1(x + i y) = (1x) + i (1y) = x + i y.

Note: Identifying the complex number x + i y with the vector ( x, y )T ∈ R2 respects the opera-

tions of vector addition and scalar multiplication, and so we are in effect reproving that R2 is a

vector space.

2.1.2. Commutativity of Addition:

(x1, y1) + (x2, y2) = (x1 x2, y1 y2) = (x2, y2) + (x1, y1).

Associativity of Addition:

(x1, y1) +h(x2, y2) + (x3, y3)

i= (x1 x2 x3, y1 y2 y3) =

h(x1, y1) + (x2, y2)

i+ (x3, y3).

Additive Identity : 0 = (1, 1), and

(x, y) + (1, 1) = (x, y) = (1, 1) + (x, y).

Additive Inverse:

−(x, y) =

1

x,1

y

!and (x, y) +

h−(x, y)

i= (1, 1) =

h−(x, y)

i+ (x, y).

Distributivity :

(c + d)(x, y) = (xc+d, yc+d) = (xc xd, yc yd) = (xc, yc) + (xd, yd) = c(x, y) + d(x, y)

ch(x1, y1) + (x2, y2)

i= ((x1 x2)

c, (y1 y2)c) = (xc

1 xc2, yc

1 yc2)

= (xc1, yc

1) + (xc2, yc

2) = c(x1, y1) + c(x2, y2).

Associativity of Scalar Multiplication:

c(d(x, y)) = c(xd, yd) = (xcd, ycd) = (cd)(x, y).

Unit for Scalar Multiplication: 1(x, y) = (x, y).

42

Note: We can uniquely identify a point (x, y) ∈ Q with the vector ( log x, log y )T ∈ R2. Then

the indicated operations agree with standard vector addition and scalar multiplication in R2,

and so Q is just a disguised version of R2.

♦ 2.1.3. We denote a typical function in F(S) by f(x) for x ∈ S.Commutativity of Addition:

(f + g)(x) = f(x) + g(x) = (f + g)(x).

Associativity of Addition:

[f +(g +h)](x) = f(x)+ (g +h)(x) = f(x)+ g(x)+h(x) = (f + g)(x)+h(x) = [(f + g)+h](x).

Additive Identity : 0(x) = 0 for all x, and (f + 0)(x) = f(x) = (0 + f)(x).Additive Inverse: (−f)(x) = −f(x) and

[f + (−f)](x) = f(x) + (−f)(x) = 0 = (−f)(x) + f(x) = [(−f) + f ](x).

Distributivity :

[(c + d)f ](x) = (c + d)f(x) = cf(x) + df(x) = (cf)(x) + (df)(x),

[c(f + g)](x) = cf(x) + cg(x) = (cf)(x) + (cg)(x).

Associativity of Scalar Multiplication:

[c(df)](x) = cdf(x) = [(cd)f ](x).

Unit for Scalar Multiplication: (1f)(x) = f(x).

2.1.4. (a) ( 1, 1, 1, 1 )T , ( 1,−1, 1,−1 )T , ( 1, 1, 1, 1 )T , ( 1,−1, 1,−1 )T . (b) Obviously not.

2.1.5. One example is f(x) ≡ 0 and g(x) = x3 − x.

2.1.6. (a) f(x) = −4x + 3; (b) f(x) = −2x2 − x + 1.

2.1.7.

(a)

x− yxy

!,

ex

cos y

!, and

13

!, which is a constant function.

(b) Their sum is

x− y + ex + 1xy + cos y + 3

!. Multiplied by −5 is

−5x + 5y − 5ex − 5−5xy − 5 cos y − 15

!.

(c) The zero element is the constant function 0 =

00

!.

♦ 2.1.8. This is the same as the space of functions F(R2, R2). Explicitly:Commutativity of Addition:

v1(x, y)v2(x, y)

!+

w1(x, y)w2(x, y)

!=

v1(x, y) + w1(x, y)v2(x, y) + w2(x, y)

!=

w1(x, y)w2(x, y)

!+

v1(x, y)v2(x, y)

!.

Associativity of Addition: u1(x, y)u2(x, y)

!+

" v1(x, y)v2(x, y)

!+

w1(x, y)w2(x, y)

!#=

u1(x, y) + v1(x, y) + w1(x, y)u2(x, y) + v2(x, y) + w2(x, y)

!

=

" u1(x, y)u2(x, y)

!+

v1(x, y)v2(x, y)

!#+

w1(x, y)w2(x, y)

!.

Additive Identity : 0 = (0, 0) for all x, y, and

v1(x, y)v2(x, y)

!+ 0 =

v1(x, y)v2(x, y)

!= 0 +

v1(x, y)v2(x, y)

!.

Additive Inverse: −

v1(x, y)v2(x, y)

!=

−v1(x, y)−v2(x, y)

!, and

v1(x, y)v2(x, y)

!+

−v1(x, y)−v2(x, y)

!= 0 =

−v1(x, y)−v2(x, y)

!+

v1(x, y)v2(x, y)

!.

43

Distributivity :

(c + d)

v1(x, y)v2(x, y)

!=

(c + d)v1(x, y)(c + d)v2(x, y)

!= c

v1(x, y)v2(x, y)

!+ d

v1(x, y)v2(x, y)

!,

c

" v1(x, y)v2(x, y)

!+

w1(x, y)w2(x, y)

!#=

cv1(x, y) + cw1(x, y)cv2(x, y) + cw2(x, y)

!= c

v1(x, y)v2(x, y)

!+ c

w1(x, y)w2(x, y)

!.

Associativity of Scalar Multiplication:

c

"d

v1(x, y)v2(x, y)

!#=

cdv1(x, y)cdv2(x, y)

!= (cd)

v1(x, y)v2(x, y)

!.

Unit for Scalar Multiplication:

1

v1(x, y)v2(x, y)

!=

v1(x, y)v2(x, y)

!.

♥ 2.1.9. We identify each sample value with the matrix entry mij = f(ih, j k). In this way, every

sampled function corresponds to a uniquely determined m × n matrix and conversely. Ad-dition of sample functions, (f + g)(ih, j k) = f(ih, j k) + g(ih, j k) corresponds to matrixaddition, mij + nij , while scalar multiplication of sample functions, cf(ih, j k), corresponds

to scalar multiplication of matrices, cmij .

2.1.10. a + b = (a1 + b1, a2 + b2, a3 + b3, . . . ), ca = (ca1, ca2, ca3, . . . ). Explicity verification ofthe vector space properties is straightforward. An alternative, smarter strategy is to iden-tify R

∞ as the space of functions f : N → R where N = 1, 2, 3, . . . is the set of naturalnumbers and we identify the function f with its sample vector f = (f(1), f(2), . . . ).

2.1.11. (i) v + (−1)v = 1v + (−1)v =“

1 + (−1)”v = 0v = 0.

(j) Let z = c0. Then z + z = c(0 + 0) = c0 = z, and so, as in the proof of (h), z = 0.

(k) Suppose c 6= 0. Then v = 1v =

1

c· c!

v =1

c(cv) =

1

c0 = 0.

♦ 2.1.12. If 0 and e0 both satisfy axiom (c), then 0 = e0 + 0 = 0 + e0 = e0.

♦ 2.1.13. Commutativity of Addition:

(v,w) + (bv, bw) = (v + bv,w + bw) = (bv, bw) + (v,w).

Associativity of Addition:

(v,w) +h(bv, bw) + (ev, ew)

i= (v + bv + ev,w + bw + ew) =

h(v,w) + (bv, bw)

i+ (ev, ew).

Additive Identity : the zero element is (0,0), and

(v,w) + (0,0) = (v,w) = (0,0) + (v,w).

Additive Inverse: −(v,w) = (−v,−w) and

(v,w) + (−v,−w) = (0,0) = (−v,−w) + (v,w).

Distributivity :(c + d)(v,w) = ((c + d)v, (c + d)w) = c(v,w) + d(v,w),

ch(v,w) + (bv, bw)

i= (cv + c bv, cv + c bw) = c(v,w) + c(bv, bw).

Associativity of Scalar Multiplication:

c(d(v,w)) = (cdv, cdw) = (cd)(v,w).

Unit for Scalar Multiplication: 1(v,w) = (1v, 1w) = (v,w).

2.1.14. Here V = C0 while W = R, and so the indicated pairs belong to the Cartesian prod-uct vector space C0 × R. The zero element is the pair 0 = (0, 0) where the first 0 denotesthe identically zero function, while the second 0 denotes the real number zero. The laws ofvector addition and scalar multiplication are

(f(x), a) + (g(x), b) = (f(x) + g(x), a + b), c(f(x), a) = (cf(x), ca).

44

2.2.1.(a) If v = ( x, y, z )T satisfies x− y + 4z = 0 and ev = ( ex, ey, ez )T also satisfies ex− ey + 4ez = 0,

so does v + ev = ( x + ex, y + ey, z + ez )T since (x + ex)− (y + ey) + 4(z + ez) = (x− y + 4z) +

(ex−ey+4ez) = 0, as does cv = ( cx, cy, cz )T since (cx)−(cy)+4(cz) = c(x−y+4z) = 0.

(b) For instance, the zero vector 0 = ( 0, 0, 0 )T does not satisfy the equation.

2.2.2. (b,c,d,g,i) are subspaces; the rest are not. Case (j) consists of the 3 coordinate axes andthe line x = y = z.

2.2.3. (a) Subspace:

-1-0.5

00.5

1

-1

-0.5

0

0.5

1

-2

0

2

-1

-0.5

0

0.5

1

(b) Not a subspace:

-1-0.50

0.51

-1-0.5

00.5

1

-10

0

10

-1-0.5

00.5

1

(c) Subspace:

-1-0.5

0

0.5

1

-1

-0.5

0

0.51

-1

-0.5

0

0.5

1

-1-0.5

0

0.5

1

-1

-0.5

0

0.51

(d) Not a subspace:

-1-0.5

00.5

1

-1

-0.5

00.5

1

-1

-0.5

0

0.5

1

-1-0.5

00.5

1

-1

-0.5

00.5

1

(e) Not a subspace:-1

-0.5

0

0.5

1

-2.5

-2.25

-2

-1.75

-1.5

0.5

0.75

1

1.25

1.5

-1-0.5

0

0.5

1

-2.5

-2.25

-2

-1.75

-1.5

(f ) Even though the cylinders are not

subspaces, their intersection is the z axis, which is a subspace:

-2-1

01

2

-2

-1

01

2

-2

-1

0

1

2

-2-1

01

2

-2

-1

01

2

2.2.4. Any vector of the form a

0B@

12−1

1CA + b

0B@

201

1CA + c

0B@

0−1

3

1CA =

0B@

a + 2b2a− c

−a + b + 3c

1CA =

0B@

xyz

1CA will

belong to W . The coefficient matrix

0B@

1 2 02 0 −1−1 1 3

1CA is nonsingular, and so for any

45

x = ( x, y, z )T ∈ R3 we can arrange suitable values of a, b, c by solving the linear system.

Thus, every vector in R3 belongs to W and so W = R

3.

2.2.5. False, with two exceptions: [0, 0] = 0 and (−∞,∞) = R.

2.2.6.(a) Yes. For instance, the set S = (x, 0 ∪ (0, y) consisting of the coordinate axes has

the required property, but is not a subspace. More generally, any (finite) collection of 2or more lines going through the origin satisfies the property, but is not a subspace.

(b) For example, S = (x, y) |x, y ≥ 0 — the positive quadrant.

2.2.7. (a,c,d) are subspaces; (b,e) are not.

2.2.8. Since x = 0 must belong to the subspace, this implies b = A0 = 0. For a homogeneoussystem, if x,y are solutions, so Ax = 0 = Ay, so are x+y since A(x+y) = Ax+ Ay = 0,as is cx since A(cx) = cAx = 0.

2.2.9. L and M are strictly lower triangular if lij = 0 = mij whenever i ≤ j. Then N = L + M

is strictly lower triangular since nij = lij + mij = 0 whenever i ≤ j, as is K = cL since

kij = c lij = 0 whenever i ≤ j.

♦ 2.2.10. Note tr(A+B) =nX

i=1

(aii + bii) = tr A+tr B and tr(cA) =nX

i=1

caii = cnX

i=1

aii = c tr A.

Thus, if tr A = tr B = 0, then tr(A + B) = 0 = tr(cA), proving closure.

2.2.11.(a) No. The zero matrix is not an element.

(b) No if n ≥ 2. For example, A =

1 00 0

!, B =

0 00 1

!satisfy det A = 0 = det B, but

det(A + B) = det

1 00 1

!= 1, so A + B does not belong to the set.

2.2.12. (d,f,g,h) are subspaces; the rest are not.

2.2.13. (a) Vector space; (b) not a vector space: (0, 0) does not belong; (c) vector space;(d) vector space; (e) not a vector space: If f is non-negative, then −1 f = −f is not (un-less f ≡ 0); (f ) vector space; (g) vector space; (h) vector space.

2.2.14. If f(1) = 0 = g(1), then (f + g)(1) = 0 and (cf)(1) = 0, so both f + g and cf be-long to the subspace. The zero function does not satisfy f0) = 1. For a subspace, a can beanything, while b = 0.

2.2.15. All cases except (e,g) are subspaces. In (g), |x | is not in C1.

2.2.16. (a) Subspace; (b) subspace; (c) Not a subspace: the zero function does not satisfythe condition; (d) Not a subspace: if f(0) = 0, f(1) = 1, and g(0) = 1, g(1) = 0, then fand g are in the set, but f + g is not; (e) subspace; (f ) Not a subspace: the zero functiondoes not satisfy the condition; (g) subspace; (h) subspace; (i) Not a subspace: the zerofunction does not satisfy the condition.

2.2.17. If u′′ = xu, v′′ = xv, are solutions, and c, d constants, then (cu + dv)′′ = cu′′ + dv′′ =cxu + dxv = x(cu + dv), and hence cu + dv is also a solution.

2.2.18. For instance, the zero function u(x) ≡ 0 is not a solution.

2.2.19.(a) It is a subspace of the space of all functions f : [a, b ]→ R

2, which is a particular instance

of Example 2.7. Note that f(t) = ( f1(t), f2(t) )T is continuously differentiable if and

46

only if its component functions f1(t) and f2(t) are. Thus, if f(t) = ( f1(t), f2(t) )T and

g(t) = ( g1(t), g2(t) )T are continuously differentiable, so are

(f + g)(t) = ( f1(t) + g1(t), f2(t) + g2(t) )T and (c f)(t) = ( cf1(t), cf2(t) )T .

(b) Yes: if f(0) = 0 = g(0), then (c f + dg)(0) = 0 for any c, d ∈ R.

2.2.20.∇ · (cv + dw) = c∇ · v + d∇ ·w = 0 whenever ∇ · v = ∇ ·w = 0 and c, d,∈ R.

2.2.21. Yes. The sum of two convergent sequences is convergent, as is any constant multiple ofa convergent sequence.

2.2.22.(a) If v,w ∈ W ∩ Z, then v,w ∈ W , so cv + dw ∈ W because W is a subspace, and

v,w ∈ Z, so cv + dw ∈ Z because Z is a subspace, hence cv + dw ∈W ∩ Z.(b) If w + z, ew + ez ∈ W + Z then c(w + z) + d( ew + ez) = (cw + d ew) + (cz + dez) ∈ W + Z,

since it is the sum of an element of W and an element of Z.(c) Given any w ∈ W and z ∈ Z, then w, z ∈ W ∪ Z. Thus, if W ∪ Z is a subspace, the

sum w + z ∈ W ∪ Z. Thus, either w + z = ew ∈ W or w + z = ez ∈ Z. In the first casez = ew −w ∈ W , while in the second w = ez − z ∈ Z. We conclude that for any w ∈ Wand z ∈ Z, either w ∈ Z or z ∈ W . Suppose W 6⊂ Z. Then we can find w ∈ W \ Z, andso for any z ∈ Z, we must have z ∈W , which proves Z ⊂W .

♦ 2.2.23. If v,w ∈ T

Wi, then v,w ∈Wi for each i and so cv + dw ∈Wi for any c, d ∈ R becauseWi is a subspace. Since this holds for all i, we conclude that cv + dw ∈ T

Wi.

♥ 2.2.24.(a) They clearly only intersect at the origin. Moreover, every v =

xy

!=

x0

!+

0y

!can

be written as a sum of vectors on the two axes.(b) Since the only common solution to x = y and x = 3y is x = y = 0, the lines only

intersect at the origin. Moreover, every v =

xy

!=

aa

!+

3bb

!, where a = − 1

2 x+ 32 y,

b = 12 x− 1

2 y, can be written as a sum of vectors on each line.

(c) A vector v = ( a, 2a, 3a )T in the line belongs to the plane if and only if a + 2(2a) +3(3a) = 14a = 0, so a = 0 and the only common element is v = 0. Moreover, every

v =

0B@

xyz

1CA =

1

14

0B@

x + 2y + 3z2(x + 2y + 3z)3(x + 2y + 3z)

1CA +

1

14

0B@

13x− 2y − 3z−2x + 10y − 6z−3x− 6y + 5z

1CA can be written as a sum

of a vector in the line and a vector in the plane.(d) If w + z = ew + ez, then w − ew = ez− z. The left hand side belongs to W , while the right

hand side belongs to Z, and so, by the first assumption, they must both be equal to 0.Therefore, w = ew, z = ez.

2.2.25.(a) (v,w) ∈ V0 ∩W0 if and only if (v,w) = (v,0) and (v,w) = (0,w), which means v =

0,w = 0, and hence (v,w) = (0,0) is the only element of the intersection. Moreover, wecan write any element (v,w) = (v,0) + (0,w).

(b) (v,w) ∈ D ∩ A if and only if v = w and v = −w, hence (v,w) = (0,0). Moreover, we

can write (v,w) = ( 12 v + 1

2 w, 12 v + 1

2 w) + ( 12 v − 1

2 w,− 12 v + 1

2 w) as the sum of anelement of D and an element of A.

2.2.26.(a) If f(−x) = f(x), ef(−x) = ef(x), then (cf + d ef)(−x) = cf(−x) + d ef(−x) = cf(x) +

d ef(x) = (cf + d ef)(x) for any c, d,∈ R, and hence it is a subspace.(b) If g(−x) = −g(x), eg(−x) = −eg(x), then (cg + deg)(−x) = cg(−x) + deg(−x) =−cg(x) − deg(x) = −(cg + deg)(x), proving it is a subspace. If f(x) is both even and

47

odd, then f(x) = f(−x) = −f(x) and so f(x) ≡ 0 for all x. Moreover, we can write any

function h(x) = f(x) + g(x) as a sum of an even function f(x) = 12

hh(x) + h(−x)

iand

an odd function g(x) = 12

hh(x)− h(−x)

i.

(c) This follows from part (b), and the uniqueness follows from Exercise 2.2.24(d).

2.2.27. If A = AT and A = −AT is both symmetric and skew-symmetric, then A = O.

Given any square matrix, write A = S + J where S = 12

“A + AT

”is symmetric and

J = 12

“A−AT

”is skew-symmetric. This verifies the two conditions for complementary

subspaces. Uniqueness of the decomposition A = S + J follows from Exercise 2.2.24(d).

♦ 2.2.28.(a) By induction, we can show that

f (n)(x) = Pn

1

x

!e−1/x = Qn(x)

e−1/x

xn,

where Pn(y) and Qn(x) = xnPn(1/x) are certain polynomials of degree n. Thus,

limx→ 0

f (n)(x) = limx→ 0

Qn(x)e−1/x

xn= Qn(0) lim

y →∞yn e−y = 0,

because the exponential e−y goes to zero faster than any power of y goes to ∞.(b) The Taylor series at a = 0 is 0 + 0x + 0x2 + · · · ≡ 0, which converges to the zero

function, not to e−1/x.

2.2.29.

(a) The Taylor series is the geometric series1

1 + x2= 1− x2 + x4 − x6 + · · · .

(b) The ratio test can be used to prove that the series converges precisely when |x | < 1.(c) Convergence of the Taylor series to f(x) for x near 0 suffices to prove analyticity of the

function at x = 0.

♥ 2.2.30.(a) If v+a,w+a ∈ A, then (v+a)+(w+a) = (v+w+a)+a ∈ A requires v+w+a = u ∈ V ,

and hence a = u− v −w ∈ A.

(b) (i) -3 -2 -1 1 2 3

-3

-2

-1

1

2

3

(ii) -3 -2 -1 1 2 3

-3

-2

-1

1

2

3

(iii) -3 -2 -1 1 2 3

-3

-2

-1

1

2

3

(c) Every subspace V ⊂ R2 is either a point (the origin), or a line through the origin, or all

of R2. Thus, the corresponding affine subspaces are the point a; a line through a, or

all of R2 since in this case a ∈ V = R

2.(d) Every vector in the plane can be written as ( x, y, z )T = ( ex, ey, ez )T + ( 1, 0, 0 )T where

( ex, ey, ez )T is an arbitrary vector in the subspace defined by ex− 2ey + 3 ex = 0.(e) Every such polynomial can be written as p(x) = q(x) + 1 where q(x) is any element of

the subspace of polynomials that satisfy q(1) = 0.

48

2.3.1.

0B@−1

23

1CA = 2

0B@

2−1

2

1CA−

0B@

5−4

1

1CA.

2.3.2.

0BBB@

−3761

1CCCA = 3

0BBB@

1−3−2

0

1CCCA+ 2

0BBB@

−2634

1CCCA+

0BBB@

−246−7

1CCCA.

2.3.3.

(a) Yes, since

0B@

1−2−3

1CA =

0B@

110

1CA− 3

0B@

011

1CA;

(b) Yes, since

0B@

1−2−1

1CA = 3

10

0B@

122

1CA+ 7

10

0B@

1−2

0

1CA− 4

10

0B@

034

1CA;

(c) No, since the vector equation

0BBB@

30−1−2

1CCCA = c1

0BBB@

1201

1CCCA+ c2

0BBB@

0−1

30

1CCCA+ c3

0BBB@

201−1

1CCCA does not have a

solution.

2.3.4. Cases (b), (c), (e) span R2.

2.3.5.

(a) The line ( 3 t, 0, t )T :

-2

0

2-1-0.500.51-1-0.500.5

1

-2

0

2

-1-0.500.5

1

(b) The plane z = − 35 x− 6

5 y:

-1-0.5

0

0.5

1-1

-0.5

00.5

1

-1

0

1

-1

-0.5

00.5

1

(c) The plane z = −x− y:

-1

-0.5

0

0.5

1

-1

-0.50

0.51

-2

-1

0

1

2-1

-0.50

0.51

2.3.6. They are the same. Indeed, since v1 = u1 + 2u2, v2 = u1 + u2, every vector v ∈ V canbe written as a linear combination v = c1v1 + c2v2 = (c1 + c2)u1 +(2c1 + c2)u2 and hencebelongs to U . Conversely, since u1 = −v1 + 2v2, u2 = v1 − v2, every vector u ∈ U can bewritten as a linear combination u = c1u1 + c2u2 = (−c1 + c2)v1 + (2c1 − c2)v2, and hencebelongs to U .

2.3.7. (a) Every symmetric matrix has the form

a bb c

!= a

1 00 0

!+ c

0 00 1

!+ b

0 11 0

!.

49

(b)

0B@

1 0 00 0 00 0 0

1CA ,

0B@

0 0 00 1 00 0 0

1CA ,

0B@

0 0 00 0 00 0 1

1CA ,

0B@

0 1 01 0 00 0 0

1CA ,

0B@

0 0 10 0 01 0 0

1CA ,

0B@

0 0 00 0 10 1 0

1CA.

2.3.8.(a) They span P(2) since ax2 +bx+c = 1

2 (a−2b+c)(x2 +1)+ 12 (a−c)(x2−1)+b(x2 +x+1).

(b) They span P(3) since ax3 +bx2 +cx+d = a(x3−1)+b(x2 +1)+c(x−1)+(a−b+c+d)1.

(c) They do not span P(3) since ax3 +bx2 +cx+d = c1x3 +c2(x2 +1)+c3(x

2−x)+c4(x+1)cannot be solved when b + c− d 6= 0.

2.3.9. (a) Yes. (b) No. (c) No. (d) Yes: cos2 x = 1− sin2 x. (e) No. (f ) No.

2.3.10. (a) sin 3x = cos“

3x− 12 π

”; (b) cos x− sin x =

√2 cos

“x + 1

4 π”,

(c) 3 cos 2x+4 sin 2x = 5 cos“

2x− tan−1 43

”, (d) cos x sin x = 1

2 sin 2x = 12 cos

“2x− 1

2 π”.

2.3.11. (a) If u1 and u2 are solutions, so is u = c1 u1 + c2 u2 since u′′ − 4u′ + 3u = c1(u′′1 −

4u′1 + 3u1) + c2(u

′′2 − 4u′

2 + 3u2) = 0. (b) span ex, e3x ; (c) 2.

2.3.12. Each is a solution, and the general solution u(x) = c1 + c2 cos x + c3 sin x is a linearcombination of the three independent solutions.

2.3.13. (a) e2x; (b) cos 2x, sin 2x; (c) e3x, 1; (d) e−x, e−3x; (e) e−x/2 cos√

32 x,

e−x/2 sin√

32 x; (f ) e5x, 1, x; (g) ex/

√2 cos

x√2

, ex/√

2 sinx√2

, e−x/√

2 cosx√2

, e−x/√

2 sinx√2

.

2.3.14. (a) If u1 and u2 are solutions, so is u = c1 u1 + c2 u2 since u′′ + 4u = c1(u′′1 + 4u1) +

c2(u′′2 + 4u2) = 0, u(0) = c1 u1(0) + c2 u2(0) = 0, u(π) = c1 u1(π) + c2 u2(π) = 0.

(b) span sin 2x

2.3.15. (a)

21

!= 2 f1(x) + f2(x) − f3(x); (b) not in the span; (c)

1− 2x−1− x

!= f1(x) −

f2(x)− f3(x); (d) not in the span; (e)

2− x

0

!= 2 f1(x)− f3(x).

2.3.16. True, since 0 = 0v1 + · · ·+ 0vn.

2.3.17. False. For example, if z =

0B@

110

1CA, u =

0B@

100

1CA, v =

0B@

010

1CA, w =

0B@

001

1CA, then z = u + v, but

the equation w = c1u + c2v + c3 z =

0B@

c1 + c3c2 + c3

0

1CA has no solution.

♦ 2.3.18. By the assumption, any v ∈ V can be written as a linear combination

v = c1v1 + · · · + cm vm = c1v1 + · · · + cn vm + 0vm+1 + · · · + 0vn

of the combined collection.

♦ 2.3.19.

(a) If v =mX

j =1

cj vj and vj =nX

i=1

aij wi, then v =nX

i=1

bi vi where bi =mX

j =1

aij cj , or, in

vector language, b = A c.(b) Every v ∈ V can be written as a linear combination of v1, . . . ,vn, and hence, by part

(a), a linear combination of w1, . . . ,wm, which shows that w1, . . . ,wm also span V .

50

♦ 2.3.20.

(a) If v =mX

i=1

ai vi, w =nX

i=1

bi vi, are two finite linear combinations, so is

cv + dw =maxm,nX

i=1

(cai + dbi)vi where we set ai = 0 if i > m and bi = 0 if i > n.

(b) The space P(∞) of all polynomials, since every polynomial is a finite linear combinationof monomials and vice versa.

2.3.21. (a) Linearly independent; (b) linearly dependent; (c) linearly dependent;(d) linearly independent; (e) linearly dependent; (f ) linearly dependent;(g) linearly dependent; (h) linearly independent; (i) linearly independent.

2.3.22. (a) The only solution to the homogeneous linear system

c1

0BBB@

1021

1CCCA+ c2

0BBB@

−23−1

1

1CCCA+ c3

0BBB@

2−2

1−1

1CCCA = 0 is c1 = c2 = c3 = 0.

(b) All but the second lie in the span. (c) a− c + d = 0.

2.3.23.(a) The only solution to the homogeneous linear system

A c = c1

0BBB@

1110

1CCCA+ c2

0BBB@

11−1

0

1CCCA+ c3

0BBB@

1−1

01

1CCCA+ c4

0BBB@

1−1

0−1

1CCCA = 0

with nonsingular coefficient matrix A =

0BBB@

1 1 1 11 1 −1 −11 −1 0 10 0 1 −1

1CCCA is c = 0.

(b) Since A is nonsingular, the inhomogeneous linear system

v = A c = c1

0BBB@

1110

1CCCA+ c2

0BBB@

11−1

0

1CCCA+ c3

0BBB@

1−1

01

1CCCA+ c4

0BBB@

1−1

0−1

1CCCA

has a solution c = A−1v for any v ∈ R4.

(c)

0BBB@

1001

1CCCA = 3

8

0BBB@

1110

1CCCA+ 1

8

0BBB@

11−1

0

1CCCA+ 3

4

0BBB@

1−1

01

1CCCA−

14

0BBB@

1−1

0−1

1CCCA

2.3.24. (a) Linearly dependent; (b) linearly dependent; (c) linearly independent; (d) linearlydependent; (e) linearly dependent; (f ) linearly independent.

2.3.25. False:0B@

1 0 00 1 00 0 1

1CA−

0B@

0 1 01 0 00 0 1

1CA−

0B@

0 0 10 1 01 0 0

1CA−

0B@

1 0 00 0 10 1 0

1CA+

0B@

0 1 00 0 11 0 0

1CA+

0B@

0 0 11 0 00 1 0

1CA = O.

2.3.26. False — the zero vector always belongs to the span.

2.3.27. Yes, when it is the zero vector.

51

2.3.28. Because x,y are linearly independent, 0 = c1u + c2v = (ac1 + cc2)x + (bc1 + dc2)y ifand only if ac1 + cc2 = 0, bc1 + dc2 = 0. The latter linear system has a nonzero solution(c1, c2) 6= 0, and so u,v are linearly dependent, if and only if the determinant of the coef-

ficient matrix is zero: det

a cb d

!= ad − bc = 0, proving the result. The full collection

x,y,u,v is linearly dependent since, for example, ax+by−u+0v = 0 is a nontrivial linearcombination.

2.3.29. The statement is false. For example, any set containing the zero element that does notspan V is linearly dependent.

♦ 2.3.30. (b) If the only solution to A c = 0 is the trivial one c = 0, then the only linear com-bination which adds up to zero is the trivial one with c1 = · · · = ck = 0, proving linearindependence. (c) The vector b lies in the span if and only if b = c1v1 + · · · + ck vk = A cfor some c, which implies that the linear system A c = b has a solution.

♦ 2.3.31.(a) Since v1, . . . ,vn are linearly independent,

0 = c1v1 + · · · + ck vk = c1v1 + · · · + ck vk + 0vk+1 + · · · + 0vn

if and only if c1 = · · · = ck = 0.

(b) This is false. For example, v1 =

11

!, v2 =

22

!, are linearly dependent, but the

subset consisting of just v1 is linearly independent.

2.3.32.(a) They are linearly dependent since (x2 − 3) + 2(2− x)− (x− 1)2 ≡ 0.

(b) They do not span P(2).

2.3.33. (a) Linearly dependent; (b) linearly independent; (c) linearly dependent; (d) linearlyindependent; (e) linearly dependent; (f ) linearly dependent; (g) linearly independent;(h) linearly independent; (i) linearly independent.

2.3.34. When x > 0, we have f(x)− g(x) ≡ 0, proving linear dependence. On the other hand, ifc1f(x) + c2g(x) ≡ 0 for all x, then at, say x = 1, we have c1 + c2 = 0 while at x = −1, wemust have −c1 + c2 = 0, and so c1 = c2 = 0, proving linear independence.

♥ 2.3.35.

(a) 0 =kX

i=1

ci pi(x) =nX

j =0

kX

i=1

ci aij xj if and only ifnX

j =0

kX

i=1

ci aij = 0, j = 0, . . . , n, or, in

matrix notation, AT c = 0. Thus, the polynomials are linearly independent if and only if

the linear system AT c = 0 has only the trivial solution c = 0 if and only if its (n+1)×k

coefficient matrix has rank AT = rank A = k.

(b) q(x) =nX

j =0

bj xj =kX

i=1

ci pi(x) if and only if AT c = b.

(c) A =

0BBBBB@

−1 0 0 1 04 −2 0 1 00 −4 0 0 11 0 1 0 01 2 0 4 −1

1CCCCCA

has rank 4 and so they are linearly dependent.

(d) q(x) is not in the span.

♦ 2.3.36. Suppose the linear combination p(x) = c0 + c1 x + c2 x2 + · · · + cn xn ≡ 0 for all x.Thus, every real x is a root of p(x), but the Fundamental Theorem of Algebra says this isonly possible if p(x) is the zero polynomial with coefficients c0 = c1 = · · · = cn = 0.

52

♥ 2.3.37.(a) If c1 f1(x) + · · · + cn fn(x) ≡ 0, then c1 f1(xi) + · · · + cn fn(xi) = 0 at all sample points,

and so c1 f1 + · · · + cn fn = 0. Thus, linear dependence of the functions implies lineardependence of their sample vectors.

(b) Sampling f1(x) = 1 and f2(x) = x2 at −1, 1 produces the linearly dependent sample

vectors f1 = f2 =

11

!.

(c) Sampling at 0, 14 π, 1

2 π, 34 π, π, leads to the linearly independent sample vectors

0BBBBBBBB@

1

1

1

1

1

1CCCCCCCCA

,

0BBBBBBBB@

1√

220

−√

22−1

1CCCCCCCCA

,

0BBBBBBBB@

0√

221√

220

1CCCCCCCCA

,

0BBBBBBBB@

1

0

−1

0

1

1CCCCCCCCA

,

0BBBBBBBB@

0

1

0

−1

0

1CCCCCCCCA

.

2.3.38.(a) Suppose c1 f1(t) + · · · + cn fn(t) ≡ 0 for all t. Then c1 f1(t0) + · · · + cn fn(t0) = 0, and

hence, by linear independence of the sample vectors, c1 = · · · = cn = 0, which proveslinear independence of the functions.

(b) c1 f1(t) + c2 f1(t) =

2c2 t + (c1 − c2)

2c2 t2 + (c1 − c2)t

!≡ 0 if and only if c2 = 0, c1 − c2 = 0, and

so c1 = c2 = 0, proving linear independence. However, at any t0, the vectors f2(t0) =(2 t0 − 1)f1(t0) are scalar multiples of each other, and hence linearly dependent.

♥ 2.3.39.(a) Suppose c1 f(x) + c2 g(x) ≡ 0 for all x for some c = ( c1, c2 )T 6= 0. Differentiating,

we find c1 f ′(x) + c2 g′(x) ≡ 0 also, and hence

0@

f(x) g(x)

f ′(x) g′(x)

1A0@

c1

c2

1A = 0 for all x.

The homogeneous system has a nonzero solution if and only if the coefficient matrix issingular, which requires its determinant W [f(x), g(x) ] = 0.

(b) This is the contrapositive of part (a), since if f, g were not linearly independent, thentheir Wronskian would vanish everywhere.

(c) Suppose c1 f(x) + c2 g(x) = c1 x3 + c2 |x |3 ≡ 0. then, at x = 1, c1 + c2 = 0, whereasat x = −1, −c1 + c2 = 0. Therefore, c1 = c2 = 0, proving linear independence. On the

other hand, W [x3, |x |3 ] = x3(3x2 sign x)− (3x2) |x |3 ≡ 0.

2.4.1. Only (a) and (c) are bases.

2.4.2. Only (b) is a basis.

2.4.3. (a)

0B@

100

1CA ,

0B@

012

1CA; (b)

0BB@

34

10

1CCA ,

0BB@

14

01

1CCA; (c)

0BBB@

−2100

1CCCA,

0BBB@

−1010

1CCCA,

0BBB@

1001

1CCCA.

2.4.4.(a) They do not span R

3 because the linear system A c = b with coefficient matrix

A =

0B@

1 3 2 40 −1 −1 −12 1 −1 3

1CA does not have a solution for all b since rank A = 2.

(b) 4 vectors in R3 are automatically linearly dependent.

53

(c) No, because if v1,v2,v3,v4 don’t span R3, no subset of them will span it either.

(d) 2, because v1 and v2 are linearly independent and span the subspace, and hence form abasis.

2.4.5.(a) They span R

3 because the linear system A c = b with coefficient matrix

A =

0B@

1 2 0 1−1 −2 −2 3

2 5 1 −1

1CA has a solution for all b since rank A = 3.

(b) 4 vectors in R3 are automatically linearly dependent.

(c) Yes, because v1,v2,v3 also span R3 and so form a basis.

(d) 3 because they span all of R3.

2.4.6.

(a) Solving the defining equation, the general vector in the plane is x =

0B@

2y + 4zyz

1CA where

y, z are arbitrary. We can write x = y

0B@

210

1CA+ z

0B@

401

1CA = (y + 2z)

0B@

2−1

1

1CA+ (y + z)

0B@

02−1

1CA

and hence both pairs of vectors span the plane. Both pairs are linearly independentsince they are not parallel, and hence both form a basis.

(b)

0B@

2−1

1

1CA = (−1)

0B@

210

1CA+

0B@

401

1CA ,

0B@

02−1

1CA = 2

0B@

210

1CA−

0B@

401

1CA;

(c) Any two linearly independent solutions, e.g.,

0B@

611

1CA ,

0B@

1012

1CA, will form a basis.

♥ 2.4.7. (a) (i) Left handed basis; (ii) right handed basis; (iii) not a basis; (iv) right handedbasis. (b) Switching two columns or multiplying a column by −1 changes the sign of thedeterminant. (c) If det A = 0, its columns are linearly dependent and hence can’t form abasis.

2.4.8.

(a)“− 2

3 , 56 , 1, 0

”T,“

13 ,− 2

3 , 0, 1”T

; dim = 2.

(b) The condition p(1) = 0 says a + b + c = 0, so p(x) = (−b− c)x2 + bx + c = b(−x2 + x) +

c(−x2 + 1). Therefore −x2 + x, −x2 + 1 is a basis, and so dim = 2.(c) ex, cos 2x, sin 2x, is a basis, so dim = 3.

2.4.9. (a)

0B@

31−1

1CA, dim = 1; (b)

0B@

201

1CA,

0B@

0−1

3

1CA, dim = 2; (c)

0BBB@

10−1

2

1CCCA,

0BBB@

0113

1CCCA,

0BBB@

1−2

11

1CCCA, dim = 3.

2.4.10. (a) We have a + bt + ct2 = c1(1 + t2) + c2(t + t2) + c3(1 + 2 t + t2) provided a = c1 + c3,

b = c2 + 2c3, c = c1 + c2 + c3. The coefficient matrix of this linear system,

0B@

1 0 10 1 21 1 1

1CA,

is nonsingular, and hence there is a solution for any a, b, c, proving that they span the spaceof quadratic polynomials. Also, they are linearly independent since the linear combinationis zero if and only if c1, c2, c3 satisfy the corresponding homogeneous linear system c1+c3 =0, c2 + 2c3 = 0, c1 + c2 + c3 = 0, and hence c1 = c2 = c3 = 0. (Or, you can use

the fact that dimP(2) = 3 and the spanning property to conclude that they form a basis.)

54

(b) 1 + 4 t + 7 t2 = 2(1 + t2) + 6(t + t2)− (1 + 2 t + t2)

2.4.11. (a) a+bt+ct2+dt3 = c1+c2(1−t)+c3(1−t)2+c4(1−t)3 provided a = c1+c2+c3+c4,

b = −c2 − 2c3 − 3c4, c = c3 + 3c4, d = −c4. The coefficient matrix

0BBB@

1 1 1 10 −1 −2 −30 0 1 30 0 0 −1

1CCCA

is nonsingular, and hence they span P(3). Also, they are linearly independent since the lin-ear combination is zero if and only if c1 = c2 = c3 = c4 = 0 satisfy the corresponding

homogeneous linear system. (Or, you can use the fact that dimP (3) = 4 and the spanning

property to conclude that they form a basis.) (b) 1 + t3 = 2− 3(1− t) + 3(1− t)2− (1− t)3.

2.4.12. (a) They are linearly dependent because 2p1 − p2 + p3 ≡ 0. (b) The dimension is 2,since p1, p2 are linearly independent and span the subspace, and hence form a basis.

2.4.13.

(a) The sample vectors

0BBBBB@

1

1

1

1

1CCCCCA

,

0BBBBB@

1√

22

0

−√

22

1CCCCCA

,

0BBBBB@

1

0

−1

0

1CCCCCA

,

0BBBBB@

1

−√

22

0√

22

1CCCCCA

are linearly independent and

hence form a basis for R4 — the space of sample functions.

(b) Sampling x produces

0BBBBB@

0141234

1CCCCCA

=1

2

0BBBBB@

1

1

1

1

1CCCCCA− 2 +

√2

8

0BBBBB@

1√

22

0

−√

22

1CCCCCA− 2−

√2

8

0BBBBB@

1

−√

22

0√

22

1CCCCCA

.

2.4.14.

(a) E11 =

1 00 0

!, E12 =

0 10 0

!, E21 =

0 01 0

!, E22 =

0 00 1

!is a basis since we

can uniquely write any

a bc d

!= aE11 + bE12 + cE21 + dE22.

(b) Similarly, the matrices Eij with a 1 in position (i, j) and all other entries 0, for

i = 1, . . . , m, j = 1, . . . , n, form a basis for Mm×n, which therefore has dimension mn.

2.4.15. k 6= −1, 2.

2.4.16. A basis is given by the matrices Eii, i = 1, . . . , n which have a 1 in the ith diagonalposition and all other entries 0.

2.4.17.

(a) E11 =

1 00 0

!, E12 =

0 10 0

!, E22 =

0 00 1

!; dimension = 3.

(b) A basis is given by the matrices Eij with a 1 in position (i, j) and all other entries 0 for

1 ≤ i ≤ j ≤ n, so the dimension is 12 n(n + 1).

2.4.18. (a) Symmetric: dim = 3; skew-symmetric: dim = 1; (b) symmetric: dim = 6; skew-

symmetric: dim = 3; (c) symmetric: dim = 12 n(n+1); skew-symmetric: dim = 1

2 n(n−1).

♥ 2.4.19.(a) If a row (column) of A adds up to a and the corresponding row (column) of B adds up

to b, then the corresponding row (column) of C = A + B adds up to c = a + b. Thus,if all row and column sums of A and B are the same, the same is true for C. Similarly,the row (column) sums of c A are c times the row (column) sums of A, and hence all thesame if A is a semi-magic square.

55

(b) A matrix A =

0B@

a b cd e fg h j

1CA is a semi-magic square if and only if

a + b + c = d + e + f = g + h + j = a + d + e = b + e + h = c + f + j.The general solution to this system is

A = e

0B@

1 −1 0−1 1 0

0 0 0

1CA+ f

0B@

1 0 −1−1 0 1

0 0 0

1CA+ g

0B@−1 1 1

1 0 01 0 0

1CA+ h

0B@

0 0 11 0 00 1 0

1CA+ j

0B@

0 1 01 0 00 0 1

1CA

= (e− g)

0B@

1 0 00 1 00 0 1

1CA+ (g + j − e)

0B@

0 1 01 0 00 0 1

1CA+ g

0B@

0 0 10 1 01 0 0

1CA+

+ f

0B@

1 0 00 0 10 1 0

1CA+ (h− f)

0B@

0 0 11 0 00 1 0

1CA ,

which is a linear combination of permutation matrices.(c) The dimension is 5, with any 5 of the 6 permutation matrices forming a basis.(d) Yes, by the same reasoning as in part (a). Its dimension is 3, with basis0

B@2 2 −1−2 1 4

3 0 0

1CA ,

0B@

2 −1 21 1 10 3 0

1CA ,

0B@−1 2 2

4 1 −20 0 3

1CA.

(e) A = c1

0B@

2 2 −1−2 1 4

3 0 0

1CA+ c2

0B@

2 −1 21 1 10 3 0

1CA+ c3

0B@−1 2 2

4 1 −20 0 3

1CA for any c1, c2, c3.

♦ 2.4.20. For instance, take v1 =

10

!, v2 =

01

!, v3 =

11

!. Then

21

!= 2v1 + v2 =

v1 + v3. In fact, there are infinitely many different ways of writing this vector as a linearcombination of v1,v2,v3.

♦ 2.4.21.(a) By Theorem 2.31, we only need prove linear independence. If 0 = c1 Av1 + · · · +

cn Avn = A(c1v1 + · · · + cn vn), then, since A is nonsingular, c1v1 + · · · + cn vn = 0,and hence c1 = · · · = cn = 0.

(b) Aei is the ith column of A, and so a basis consists of the column vectors of the matrix.

♦ 2.4.22. Since V 6= 0, at least one vi 6= 0. Let vi16= 0 be the first nonzero vector in the list

v1, . . . ,vn. Then, for each k = i1 + 1, . . . , n− 1, suppose we have selected linearly indepen-dent vectors vi1

, . . . ,vij

from among v1, . . . ,vk. If vi1, . . . ,vij

,vk+1 form a linearly inde-

pendent set, we set vij+1= vk+1; otherwise, vk+1 is a linear combination of vi1

, . . . ,vij

,

and is not needed in the basis. The resulting collection vi1, . . . ,vim

forms a basis for Vsince they are linearly independent by design, and span V since each vi either appears inthe basis, or is a linear combination of the basis elements that were selected before it. Wehave dim V = n if and only if v1, . . . ,vn are linearly independent and so form a basis forV .

♦ 2.4.23. This is a special case of Exercise 2.3.31(a).

♦ 2.4.24.(a) m ≤ n as otherwise v1, . . . ,vm would be linearly dependent. If m = n then v1, . . . ,vn

are linearly independent and hence, by Theorem 2.31 span all of Rn. Since every vector

in their span also belongs to V , we must have V = Rn.

(b) Starting with the basis v1, . . . ,vm of V with m < n, we choose any vm+1 ∈ Rn \ V .

Since vm+1 does not lie in the span of v1, . . . ,vm, the vectors v1, . . . ,vm+1 are linearly

independent and span an m + 1 dimensional subspace of Rn. Unless m + 1 = n we can

56

then choose another vector vm+2 not in the span of v1, . . . ,vm+1, and so v1, . . . ,vm+2are also linearly independent. We continue on in this fashion until we arrive at n lin-early independent vectors v1, . . . ,vn which necessarily form a basis of R

n.

(c) (i)“

1, 1, 12

”T, ( 1, 0, 0 )T , ( 0, 1, 0 )T ; (ii) ( 1, 0,−1 )T , ( 0, 1,−2 )T , ( 1, 0, 0 )T .

♦ 2.4.25.(a) If dim V = ∞, then the inequality is trivial. Also, if dim W = ∞, then one can find

infinitely many linearly independent elements in W , but these are also linearly indepen-dent as elements of V and so dim V = ∞ also. Otherwise, let w1, . . . ,wn form a basisfor W . Since they are linearly independent, Theorem 2.31 implies n ≤ dim V .

(b) Since w1, . . . ,wn are linearly independent, if n = dim V , then by Theorem 2.31, theyform a basis for V . Thus every v ∈ V can be written as a linear combination ofw1, . . . ,wn, and hence, since W is a subspace, v ∈W too. Therefore, W = V .

(c) Example: V = C0[a, b ] and W = P(∞).

♦ 2.4.26. (a) Every v ∈ V can be uniquely decomposed as v = w + z where w ∈ W, z ∈ Z. Writew = c1w1 + . . . + cj wj and z = d1 z1 + · · ·+ dk zk. Then v = c1w1 + . . . + cj wj + d1 z1 +

· · · + dk zk, proving that w1, . . . ,wj , z1, . . . , zk span V . Moreover, by uniqueness, v = 0 if

and only if w = 0 and z = 0, and so the only linear combination that sums up to 0 ∈ V isthe trivial one c1 = · · · = cj = d1 = · · · = dk = 0, which proves linear independence of the

full collection. (b) This follows immediately from part (a): dim V = j+k = dim W +dim Z.

♦ 2.4.27. Suppose the functions are linearly independent. This means that for every 0 6= c =

( c1, c2, . . . , cn )T ∈ Rn, there is a point x

c∈ R such that

nX

i=1

ci fi(xc) 6= 0. The as-

sumption says that 0 6= Vx1,...,xm

for all choices of sample points. Recursively define the

following sample points. Choose x1 so that f1(x1) 6= 0. (This is possible since if f1(x) ≡ 0,then the functions are linearly dependent.) Thus Vx1

( Rm since e1 6∈ Vx1

. Then, for eachm = 1, 2, . . . , given x1, . . . , xm, choose 0 6= c0 ∈ Vx1,...,xm

, and set xm+1 = xc0

. Then

c0 6∈ Vx1,...,xm+1( Vx1,...,xm

and hence, by induction, dim Vm ≤ n − m. In particular,

dim Vx1,...,xn

= 0, so Vx1,...,xn

= 0, which contradicts our assumption and proves the

result. Note that the proof implies we only need check linear dependence at all possible col-lections of n sample points to conclude that the functions are linearly dependent.

2.5.1.

(a) Range: all b =

b1b2

!such that 3

4 b1 + b2 = 0; kernel spanned by

12

1

!.

(b) Range: all b =

b1b2

!such that 2b1 + b2 = 0; kernel spanned by

0B@

110

1CA,

0B@−2

01

1CA.

(c) Range: all b =

0B@

b1b2b3

1CA such that −2b1 + b2 + b3 = 0; kernel spanned by

0BBB@

− 54

− 78

1

1CCCA.

(d) Range: all b = ( b1, b2, b3, b4 )T such that −2b1 − b2 + b3 = 2b1 + 3b2 + b4 = 0;

kernel spanned by

0BBB@

1110

1CCCA,

0BBB@

−1001

1CCCA.

57

2.5.2. (a)

0BB@− 5

2

01

1CCA,

0BB@

12

10

1CCA: plane; (b)

0BBB@

1438

1

1CCCA: line; (c)

0B@

201

1CA,

0B@−3

10

1CA: plane;

(d)

0B@−1−2

1

1CA: line; (e)

0B@

000

1CA: point; (f )

0BB@

13531

1CCA: line.

2.5.3.

(a) Kernel spanned by

0BBB@

3100

1CCCA; range spanned by

0B@

120

1CA,

0B@

201

1CA,

0B@

02−3

1CA;

(b) compatibility: − 12 a + 1

4 b + c = 0.

2.5.4. (a) b =

0B@−1

2−1

1CA; (b) x =

0B@

1 + t2 + t3 + t

1CA where t is arbitrary.

2.5.5. In each case, the solution is x = x? + z, where x? is the particular solution and z belongsto the kernel:

(a) x? =

0B@

100

1CA, z = y

0B@

110

1CA+ z

0B@−3

01

1CA; (b) x? =

0B@

1−1

0

1CA, z = z

0BB@− 2

7171

1CCA;

(c) x? =

0BBB@

− 79

29109

1CCCA, z = z

0B@

221

1CA; (d) x? =

0BB@

56

1

− 23

1CCA, z = 0; (e) x? =

−1

0

!, z = v

21

!;

(f ) x? =

0BBBB@

11212

00

1CCCCA

, z = r

0BBBB@

− 132

− 32

10

1CCCCA

+ s

0BBBB@

− 32

− 12

01

1CCCCA

; (g) x? =

0BBB@

3200

1CCCA, z = z

0BBB@

6210

1CCCA+ w

0BBB@

−4−1

01

1CCCA.

2.5.6. The ith entry of A ( 1, 1, . . . , 1 )T is ai1 + . . . + ain which is n times the average of the en-

tries in the ith row. Thus, A ( 1, 1, . . . , 1 )T = 0 if and only if each row of A has average 0.

2.5.7. The kernel has dimension n−1, with basis −rk−1 e1+ek =“−rk−1, 0, . . . , 0, 1, 0, . . . , 0

”T

for k = 2, . . . n. The range has dimension 1, with basis (1, rn, r2n . . . , r(n−1)n)T .

♦ 2.5.8. (a) If w = P w, then w ∈ rng P . On the other hand, if w ∈ rng P , then w = P v for

some v. But then P w = P 2v = P v = w. (b) Given v, set w = P v. Then v = w + z

where z = v−w ∈ ker P since P z = P v− P w = P v− P 2v = P v− P v = 0. Moreover, ifw ∈ ker P ∩rng P , then 0 = P w = w, and so ker P ∩rng P = 0, proving complementarity.

2.5.9. False. For example, if A =

1 1−1 −1

!then

11

!is in both ker A and rng A.

♦ 2.5.10. Let r1, . . . , rm+k be the rows of C, so r1, . . . , rm are the rows of A. For v ∈ ker C, the

ith entry of C v = 0 is ri v = 0, but then this implies Av = 0 and so v ∈ ker A. As an

example, A = ( 1 0 ) has kernel spanned by

10

!, while C =

1 00 1

!has ker C = 0.

58

♦ 2.5.11. If b = Ax ∈ rng A, then b = C z where z =

x0

!, and so b ∈ rng C. As an example,

A =

00

!has rng A = 0, while the range of C =

0 10 0

!is the x axis.

2.5.12. x?1 =

−232

!, x?

2 =

−112

!; x = x?

1 + 4x?2 =

−672

!.

2.5.13. x? = 2x?1 + x?

2 =

0B@−1

33

1CA.

2.5.14.(a) By direct matrix multiplication: Ax?

1 = Ax?2 =

0B@

1−3

5

1CA.

(b) The general solution is x = x?1 + t(x?

2 − x?1) = (1− t)x?

1 + tx?2 =

0B@

110

1CA+ t

0B@−4

2−2

1CA.

2.5.15. 5 meters.

2.5.16. The mass will move 6 units in the horizontal direction and −6 units in the vertical di-rection.

2.5.17. x = c1x?1 + c2x?

2 where c1 = 1− c2.

2.5.18. False: in general, (A + B)x? = (A + B)x?1 + (A + B)x?

2 = c + d + Bx?1 + Ax?

2, and thethird and fourth terms don’t necessarily add up to 0.

♦ 2.5.19. rng A = Rn, and so A must be a nonsingular matrix.

♦ 2.5.20.(a) If Axi = ei, then xi = A−1ei which, by (2.13), is the ith column of the matrix A−1.

(b) The solutions to Axi = ei in this case are x1 =

0BB@

12

2

− 12

1CCA, x2 =

0BB@

− 12

− 1

− 1

1CCA, x3 =

0BB@

12

− 112

1CCA,

which are the columns of A−1 =

0BB@

12 − 1

212

2 −1 −1

− 12

12

12

1CCA.

2.5.21.

(a) range:

12

!; corange:

1−3

!; kernel:

31

!; cokernel:

−2

1

!.

(b) range:

0B@

012

1CA,

0B@−8−1

6

1CA; corange:

0B@

12−1

1CA,

0B@

00−8

1CA; kernel:

0B@−2

10

1CA; cokernel:

0B@

1−2

1

1CA.

(c) range:

0B@

112

1CA,

0B@

103

1CA; corange:

0BBB@

1121

1CCCA,

0BBB@

0−1−3

2

1CCCA; kernel:

0BBB@

1−3

10

1CCCA,

0BBB@

−3201

1CCCA; cokernel:

0B@−3

11

1CA.

59

(d) range:

0BBBBB@

10231

1CCCCCA

,

0BBBBB@

−33−3−3

0

1CCCCCA

,

0BBBBB@

1−2

033

1CCCCCA

; corange:

0BBBBB@

1−3

221

1CCCCCA

,

0BBBBB@

03−6

0−2

1CCCCCA

,

0BBBBB@

00004

1CCCCCA

;

kernel:

0BBBBB@

42100

1CCCCCA

,

0BBBBB@

−20010

1CCCCCA

; cokernel:

0BBBBB@

−2−1

100

1CCCCCA

,

0BBBBB@

210−1

1

1CCCCCA

.

2.5.22.

0B@−1

2−3

1CA,

0B@

012

1CA,

0B@−3

10

1CA, which are its first, third and fourth columns;

Second column:

0B@

2−4

6

1CA = 2

0B@−1

2−3

1CA; fifth column:

0B@

5−4

8

1CA = −2

0B@−1

2−3

1CA+

0B@

012

1CA−

0B@−3

10

1CA.

2.5.23. range:

0B@

12−3

1CA,

0B@

041

1CA; corange:

0B@

1−3

0

1CA,

0B@

004

1CA; second column:

0B@−3−6

9

1CA = −3

0B@

12−3

1CA;

second and third rows:

0B@

2−6

4

1CA = 2

0B@

1−3

0

1CA+

0B@

004

1CA,

0B@−3

91

1CA = −3

0B@

1−3

0

1CA+ 1

4

0B@

004

1CA.

2.5.24.(i) rank = 1; dim rng A = dim corng A = 1, dim ker A = dim coker A = 1;

kernel basis:

−2

1

!; cokernel basis:

21

!; compatibility conditions: 2b1 + b2 = 0;

example: b =

1−2

!, with solution x =

10

!+ z

−2

1

!.

(ii) rank = 1; dim rng A = dim corng A = 1, dimker A = 2, dim coker A = 1; kernel basis:0B@

13

10

1CA,

0B@

23

01

1CA; cokernel basis:

21

!; compatibility conditions: 2b1 + b2 = 0;

example: b =

3−6

!, with solution x =

0B@

100

1CA+ y

0B@

13

10

1CA+ z

0B@

23

01

1CA.

(iii) rank = 2; dim rng A = dim corng A = 2, dimker A = 0, dim coker A = 1;

kernel: 0; cokernel basis:

0BBB@

− 2013313

1

1CCCA; compatibility conditions: − 20

13 b1 + 313 b2 + b3 = 0;

example: b =

0B@

1−2

2

1CA, with solution x =

0B@

100

1CA.

(iv) rank = 2; dim rng A = dim corng A = 2, dimker A = dim coker A = 1;

kernel basis:

0B@−2−1

1

1CA; cokernel basis:

0B@−2

11

1CA; compatibility conditions:

−2b1 + b2 + b3 = 0; example: b =

0B@

213

1CA, with solution x =

0B@

100

1CA+ z

0B@−2−1

1

1CA.

(v) rank = 2; dim rng A = dim corng A = 2, dim ker A = 1, dim coker A = 2; kernel

60

basis:

0B@−1−1

1

1CA; cokernel basis:

0BBBBB@

− 94

14

1

0

1CCCCCA

,

0BBBBB@

14

− 14

0

1

1CCCCCA

; compatibility: − 94 b1 + 1

4 b2 + b3 = 0,

14 b1 − 1

4 b2 + b4 = 0; example: b =

0BBB@

2631

1CCCA, with solution x =

0B@

100

1CA+ z

0B@−1−1

1

1CA.

(vi) rank = 3; dim rng A = dim corng A = 3, dimker A = dim coker A = 1; kernel basis:0BBBBB@

134138

− 72

1

1CCCCCA

; cokernel basis:

0BBB@

−1−1

11

1CCCA; compatibility conditions: −b1 − b2 + b3 + b4 = 0;

example: b =

0BBB@

1313

1CCCA, with solution x =

0BBBBB@

1

0

0

0

1CCCCCA

+ w

0BBBBB@

134138

− 72

1

1CCCCCA

.

(vii) rank = 4; dim rng A = dim corng A = 4, dimker A = 1, dim coker A = 0; kernel basis:0BBBBB@

−21000

1CCCCCA

; cokernel is 0; no conditions;

example: b =

0BBB@

213−3

1CCCA, with x =

0BBBBB@

10000

1CCCCCA

+ y

0BBBBB@

−21000

1CCCCCA

.

2.5.25. (a) dim = 2; basis:

0B@

12−1

1CA,

0B@

220

1CA; (b) dim = 1; basis:

0B@

11−1

1CA;

(c) dim = 3; basis:

0BBB@

1010

1CCCA,

0BBB@

1001

1CCCA,

0BBB@

2210

1CCCA; (d) dim = 3; basis:

0BBB@

10−3

2

1CCCA,

0BBB@

012−3

1CCCA,

0BBB@

1−3−8

7

1CCCA;

(e) dim = 3; basis:

0BBBBB@

11−1

11

1CCCCCA

,

0BBBBB@

2−1

221

1CCCCCA

,

0BBBBB@

13−1

21

1CCCCCA

.

2.5.26. It’s the span of

0BBB@

1100

1CCCA,

0BBB@

−3010

1CCCA,

0BBB@

0231

1CCCA,

0BBB@

04−1−1

1CCCA; the dimension is 3.

2.5.27. (a)

0BBB@

2010

1CCCA,

0BBB@

0−1

01

1CCCA; (b)

0BBB@

1110

1CCCA,

0BBB@

0−1

01

1CCCA; (c)

0BBB@

−1301

1CCCA.

61

2.5.28. First method:

0BBB@

1021

1CCCA,

0BBB@

23−4

5

1CCCA; second method:

0BBB@

1021

1CCCA,

0BBB@

03−8

3

1CCCA. The first vectors are the

same, while

0BBB@

23−4

5

1CCCA = 2

0BBB@

1021

1CCCA+

0BBB@

03−8

3

1CCCA;

0BBB@

03−8

3

1CCCA = −2

0BBB@

1021

1CCCA+

0BBB@

23−4

5

1CCCA.

2.5.29. Both sets are linearly independent and hence span a three-dimensional subspace of R4.

Moreover, w1 = v1 + v3,w2 = v1 + v2 + 2v3,w3 = v1 + v2 + v3 all lie in the span ofv1,v2,v3 and hence, by Theorem 2.31(d) also form a basis for the subspace.

2.5.30.(a) If A = AT , then ker A = Ax = 0 = AT x = 0 = coker A, and rng A = Ax =

AT x = corng A.

(b) ker A = coker A has basis ( 2,−1, 1 )T ; rng A = corng A has basis ( 1, 2, 0 )T , ( 2, 6, 2 )T .(c) No. For instance, if A is any nonsingular matrix, then ker A = coker A = 0 and

rng A = corng A = R3.

2.5.31.(a) Yes. This is our method of constructing the basis for the range, and the proof is out-

lined in the text.

(b) No. For example, if A =

0BBB@

1 0 0 01 0 0 00 1 0 00 0 1 0

1CCCA, then U =

0BBB@

1 0 0 00 1 0 00 0 1 00 0 0 0

1CCCA and the first three

rows of U form a basis for the three-dimensional corng U = corng A. but the first threerows of A only span a two-dimensional subspace.

(c) Yes, since ker U = ker A.(d) No, since coker U 6= coker A in general. For the example in part (b), coker A has basis

(−1, 1, 0, 0 )T while coker A has basis ( 0, 0, 0, 1 )T .

2.5.32. (a) Example:

0 01 0

!. (b) No, since then the first r rows of U are linear combina-

tions of the first r rows of A. Hence these rows span corng A, which, by Theorem 2.31c,implies that they form a basis for the corange.

2.5.33. Examples: any symmetric matrix; any permutation matrix since the row echelon form is

the identity. Yet another example is the complex matrix

0B@

0 0 11 i i0 i i

1CA.

♦ 2.5.34. The rows r1, . . . , rm of A span the corange. Reordering the rows — in particular inter-changing two — will not change the span. Also, multiplying any of the rows by nonzeroscalars, eri = ai ri, for ai 6= 0, will also span the same space, since

v =nX

i=1

ci ri =nX

i=1

ci

ai

eri.

2.5.35. We know rng A ⊂ Rm is a subspace of dimension r = rank A. In particular, rng A = R

m

if and only if it has dimension m = rank A.

2.5.36. This is false. If A =

1 11 1

!then rng A is spanned by

11

!whereas the range of its

62

row echelon form U =

1 10 0

!is spanned by

10

!.

♦ 2.5.37.(a) Method 1: choose the nonzero rows in the row echelon form of A. Method 2: choose the

columns of AT that correspond to pivot columns of its row echelon form.

(b) Method 1:

0B@

124

1CA,

0B@

3−1

5

1CA,

0B@

2−4

2

1CA. Method 2:

0B@

124

1CA,

0B@

0−7−7

1CA,

0B@

002

1CA. Not the same.

♦ 2.5.38. If v ∈ ker A then Av = 0 and so BAv = B0 = 0, so v ∈ ker(BA). The first statementfollows from setting B = A.

♦ 2.5.39. If v ∈ rng AB then v = ABx for some vector x. But then v = Ay where y = Bx, andso v ∈ rng A. The first statement follows from setting B = A.

2.5.40. First note that BA and AC also have size m× n. To show rank A = rank BA, we provethat ker A = ker BA, and so rank A = n − dim ker A = n − dimker BA = rank BA.Indeed, if v ∈ ker A, then Av = 0 and hence BAv = 0 so v ∈ ker BA. Conversely, if v ∈ker BA then BAv = 0. Since B is nonsingular, this implies Av = 0 and hence v ∈ ker A,proving the first result. To show rank A = rank AC, we prove that rng A = rng AC, andso rank A = dim rng A = dim rng AC = rank AC. Indeed, if b ∈ rng AC, then b = AC xfor some x and so b = Ay where y = C x, and so b ∈ rng A. Conversely, if b ∈ rng Athen b = Ay for some y and so b = AC x where x = C−1y, so b ∈ rng AC, proving thesecond result. The final equality is a consequence of the first two: rank A = rank BA =rank(BA)C.

♦ 2.5.41. (a) Since they are spanned by the columns, the range of ( A B ) contains the range ofA. But since A is nonsingular, rng A = R

n, and so rng ( A B ) = Rn also, which proves

rank ( A B ) = n. (b) Same argument, using the fact that the corange is spanned by therows.

2.5.42. True if the matrices have the same size, but false in general.

♦ 2.5.43. Since we know dim rng A = r, it suffices to prove that w1, . . . ,wr are linearly indepen-dent. Given

0 = c1w1 + · · ·+ cr wr = c1 Av1 + · · ·+ cr Avr = A(c1v1 + · · ·+ cr vr),

we deduce that c1v1 + · · ·+ cr vr ∈ ker A, and hence can be written as a linear combinationof the kernel basis vectors:

c1v1 + · · ·+ cr vr = cr+1vr+1 + · · ·+ cn vn.

But v1, . . . ,vn are linearly independent, and so c1 = · · · = cr = cr+1 = · · · = cn = 0, whichproves linear independence of w1, . . . ,wr.

♦ 2.5.44.(a) Since they have the same kernel, their ranks are the same. Choose a basis v1, . . . ,vn of

Rn such that vr+1, . . . ,vn form a basis for ker A = ker B. Then w1 = Av1, . . . ,wr =

Avr form a basis for rng A, while y1 = Bv1, . . . ,yr = Bvr form a basis for rng B.Let M be any nonsingular m × m matrix such that M wj = yj , j = 1, . . . , r, which

exists since both sets of vectors are linearly independent. We claim M A = B. Indeed,M Avj = Bvj , j = 1, . . . , r, by design, while M Avj = 0 = Bvj , j = r + 1, . . . , n,

since these vectors lie in the kernel. Thus, the matrices agree on a basis of Rn which is

enough to conclude that M A = B.(b) If the systems have the same solutions x? + z where z ∈ ker A = ker B, then B x =

M Ax = M b = c. Since M can be written as a product of elementary matrices, we

conclude that one can get from the augmented matrix“

A | b”

to the augmented matrix

63

“B | c

”by applying the elementary row operations that make up M .

♦ 2.5.45. (a) First, W ⊂ rng A since every w ∈ W can be written as w = Av for some v ∈V ⊂ R

n, and so w ∈ rng A. Second, if w1 = Av1 and w2 = Av2 are elements of W , thenso is cw1 + dw2 = A(cv1 + dv2) for any scalars c, d because cv1 + dv2 ∈ V , provingthat W is a subspace. (b) First, using Exercise 2.4.25, dim W ≤ r = dim rng A since it isa subspace of the range. Suppose v1, . . . ,vk form a basis for V , so dim V = k. Let w =Av ∈ W . We can write v = c1v1 + · · · + ck vk, and so, by linearity, w = c1 Av1 + · · · +ck Avk. Therefore, the k vectors w1 = Av1, . . . ,wk = Avk span W , and therefore, byProposition 2.33, dim W ≤ k.

♦ 2.5.46.(a) To have a left inverse requires an n×m matrix B such that BA = I . Suppose dim rng A =

rank A < n. Then, according to Exercise 2.5.45, the subspace W = Bv |v ∈ rng A has dim W ≤ dim rng A < n. On the other hand, w ∈ W if and only if w = Bv wherev ∈ rng A, and so v = Ax for some x ∈ R

n. But then w = Bv = BAx = x, andtherefore W = R

n since every vector x ∈ Rn lies in it; thus, dim W = n, contradicting

the preceding result. We conclude that having a left inverse implies rank A = n. (Therank can’t be larger than n.)

(b) To have a right inverse requires an m×n matrix C such that AC = I . Suppose dim rng A =rank A < m and hence rng A ( R

m. Choose y ∈ Rm \ rng A. Then y = AC y = Ax,

where x = C y. Therefore, y ∈ rng A, which is a contradiction. We conclude that havinga right inverse implies rank A = m.

(c) By parts (a–b), having both inverses requires m = rank A = n and A must be squareand nonsingular.

2.6.1. (a) (b) (c)

(d) (e) or, equivalently,

2.6.2. (a)

(b) ( 1, 1, 1, 1, 1, 1, 1 )T is a basis for the kernel. The cokernel is trivial, containing only thezero vector, and so has no basis. (c) Zero.

64

2.6.3. (a)

0BBB@

−1 0 1 00 −1 1 00 1 0 −10 0 1 −1

1CCCA; (b)

0BBBBB@

−1 1 0 0−1 0 0 1

1 0 −1 00 1 0 −10 0 −1 1

1CCCCCA

; (c)

0BBBBBBB@

−1 0 1 0 0−1 1 0 0 0

0 −1 0 1 00 −1 0 0 10 0 1 −1 00 0 0 1 −1

1CCCCCCCA

;

(d)

0BBBBBBBBBB@

1 −1 0 0 01 0 −1 0 00 −1 0 1 00 −1 0 0 10 0 1 −1 00 0 −1 0 10 0 0 1 −1

1CCCCCCCCCCA

; (e)

0BBBBBBB@

−1 0 0 1 0 01 0 0 0 −1 00 1 −1 0 0 00 −1 0 0 0 10 0 1 0 0 −10 0 0 −1 1 0

1CCCCCCCA

;

(f )

0BBBBBBBBBBBB@

1 −1 0 0 0 01 0 −1 0 0 00 1 0 −1 0 0−1 0 0 1 0 0

0 0 1 0 0 −10 0 −1 0 1 00 0 0 −1 0 10 0 0 0 −1 1

1CCCCCCCCCCCCA

.

2.6.4. (a) 1 circuit:

0BBB@

0−1−1

1

1CCCA; (b) 2 circuits:

0BBBBB@

−11010

1CCCCCA

,

0BBBBB@

0−1−1

01

1CCCCCA

; (c) 2 circuits:

0BBBBBBB@

−111010

1CCCCCCCA

,

0BBBBBBB@

00−1

101

1CCCCCCCA

;

(d) 3 circuits:

0BBBBBBBBBB@

−1110100

1CCCCCCCCCCA

,

0BBBBBBBBBB@

1−1

0−1

010

1CCCCCCCCCCA

,

0BBBBBBBBBB@

00−1

1001

1CCCCCCCCCCA

; (e) 2 circuits:

0BBBBBBB@

001110

1CCCCCCCA

,

0BBBBBBB@

110001

1CCCCCCCA

;

(f ) 3 circuits:

0BBBBBBBBBBBB@

10110000

1CCCCCCCCCCCCA

,

0BBBBBBBBBBBB@

−11−1

01010

1CCCCCCCCCCCCA

,

0BBBBBBBBBBBB@

00001101

1CCCCCCCCCCCCA

.

♥ 2.6.5. (a)

0BBBBB@

1 −1 0 01 0 −1 01 0 0 −10 1 −1 00 1 0 −1

1CCCCCA

; (b) rank = 3; (c) dim rng A = dim corng A = 3,

dim ker A = 1, dim coker A = 2; (d) kernel:

0BBB@

1111

1CCCA; cokernel:

0BBBBB@

1−1

010

1CCCCCA

,

0BBBBB@

10−1

01

1CCCCCA

;

65

(e) b1 − b2 + b4 = 0, b1 − b3 + b5 = 0; (f ) example: b =

0BBBBB@

11100

1CCCCCA

; x =

0BBB@

1 + tttt

1CCCA.

♦ 2.6.6.(a)

0BBBBBBBBBBBBBBBBBBBBBB@

1 −1 0 0 0 0 0 01 0 −1 0 0 0 0 01 0 0 −1 0 0 0 00 1 0 0 −1 0 0 00 1 0 0 0 −1 0 00 0 1 0 −1 0 0 00 0 1 0 0 0 −1 00 0 0 1 0 −1 0 00 0 0 1 0 0 −1 00 0 0 0 1 0 0 −10 0 0 0 0 1 0 −10 0 0 0 0 0 1 −1

1CCCCCCCCCCCCCCCCCCCCCCA

Cokernel basis: v1 =

0BBBBBBBBBBBBBBBBBBBBBB@

−110−1

01000000

1CCCCCCCCCCCCCCCCCCCCCCA

,v2 =

0BBBBBBBBBBBBBBBBBBBBBB@

−1010−1

0010000

1CCCCCCCCCCCCCCCCCCCCCCA

,v3 =

0BBBBBBBBBBBBBBBBBBBBBB@

0−1

1000−1

01000

1CCCCCCCCCCCCCCCCCCCCCCA

,v4 =

0BBBBBBBBBBBBBBBBBBBBBB@

000−1

10000−1

10

1CCCCCCCCCCCCCCCCCCCCCCA

,v5 =

0BBBBBBBBBBBBBBBBBBBBBB@

00000−1

100−1

01

1CCCCCCCCCCCCCCCCCCCCCCA

.

These vectors represent the circuits around 5 of the cube’s faces.

(b) Examples:

0BBBBBBBBBBBBBBBBBBBBBB@

0000000−1

10−1

1

1CCCCCCCCCCCCCCCCCCCCCCA

= v1 − v2 + v3 − v4 + v5,

0BBBBBBBBBBBBBBBBBBBBBB@

01−1−1

110−1

0000

1CCCCCCCCCCCCCCCCCCCCCCA

= v1 − v2,

0BBBBBBBBBBBBBBBBBBBBBB@

0−1

11−1

0−1

011−1

0

1CCCCCCCCCCCCCCCCCCCCCCA

= v3 − v4.

♥ 2.6.7.

(a) Tetrahedron:

0BBBBBBB@

1 −1 0 01 0 −1 01 0 0 −10 1 −1 00 1 0 −10 0 1 −1

1CCCCCCCA

66

number of circuits = dim coker A = 3, number of faces = 4;

(b) Octahedron: 0BBBBBBBBBBBBBBBBBBBBBB@

1 −1 0 0 0 01 0 −1 0 0 01 0 0 −1 0 01 0 0 0 −1 00 1 −1 0 0 00 1 0 0 −1 00 1 0 0 0 −10 0 1 −1 0 00 0 1 0 0 −10 0 0 1 −1 00 0 0 1 0 −10 0 0 0 1 −1

1CCCCCCCCCCCCCCCCCCCCCCA

number of circuits = dim coker A = 7, number of faces = 8.

(c) Dodecahedron:0BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB@

1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 00 0 0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0−1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0 −1 0 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 0 0 0 −1 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 0 −1 0 0 0 0 0 00 0 0 0 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 0 00 0 0 0 0 0 0 0 1 0 0 0 0 0 −1 0 0 0 0 00 0 0 0 0 0 0 0 0 1 0 0 0 0 −1 0 0 0 0 00 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 1 0 0 0 0 −1 0 0 0 00 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 −1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 −1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 −1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 −10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −1 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 1

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCA

number of circuits = dim coker A = 11, number of faces = 12.

67

(d) Icosahedron:

0BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB@

1 −1 0 0 0 0 0 0 0 0 0 01 0 −1 0 0 0 0 0 0 0 0 01 0 0 −1 0 0 0 0 0 0 0 01 0 0 0 −1 0 0 0 0 0 0 01 0 0 0 0 −1 0 0 0 0 0 00 1 −1 0 0 0 0 0 0 0 0 00 1 0 0 0 0 −1 0 0 0 0 00 1 0 0 0 0 0 0 0 0 −1 00 0 1 −1 0 0 0 0 0 0 0 00 0 1 0 0 0 −1 0 0 0 0 00 0 1 0 0 0 0 −1 0 0 0 00 0 0 1 −1 0 0 0 0 0 0 00 0 0 1 0 0 0 −1 0 0 0 00 0 0 1 0 0 0 0 −1 0 0 00 0 0 0 1 −1 0 0 0 0 0 00 0 0 0 1 0 0 0 −1 0 0 00 0 0 0 1 0 0 0 0 −1 0 00 −1 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 −1 0 00 0 0 0 0 1 0 0 0 0 −1 00 0 0 0 0 0 1 −1 0 0 0 00 0 0 0 0 0 1 0 0 0 0 −10 0 0 0 0 0 0 1 −1 0 0 00 0 0 0 0 0 0 1 0 0 0 −10 0 0 0 0 0 0 0 1 −1 0 00 0 0 0 0 0 0 0 1 0 0 −10 0 0 0 0 0 0 0 0 1 −1 00 0 0 0 0 0 0 0 0 1 0 −10 0 0 0 0 0 −1 0 0 0 1 00 0 0 0 0 0 0 0 0 0 1 −1

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCA

number of circuits = dim coker A = 19, number of faces = 20.

♥ 2.6.8.

(a) (i)

0B@−1 1 0 0

0 1 −1 00 1 0 −1

1CA, (ii)

0BBB@

−1 1 0 0 00 −1 1 0 00 0 −1 1 00 1 0 0 −1

1CCCA,

(iii)

0BBBBB@

−1 1 0 0 0 00 1 −1 0 0 00 0 1 −1 0 00 1 0 0 −1 00 1 0 0 0 −1

1CCCCCA

, (iv)

0BBBBB@

−1 1 0 0 0 00 −1 1 0 0 00 0 −1 1 0 00 1 0 0 −1 00 0 1 0 0 −1

1CCCCCA

.

(b)

0BBB@

−1 1 0 0 00 −1 1 0 00 0 −1 1 00 0 0 −1 1

1CCCA,

0BBB@

−1 1 0 0 00 −1 1 0 00 0 −1 1 00 1 0 0 −1

1CCCA,

0BBB@

−1 1 0 0 00 1 −1 0 00 1 0 −1 00 1 0 0 −1

1CCCA.

68

(c) Let m denote the number of edges. Since the graph is connected, its incidence matrixA has rank n − 1. There are no circuits if and only if coker A = 0, which implies0 = dim coker A = m− (n− 1), and so m = n− 1.

♥ 2.6.9.

(a)

(b)

0B@

1 −1 01 0 −10 1 −1

1CA,

0BBBBBBB@

1 −1 0 01 0 −1 01 0 0 −10 1 −1 00 1 0 −10 0 1 −1

1CCCCCCCA

,

0BBBBBBBBBBBBBBBBB@

1 −1 0 0 01 0 −1 0 01 0 0 −1 01 0 0 0 −10 1 −1 0 00 1 0 −1 00 1 0 0 −10 0 1 −1 00 0 1 0 −10 0 0 1 −1

1CCCCCCCCCCCCCCCCCA

.

(c) 12 n(n− 1); (d) 1

2 (n− 1)(n− 2).

♥ 2.6.10.

(a)

(b)

0BBBBBBB@

1 0 −1 0 01 0 0 −1 01 0 0 0 −10 1 −1 0 00 1 0 −1 00 1 0 0 −1

1CCCCCCCA

,

0BBBBBBBBBBBB@

1 0 −1 0 0 01 0 0 −1 0 01 0 0 0 −1 01 0 0 0 0 −10 1 −1 0 0 00 1 0 −1 0 00 1 0 0 −1 00 1 0 0 0 −1

1CCCCCCCCCCCCA

,

0BBBBBBBBBBBBBB@

1 0 0 −1 0 01 0 0 0 −1 01 0 0 0 0 −10 1 0 −1 0 00 1 0 0 −1 00 1 0 0 0 −10 0 1 −1 0 00 0 1 0 −1 00 0 1 0 0 −1

1CCCCCCCCCCCCCCA

.

(c) m n; (d) (m− 1)(n− 1).

69

♥ 2.6.11.

(a) A =

0BBBBBBBBBB@

1 −1 0 0 0 0 0 0−1 0 0 1 0 0 0 0

0 −1 0 1 0 0 0 00 0 −1 0 0 1 0 00 0 0 0 −1 0 1 00 0 0 0 −1 0 0 10 0 0 0 0 0 1 −1

1CCCCCCCCCCA

.

(b) The vectors v1 =

0BBBBBBBBBBBB@

11010000

1CCCCCCCCCCCCA

, v2 =

0BBBBBBBBBBBB@

00100100

1CCCCCCCCCCCCA

, v3 =

0BBBBBBBBBBBB@

00001011

1CCCCCCCCCCCCA

form a basis for ker A.

(c) The entries of each vi are indexed by the vertices. Thus the nonzero entries in v1 cor-respond to the vertices 1,2,4 in the first connected component, v2 to the vertices 3,6 inthe second connected component, and v3 to the vertices 5,7,8 in the third connectedcomponent.

(d) Let A have k connected components. A basis for ker A consists of the vectors v1, . . . ,vk

where vi has entries equal to 1 if the vertex lies in the ith connected component of thegraph and 0 if it doesn’t. To prove this, suppose Av = 0. If edge #` connects vertex ato vertex b, then the `th component of the linear system is va − vb = 0. Thus, va = vbwhenever the vertices are connected by an edge. If two vertices are in the same con-nected component, then they can be connected by a path, and the values va = vb = · · ·at each vertex on the path must be equal. Thus, the values of va on all vertices in theconnected component are equal, and hence v = c1v1 + · · · + ckvk can be written as alinear combination of the basis vectors, with ci being the common value of the entries

va corresponding to vertices in the ith connected component. Thus, v1, . . . ,vk span thekernel. Moreover, since the coefficients ci coincide with certain entries va of v, the onlylinear combination giving the zero vector is when all ci are zero, proving their linear in-dependence.

♦ 2.6.12. If the incidence matrix has rank r, then # circuits

= dim coker A = n− r = dim ker A ≥ 1,

since ker A always contains the vector ( 1, 1, . . . , 1 )T .

2.6.13. Changing the direction of an edge is the same as multiplying the corresponding row ofthe incidence matrix by −1. The dimension of the cokernel, being the number of indepen-dent circuits, does not change. Each entry of a cokernel vector that corresponds to an edgethat has been reversed is multiplied by −1. This can be realized by left multiplying theincidence matrix by a diagonal matrix whose diagonal entries are −1 is the correspondingedge has been reversed, and +1 if it is unchanged.

♥ 2.6.14.(a) Note that P permutes the rows of A, and corresponds to a relabeling of the vertices of

the digraph, while Q permutes its columns, and so corresponds to a relabeling of theedges.

(b) (i),(ii),(v) represent equivalent digraphs; none of the others are equivalent.(c) v = (v1, . . . , vm) ∈ coker A if and only if bv = P v = ( vπ(1) . . . vπ(m) ) ∈ coker B. Indeed,

bvT B = (P v)T P AQ = vT AQ = 0 since, according to Exercise 1.6.14, P T = P−1 is theinverse of the permutation matrix P .

70

2.6.15. False. For example, any two inequivalent trees, cf. Exercise 2.6.8, with the same num-ber of nodes have incidence matrices of the same size, with trivial cokernels: coker A =coker B = 0. As another example, the incidence matrices

A =

0BBBBB@

1 −1 0 0 00 1 −1 0 0−1 0 1 0 0

1 0 0 −1 01 0 0 0 −1

1CCCCCA

and B =

0BBBBB@

1 −1 0 0 00 1 −1 0 0−1 0 1 0 0

1 0 0 −1 00 1 0 0 −1

1CCCCCA

both have cokernel basis ( 1, 1, 1, 0, 0 )T , but do not represent equivalent digraphs.

2.6.16.(a) If the first k vertices belong to one component and the last n−k to the other, then there

is no edge between the two sets of vertices and so the entries aij = 0 whenever i =

1, . . . , k, j = k + 1, . . . , n, or when i = k + 1, . . . , n, j = 1, . . . , k, which proves that A hasthe indicated block form.

(b) The graph consists of two disconnected triangles. If we use 1, 2, 3 to label the vertices inone triangle and 4, 5, 6 for those in the second, the resulting incidence matrix has the in-

dicated block form

0BBBBBBB@

1 −1 0 0 0 00 1 −1 0 0 0−1 0 1 0 0 0

0 0 0 1 −1 00 0 0 0 1 −10 0 0 −1 0 1

1CCCCCCCA

, with each block a 3× 3 submatrix.

71

Solutions — Chapter 3

3.1.1. Bilinearity:

〈 cu + dv ,w 〉 = (cu1 + dv1)w1 − (cu1 + dv1)w2 − (cu2 + dv2)w1 + b(cu2 + dv2)w2

= c(u1w1 − u1w2 − u2w1 + bu2w2) + d(v1w1 − v1w2 − v2w1 + bv2w2)

= c 〈u ,w 〉+ d 〈v ,w 〉,〈u , cv + dw 〉 = u1 (cv1 + dw1)− u1 (cv2 + dw2)− u2 (cv1 + dw1) + bu2 (cv2 + dw2)

= c(u1 v1 − u1 v2 − u2 v1 + bu2 v2) + d(u1w1 − u1w2 − u2w1 + bu2w2)

= c 〈u ,v 〉+ d 〈u ,w 〉.

Symmetry:

〈v ,w 〉 = v1w1 − v1w2 − v2w1 + bv2w2 = w1 v1 − w1 v2 − w2 v1 + bw2 v2 = 〈w ,v 〉.

To prove positive definiteness, note

〈v ,v 〉 = v21 − 2v1 v2 + bv22 = (v1 − v2)2 + (b− 1)v22 > 0 for all v = ( v1, v2 )T 6= 0

if and only if b > 1. (If b = 1, the formula is only positive semi-definite, since v = ( 1, 1 )T

gives 〈v ,v 〉 = 0, for instance.)

3.1.2. (a), (f ) and (g) define inner products; the others don’t.

3.1.3. It is not positive definite, since if v = ( 1,−1 )T , say, 〈v ,v 〉 = 0.

3.1.4.(a) Bilinearity:

〈 cu + dv ,w 〉 = (cu1 + dv1)w1 + 2(cu2 + dv2)w2 + 3(cu3 + dv3)w3

= c(u1w1 + 2u2w2 + 3u3w3) + d(v1w1 + 2v2w2 + 3v3w3)

= c 〈u ,w 〉+ d 〈v ,w 〉,〈u , cv + dw 〉 = u1 (cv1 + dw1) + 2u2 (cv2 + dw2) + 3u3 (cv3 + dw3)

= c(u1 v1 + 2u2 v2 + 3u3 v3) + d(u1w1 + 2u2w2 + 3u3w3)

= c 〈u ,v 〉+ d 〈u ,w 〉.

Symmetry:

〈v ,w 〉 = v1w1 + 2v2w2 + 3v3w3 = w1 v1 + 2w2 v2 + 3w3 v3 = 〈w ,v 〉.

Positivity:

〈v ,v 〉 = v21 + 2v22 + 3v23 > 0 for all v = ( v1, v2, v3 )T 6= 0

because it is a sum of non-negative terms, at least one of which is strictly positive.

72

(b) Bilinearity:〈 cu + dv ,w 〉 = 4(cu1 + dv1)w1 + 2(cu1 + dv1)w2 + 2(cu2 + dv2)w1 +

+ 4(cu2 + dv2)w2 + (cu3 + dv3)w3

= c(4u1w1 + 2u1w2 + 2u2w1 + 4u2w2 + u3w3) +

+ d(4v1w1 + 2v1w2 + 2v2w1 + 4v2w2 + v3w3)

= c 〈u ,w 〉+ d 〈v ,w 〉,〈u , cv + dw 〉 = 4u1 (cv1 + dw1) + 2u1 (cv2 + dw2) + 2u2 (cv1 + dw1) +

+ 4u2 (cv2 + dw2) + u3 (cv3 + dw3)

= c(4u1 v1 + 2u1 v2 + 2u2 v1 + 4u2 v2 + u3 v3) +

+ d(4u1w1 + 2u1w2 + 2u2w1 + 4u2w2 + u3w3)

= c 〈u ,v 〉+ d 〈u ,w 〉.Symmetry:

〈v ,w 〉 = 4v1w1 + 2v1w2 + 2v2w1 + 4v2w2 + v3w3

= 4w1 v1 + 2w1 v2 + 2w2 v1 + 4w2 v2 + w3 v3 = 〈w ,v 〉.Positivity:

〈v ,v 〉 = 4v21 +4v1 v2 +4v22 + v23 = (2v1 + v2)2 +3v22 + v23 > 0 for all v = ( v1, v2, v3 )T 6= 0,

because it is a sum of non-negative terms, at least one of which is strictly positive.(c) Bilinearity:〈 cu + dv ,w 〉 = 2(cu1 + dv1)w1 − 2(cu1 + dv1)w2 − 2(cu2 + dv2)w1 +

+ 3(cu2 + dv2)w2 − (cu2 + dv2)w3 − (cu3 + dv3)w2 + 2(cu3 + dv3)w3

= c(2u1w1 − 2u1w2 − 2u2w1 + 3u2w2 − u2w3 − u3w2 + 2u3w3) +

+ d(2v1w1 − 2v1w2 − 2v2w1 + 3v2w2 − v2w3 − v3w2 + 2v3w3)

= c 〈u ,w 〉+ d 〈v ,w 〉,〈u , cv + dw 〉 = 2u1 (cv1 + dw1)− 2u1 (cv2 + dw2)− 2u2 (cv1 + dw1) +

+ 3u2 (cv2 + dw2)− u2 (cv3 + dw3)− u3 (cv2 + dw2) + 2u3 (cv3 + dw3)

= c(2u1 v1 − 2u1 v2 − 2u2 v1 + 3u2 v2 − u2 v3 − u3 v2 + 2u3 v3) +

+ d(2u1w1 − 2u1w2 − 2u2w1 + 3u2w2 − u2w3 − u3w2 + 2u3w3)

= c 〈u ,v 〉+ d 〈u ,w 〉.Symmetry:〈v ,w 〉 = 2v1w1 − 2v1w2 − 2v2w1 + 3v2w2 − v2w3 − v3w2 + 2v3w3

= 2w1 v1 − 2w1 v2 − 2w2 v1 + 3w2 v2 − w2 v3 − w3 v2 + 2w3 v3 = 〈w ,v 〉.Positivity:

〈v ,v 〉 = 2v21 − 4v1 v2 + 3v22 − 2v2 v3 + 2v23 = 2(v1 − v2)2 + (v2 − v3)

2 + v23 > 0

for all v = ( v1, v2, v3 )T 6= 0, because it is a sum of non-negative terms, at least one ofwhich is strictly positive.

3.1.5. (a) ( cos t, sin t )T ,

cos t√

2,

sin t√5

!T

,

cos t+

sin t√3,

sin t√5

!T

.

-1 -0.5 0.5 1

-1

-0.5

0.5

1

-1 -0.5 0.5 1

-1

-0.5

0.5

1

-1 -0.5 0.5 1

-1

-0.5

0.5

1

73

(b) Note: By elementary analytical geometry, any quadratic equation of the form

ax2 + bxy + cy2 = 1 defines an ellipse provided a > 0 and b2 − 4ac < 0.Case (b): The equation 2v2

1 + 5v22 = 1 defines an ellipse with semi-axes 1√2, 1√

5.

Case (c): The equation v21−2v1 v2+4v22 = 1 also defines an ellipse by the preceding remark.

♦ 3.1.6.(a) The vector v = (x, y )T can be viewed as the hypotenuse of a right triangle with side

lengths x, y, and so by Pythagoras, ‖v ‖2 = x2 + y2.

(b) First, the projection p = (x, y, 0 )T of v = (x, y, z )T onto the xy plane ha length

‖p ‖ =qx2 + y2 by Pythagoras, as in part (a). Second, the right triangle formed by

0,p and v has side lengths ‖p ‖ and z, and, again by Pythagoras,

‖v ‖2 = ‖p ‖2 + z2 = x2 + y2 + z2.

♦ 3.1.7. ‖ cv ‖ =q〈 cv , cv 〉 =

qc2 〈v ,v 〉 = | c | ‖v ‖.

3.1.8. By bilinearity and symmetry,

〈 av + bw , cv + dw 〉 = a〈v , cv + dw 〉+ b〈w , cv + dw 〉= ac〈v ,v 〉+ ad〈v ,w 〉+ bc〈w ,v 〉+ bd〈w ,w 〉= ac‖v ‖2 + (ad+ bc)〈v ,w 〉+ bd‖w ‖2.

♦ 3.1.9. If we know the first bilinearity property and symmetry, then the second follows:

〈u , cv + dw 〉 = 〈 cv + dw ,u 〉 = c 〈v ,u 〉+ d 〈w ,u 〉 = c 〈u ,v 〉+ d 〈u ,w 〉.

♦ 3.1.10.(a) Choosing v = x, we have 0 = 〈x ,x 〉 = ‖x ‖2, and hence x = 0.(b) Rewrite the condition as 0 = 〈x ,v 〉 − 〈y ,v 〉 = 〈x− y ,v 〉 for all v ∈ V . Now use part

(a) to conclude that x− y = 0 and so x = y.(c) If v is any element of V , then we can write v = c1v1+· · ·+cn vn as a linear combination

of the basis elements, and so, by bilinearity, 〈x ,v 〉 = c1 〈x ,v1 〉 + · · · + cn 〈x ,vn 〉 = 0.Since this holds for all v ∈ V , the result in part (a) implies x = 0.

♦ 3.1.11.

(a) ‖u + v ‖2 − ‖u− v ‖2 = 〈u + v ,u + v 〉 − 〈u− v ,u− v 〉=“〈u ,u 〉+ 2 〈u ,v 〉+ 〈v ,v 〉

”−“〈u ,u 〉 − 2 〈u ,v 〉+ 〈v ,v 〉

”= 4 〈u ,v 〉.

(b) 〈v ,w 〉 = 14

h(v1 + w1)

2 − 3(v1 + w1)(v2 + w2) + 5(v2 + w2)2i−

− 14

h(v1 − w1)

2 − 3(v1 − w1)(v2 − w2) + 5(v2 − w2)2i

= v1w1 − 32 v1w2 − 3

2 v2w1 + 5 v2w2.

3.1.12.(a) ‖x + y ‖2 + ‖x− y ‖2 =

“‖x ‖2 + 2 〈x ,y 〉+ ‖y ‖2

”+“‖x ‖2 − 2 〈x ,y 〉+ ‖y ‖2

”

= 2 ‖x ‖2 + 2 ‖y ‖2.(b) The sum of the squared lengths of the diagonals in a parallelogram equals the sum of

the squared lengths of all four sides:

74

‖x ‖

‖x ‖

‖y ‖

‖y ‖

‖x + y ‖

‖x− y ‖

3.1.13. By Exercise 3.1.12, ‖v ‖2 = 12

“‖u + v ‖2 + ‖u− v ‖2

”− ‖u ‖2 = 17, so ‖v ‖ =

√17 .

The answer is the same in all norms coming from inner products.

3.1.14. Using (3.2), v · (Aw) = vTAw = (AT v)T w = (AT v) ·w.♦ 3.1.15. First, if A is symmetric, then

(Av) ·w = (Av)T w = vTAT

w = vTAw = v · (Aw).

To prove the converse, note that Aej gives the jth column of A, and so

aij = ei · (Aej) = (Aei) · ej = aji for all i, j. Hence A = AT .

3.1.16. The inner product axioms continue to hold when restricted to vectors in W since theyhold for all vectors in V , including those in W .

3.1.17. Bilinearity:〈〈〈 cu + dv ,w 〉〉〉 = 〈 cu + dv ,w 〉+ 〈〈 cu + dv ,w 〉〉

= c 〈u ,w 〉+ d 〈v ,w 〉+ c 〈〈u ,w 〉〉+ d 〈〈v ,w 〉〉 = c 〈〈〈u ,w 〉〉〉+ d 〈〈〈v ,w 〉〉〉,〈〈〈u , cv + dw 〉〉〉 = 〈u , cv + dw 〉+ 〈〈u , cv + dw 〉〉

= c 〈u ,v 〉+ d 〈u ,w 〉+ c 〈〈u ,v 〉〉+ d 〈〈u ,w 〉〉 = c 〈〈〈u ,v 〉〉〉+ d 〈〈〈u ,w 〉〉〉.Symmetry:

〈〈〈v ,w 〉〉〉 = 〈v ,w 〉+ 〈〈v ,w 〉〉 = 〈w ,v 〉+ 〈〈w ,v 〉〉 = 〈〈〈w ,v 〉〉〉.Positivity: 〈〈〈v ,v 〉〉〉 = 〈v ,v 〉+ 〈〈v ,v 〉〉 > 0 for all v 6= 0 since both terms are positive.

♦ 3.1.18. Bilinearity:〈〈〈 c (v,w) + d (ev, ew) , (bv, bw) 〉〉〉 = 〈〈〈 (cv + d ev, cw + d ew) , (bv, bw) 〉〉〉

= 〈 cv + d ev , bv 〉+ 〈〈 cw + d ew , bw 〉〉= c 〈v , bv 〉+ d 〈 ev , bv 〉+ c 〈w , bw 〉+ d 〈 ew , bw 〉= c 〈〈〈 (v,w) , (bv, bw) 〉〉〉+ d 〈〈〈 (ev, ew) , (bv, bw) 〉〉〉,

〈〈〈 (v,w) , c (ev, ew) + d (bv, bw) 〉〉〉 = 〈〈〈 (v,w) , (c ev + d bv, c ew + d bw) 〉〉〉= 〈v , c ev + d bv 〉+ 〈〈w , c ew + d bw 〉〉= c 〈v , ev 〉+ d 〈v , bv 〉+ c 〈w , ew 〉+ d 〈w , bw 〉= c 〈〈〈 (v,w) , (ev, ew) 〉〉〉+ d 〈〈〈 (v,w) , (bv, bw) 〉〉〉.

Symmetry:

〈〈〈 (v,w) , (ev, ew) 〉〉〉 = 〈v , ev 〉+ 〈〈w , ew 〉〉 = 〈 ev ,v 〉+ 〈〈 ew ,w 〉〉 = 〈〈〈 (ev, ew) , (v,w) 〉〉〉.Positivity:

〈〈〈 (v,w) , (v,w) 〉〉〉 = 〈v ,v 〉+ 〈〈w ,w 〉〉 > 0

75

for all (v,w) 6= (0,0), since both terms are non-negative and at least one is positive be-cause either v 6= 0 or w 6= 0.

3.1.19.(a) 〈 1 , x 〉 = 1

2 , ‖ 1 ‖ = 1, ‖x ‖ = 1√3;

(b) 〈 cos 2πx , sin 2πx 〉 = 0, ‖ cos 2πx ‖ = 1√2, ‖ sin 2πx ‖ = 1√

2;

(c) 〈x , ex 〉 = 1, ‖x ‖ = 1√3, ‖ ex ‖ =

r12 (e2 − 1) ;

(d)

*(x+ 1)2 ,

1

x+ 1

+=

3

2,‚‚‚ (x+ 1)2

‚‚‚ =

s31

5,

‚‚‚‚‚1

x+ 1

‚‚‚‚‚ =1√2.

3.1.20. (a) 〈 f , g 〉 = 34 , ‖ f ‖ = 1√

3, ‖ g ‖ =

r2815 ; (b) 〈 f , g 〉 = 0, ‖ f ‖ =

r23 , ‖ g ‖ =

r5615 ;

(c) 〈 f , g 〉 = 815 , ‖ f ‖ = 1

2 , ‖ g ‖ =

r76 .

3.1.21. All but (b) are inner products. (b) is not because it fails positivity: for instance,Z 1

−1(1− x)2 x dx = − 4

3 .

3.1.22. If f(x) is any nonzero function that satisfies f(x) = 0 for all 0 ≤ x ≤ 1 then 〈 f , f 〉 = 0.

An example is the function f(x) =

(x, −1 ≤ x ≤ 0,

0, 0 ≤ x ≤ 1.However, if the function f ∈ C0[0, 1]

is only considered on [0, 1], its values outside the interval are irrelevant, and so the positiv-ity is unaffected. The formula does define an inner product on the subspace of polynomialfunctions because 〈 p , p 〉 = 0 if and only if p(x) = 0 for all 0 ≤ x ≤ 1, which impliesp(x) ≡ 0 for all x since only the zero polynomial can vanish on an interval.

3.1.23.(a) No — positivity doesn’t hold since if f(0) = f(1) = 0 then 〈 f , f 〉 = 0 even if f(x) 6= 0

for any 0 < x < 1;(b) Yes. Bilinearity and symmetry are readily established. As for positivity,

〈 f , f 〉 = f(0)2+f(1)2+Z 1

0f(x)2 dx ≥ 0 is a sum of three non-negative quantities, and is

equal to 0 if and only if all three terms vanish, so f(0) = f(1) = 0 andZ 1

0f(x)2 dx = 0

which, by continuity, implies f(x) ≡ 0 for all 0 ≤ x ≤ 1.

3.1.24. No. For example, on [−1, 1], ‖ 1 ‖ =√

2 , but ‖ 1 ‖2 = 2 6= ‖ 12 ‖ =√

2.

♦ 3.1.25. Bilinearity:

〈 cf + dg , h 〉 =Z b

a

hcf(x) + dg(x)h(x) + cf(x) + dg(x)′h′(x)

idx

=Z b

a

hcf(x)h(x) + dg(x)h(x) + cf ′(x)h′(x) + dg′(x)h′(x)

idx

= cZ b

a

hf(x)h(x) + f ′(x)h′(x)

idx+ d

Z b

a

hg(x)h(x) + g′(x)h′(x)

idx

= c〈 f , h 〉+ d〈 g , h 〉.

〈 f , cg + dh 〉 =Z b

a

hf(x)cg(x) + dh(x)+ f ′(x)cg(x) + dh(x)′

idx

76

=Z b

a

hcf(x)g(x) + df(x)h(x) + cf ′(x)g′(x) + df ′(x)h′(x)

idx

= cZ b

a

hf(x)g(x) + f ′(x)g′(x)

idx+ d

Z b

a

hf(x)h(x) + f ′(x)h′(x)

idx

= c〈 f , g 〉+ d〈 f , h 〉.Symmetry:

〈 f , g 〉 =Z b

a

hf(x)g(x) + f ′(x)g′(x)

idx =

Z b

a

hg(x)f(x) + g′(x)f ′(x)

idx = 〈 g , f 〉.

Positivity

〈 f , f 〉 =Z b

a

hf(x)2 + f ′(x)2

idx > 0 for all f 6≡ 0,

since the integrand is non-negative, and, by continuity, the integral is zero if and only ifboth f(x) ≡ 0 and f ′(x) ≡ 0 for all a ≤ x ≤ b.

3.1.26.(a) No, because if f(x) is any constant function, then 〈 f , f 〉 = 0, and so positive definite-

ness does not hold.(b) Yes. To prove the first bilinearity condition:

〈 cf + dg , h 〉 =Z 1

−1

hcf ′(x) + dg′(x)

ih′(x) dx

= cZ 1

−1f ′(x)h′(x) dx+ d

Z 1

−1g′(x)h′(x) dx = c〈 f , h 〉+ d〈 g , h 〉.

The second has a similar proof, or follows from symmetry, cf. Exercise 3.1.9. To provesymmetry:

〈 f , g 〉 =Z 1

−1f ′(x) g′(x) dx =

Z 1

−1g′(x) f ′(x) dx = 〈 g , f 〉.

As for positivity, 〈 f , f 〉 =Z 1

−1f ′(x)2 dx ≥ 0. Moreover, since f ′ is continuous,

〈 f , f 〉 = 0 if and only if f ′(x) ≡ 0 for all x, and so f(x) ≡ c is constant. But the onlyconstant function in W is the zero function, and so 〈 f , f 〉 > 0 for all 0 6= f ∈W .

♦ 3.1.27. Suppose h(x0) = k > 0 for some a < x0 < b. Then, by continuity, h(x) ≥ 12k for

a < x0 − δ < x < x0 + δ < b for some δ > 0. But then, since h(x) ≥ 0 everywhere,Z b

ah(x) dx ≥

Z x0+δ

x0−δh(x) dx ≥ k δ > 0,

which is a contradiction. A similar contradiction can be shown when h(x0) = k < 0 forsome a < x0 < b. Thus h(x) = 0 for all a < x < b, which, by continuity, also impliesh(a) = h(b) = 0. The function in (3.14) gives a discontinuous counterexample.

♦ 3.1.28.(a) To prove the first bilinearity condition:

〈 cf + dg , h 〉 =Z b

a

hcf(x) + dg(x)

ih(x)w(x) dx

= cZ b

af(x)h(x)w(x) dx+ d

Z b

ag(x)h(x)w(x) dx = c〈 f , h 〉+ d〈 g , h 〉.

The second has a similar proof, or follows from symmetry, cf. Exercise 3.1.9. To provesymmetry:

〈 f , g 〉 =Z b

af(x) g(x)w(x) dx =

Z b

ag(x) f(x)w(x) dx = 〈 g , f 〉.

As for positivity, 〈 f , f 〉 =Z b

af(x)2 w(x) dx ≥ 0. Moreover, since w(x) > 0 and the inte-

77

grand is continuous, Exercise 3.1.27 implies that 〈 f , f 〉 = 0 if and only if f(x)2 w(x) ≡0 for all x, and so f(x) ≡ 0.

(b) If w(x0) < 0, then, by continuity, w(x) < 0 for x0 − δ ≤ x ≤ x0 + δ for some δ > 0. Nowchoose f(x) 6≡ 0 so that f(x) = 0 whenever |x− x0 | > δ. Then

〈 f , f 〉 =Z b

af(x)2 w(x) dx =

Z x0+δ

x0−δf(x)2 w(x) dx < 0, violating positivity.

(c) Bilinearity and symmetry continue to hold. The positivity argument says that 〈 f , f 〉 =0 implies that f(x) = 0 whenever w(x) > 0. By continuity, f(x) ≡ 0, provided w(x) 6≡ 0on any open subinterval a ≤ c < x < d ≤ b, and so under this assumption it remains aninner product. However, if w(x) ≡ 0 on a subinterval, then positivity is violated.

♥ 3.1.29.(a) If f(x0, y0) = k > 0 then, by continuity, f(x, y) ≥ 1

2 k for (x, y) ∈ D = ‖x− x0 ‖ ≤ δfor some δ > 0. But then

ZZ

Ωf(x, y) dx dy ≥

ZZ

Df(x, y) dx dy ≥ 1

2 πk δ2 > 0.

(b) Bilinearity:

〈 cf + dg , h 〉 =ZZ

Ω

hcf(x, y) + dg(x, y)h(x, y)

idx dy

= cZZ

Ωf(x, y)h(x, y) dx dy + d

ZZ

Ω

hg(x, y)h(x, y)

idx dy = c〈 f , h 〉+ d〈 g , h 〉.

The second bilinearity conditions follows from the first and symmetry:

〈 f , g 〉 =ZZ

Ωf(x, y)g(x, y) dx dy =

ZZ

Ωg(x, y)f(x, y) dx dy = 〈 g , f 〉.

Positivity; using part (a),

〈 f , f 〉 =ZZ

Ω[f(x, y) ]2 dx dy > 0 for all f 6≡ 0.

The formula for the norm is ‖ f ‖ =

sZZ

Ω[f(x, y) ]2 dx dy .

3.1.30. (a) 〈 f , g 〉 = 23 , ‖ f ‖ = 1, ‖ g ‖ =

q2845 ; (b) 〈 f , g 〉 = 1

2 π, ‖ f ‖ =√π , ‖ g ‖ =

qπ3 .

♥ 3.1.31.(a) To prove the first bilinearity condition:

〈〈 c f + dg ,h 〉〉 =Z 1

0

h(cf1(x) + dg1(x))h1(x) + (cf2(x) + dg2(x))h2(x)

idx

= cZ 1

0

hf1(x)h1(x) + f2(x)h2(x)

idx+ d

Z 1

0

hg1(x)h1(x) + g2(x)h2(x)

idx

= c〈〈 f ,h 〉〉+ d〈〈g ,h 〉〉.The second has a similar proof, or follows from symmetry, cf. Exercise 3.1.9. To provesymmetry:

〈〈 f ,g 〉〉 =Z 1

0

hf1(x)g1(x) + f2(x)g2(x)

idx =

Z 1

0

hg1(x)f1(x) + g2(x)f2(x)

idx = 〈〈g , f 〉〉.

As for positivity, 〈〈 f , f 〉〉 =Z 1

0

hf1(x)

2 + f2(x)2idx ≥ 0, since the integrand is a non-

negative function. Moreover, since f1(x) and f2(x) are continuous, so is f1(x)2 + f2(x)

2,

and hence 〈〈 f , f 〉〉 = 0 if and only if f1(x)2 + f2(x)

2 = 0 for all x, and so f(x) =

( f1(x), f2(x) )T ≡ 0.

(b) First bilinearity:

〈〈 c f + dg ,h 〉〉 =Z 1

0〈 c f(x) + dg(x) ,h(x) 〉 dx

= cZ 1

0〈 f(x) ,h(x) 〉 dx+ d

Z 1

0〈g(x) ,h(x) 〉 dx = c〈〈 f ,h 〉〉+ d〈〈g ,h 〉〉.

78

Symmetry:

〈〈 f ,g 〉〉 =Z 1

0〈 f(x) ,g(x) 〉 dx =

Z 1

0〈g(x) , f(x) 〉 dx = 〈〈g , f 〉〉.

Positivity: 〈〈 f , f 〉〉 =Z 1

0‖ f(x) ‖2 dx ≥ 0 since the integrand is non-negative. Moreover,

〈 f , f 〉 = 0 if and only if ‖ f(x) ‖2 = 0 for all x, and so, in view of the continuity of

‖ f(x) ‖2 we conclude that f(x) ≡ 0.

(c) This follows because 〈v ,w 〉 = v1w1−v1w2−v2w1+3v2w2vTKw for K =

1 −1−1 3

!> 0

defines an inner product on R2.

3.2.1.(a) |v1 · v2 | = 3 ≤ 5 =

√5√

5 = ‖v1 ‖ ‖v2 ‖; angle: cos−1 35 ≈ .9273;

(b) |v1 · v2 | = 1 ≤ 2 =√

2√

2 = ‖v1 ‖ ‖v2 ‖; angle: 23 π ≈ 2.0944;

(c) |v1 · v2 | = 0 ≤ 2√

6 =√

2√

12 = ‖v1 ‖ ‖v2 ‖; angle: 12 π ≈ 1.5708;

(d) |v1 · v2 | = 3 ≤ 3√

2 =√

3√

6 = ‖v1 ‖ ‖v2 ‖; angle: 34 π ≈ 2.3562;

(e) |v1 · v2 | = 4 ≤ 2√

15 =√

10√

6 = ‖v1 ‖ ‖v2 ‖; angle: cos−1„− 2√

15

«≈ 2.1134.

3.2.2. (a) 13 π; (b) 0, 1

3 π,12 π,

23 π, or π, depending upon whether −1 appears 0, 1, 2, 3 or 4 times

in the second vector.

3.2.3. The side lengths are all equal to

‖ (1, 1, 0)− (0, 0, 0) ‖ = ‖ (1, 1, 0)− (1, 0, 1) ‖ = ‖ (1, 1, 0)− (0, 1, 1) ‖ = · · · =√

2 .

The edge angle is 13 π = 60. The center angle is cos θ = − 1

3 , so θ = 1.9106 = 109.4712.

3.2.4.(a) |v ·w | = 5 ≤ 7.0711 =

√5√

10 = ‖v ‖ ‖w ‖.(b) |v ·w | = 11 ≤ 13.0767 = 3

√19 = ‖v ‖ ‖w ‖.

(c) |v ·w | = 22 ≤ 23.6432 =√

13√

43 = ‖v ‖ ‖w ‖.3.2.5.

(a) |v ·w | = 6 ≤ 6.4807 =√

14√

3 = ‖v ‖ ‖w ‖.(b) | 〈v ,w 〉 | = 11 ≤ 11.7473 =

√23√

6 = ‖v ‖ ‖w ‖.(c) | 〈v ,w 〉 | = 19 ≤ 19.4936 =

√38√

10 = ‖v ‖ ‖w ‖.

3.2.6. Set v = ( a, b )T ,w = ( cos θ, sin θ )T , so that Cauchy–Schwarz gives

|v ·w | = | a cos θ + b sin θ | ≤qa2 + b2 = ‖v ‖ ‖w ‖.

3.2.7. Set v = ( a1, . . . , an )T ,w = ( 1, 1, . . . , 1 )T , so that Cauchy–Schwarz gives

|v ·w | = | a1 + a2 + · · · + an | ≤√nqa21 + a2

2 + · · ·+ a2n = ‖v ‖ ‖w ‖.

Equality holds if and only if v = aw, i.e., a1 = a2 = · · · = an.

♦ 3.2.8. Using (3.20), ‖v −w ‖2 = 〈v −w ,v −w 〉 = ‖v ‖2 − 2 〈v ,w 〉+ ‖w ‖2= ‖v ‖2 + ‖w ‖2 − 2 ‖v ‖ ‖w ‖ cos θ.

♦ 3.2.9. Since a ≤ | a | for any real number a, so 〈v ,w 〉 ≤ | 〈v ,w 〉 | ≤ ‖v ‖ ‖w ‖.

♦ 3.2.10. Expanding ‖v + w ‖2 − ‖v −w ‖2 = 4 〈v ,w 〉 = 4 ‖v ‖ ‖w ‖ cos θ.

79

‖v ‖

‖w ‖

θ

‖v + w ‖

‖v −w ‖

♥ 3.2.11.(a) It is not an inner product. Bilinearity holds, but symmetry and positivity do not.(b) Assuming v,w 6= 0, we compute

sin2 θ = 1− cos2 θ =‖v ‖2 ‖w ‖2 − 〈v ,w 〉

‖v ‖2 ‖w ‖2 =(v ×w)2

‖v ‖2 ‖w ‖2 .

The result follows by taking square roots of both sides, where the sign is fixed by theorientation of the angle.

(c) By (b), v × w = 0 if and only if sin θ = 0 or v = 0 or w = 0, which implies that they

are parallel vectors. Alternative proof: v × w = det

v1 w1v2 w2

!= 0 if and only if the

columns v,w of the matrix are linearly dependent, and hence parallel vectors.(d) The parallelogram has side length ‖v ‖ and height ‖w ‖ | sin θ |, so its area is

‖v ‖ ‖w ‖ | sin θ | = |v ×w |.3.2.12.

(a) | 〈 f , g 〉 | = 1 ≤ 1.03191 =q

13

q12 e

2 − 12 = ‖ f ‖ ‖ g ‖;

(b) | 〈 f , g 〉 | = 2/e = .7358 ≤ 1.555 =q

23

q12 (e2 − e−2) = ‖ f ‖ ‖ g ‖;

(c) | 〈 f , g 〉 | = 12 = .5 ≤ .5253 =

√2− 5e−1

√e− 1 = ‖ f ‖ ‖ g ‖.

3.2.13. (a) 12 π, (b) cos−1 2

√2√π

= .450301, (c) 12 π.

3.2.14. (a) | 〈 f , g 〉 | = 23 ≤

q2845 = ‖ f ‖ ‖ g ‖; (b) | 〈 f , g 〉 | = π

2 ≤ π√3

= ‖ f ‖ ‖ g ‖.

3.2.15. (a) a = − 43 ; (b) no.

3.2.16. All scalar multiples of“

12 ,− 7

4 , 1”T

.

3.2.17. 3.2.15: a = 0. 3.2.16: all scalar multiples of“

16 ,− 21

24 , 1”T

.

3.2.18. All vectors in the subspace spanned by

0BBB@

1−2

10

1CCCA,

0BBB@

2−3

01

1CCCA, so v = a

0BBB@

1−2

10

1CCCA+ b

0BBB@

2−3

01

1CCCA.

3.2.19. (−3, 0, 0, 1 )T ,“

0,− 32 , 0, 1

”T, ( 0, 0, 3, 1 )T .

80

3.2.20. For example, u = ( 1, 0, 0 )T , v = ( 0, 1, 0 )T , w = ( 0, 1, 1 )T are linearly independent,

whereas u = ( 1, 0, 0 )T , v = w = ( 0, 1, 0 )T are linearly dependent.

3.2.21. (a) All solutions to a+ b = 1; (b) all solutions to a+ 3b = 2.

3.2.22. Only the zero vector satisfies 〈0 ,0 〉 = ‖0 ‖2 = 0.

♦ 3.2.23. Choose v = w; then 0 = 〈w ,w 〉 = ‖w ‖2, and hence w = 0.

3.2.24. 〈v + w ,v −w 〉 = ‖v ‖2 − ‖w ‖2 = 0 provided ‖v ‖ = ‖w ‖. They can’t both be

unit vectors: ‖v + w ‖2 = ‖v ‖2 + 2 〈v ,w 〉 + ‖w ‖2 = 2 + 2 〈v ,w 〉 = 1 if and only if

〈v ,w 〉 = − 12 , while ‖v −w ‖2 = 2 − 2 〈v ,w 〉 = 1 if and only if 〈v ,w 〉 = 1

2 , and soθ = 60.

♦ 3.2.25. If 〈v ,x 〉 = 0 = 〈v ,y 〉, then 〈v , cx + dy 〉 = c 〈v ,x 〉 + d 〈v ,y 〉 = 0 for c, d,∈ R,proving closure.

3.2.26. (a) 〈 p1 , p2 〉 =Z 1

0

“x− 1

2

”dx = 0, 〈 p1 , p3 〉 =

Z 1

0

“x2 − x+ 1

6

”dx = 0,

〈 p2 , p3 〉 =Z 1

0

“x− 1

2

”“x2 − x+ 1

6

”dx = 0.

(b) For n 6= m,

〈 sinnπx , sinmπx 〉 =Z 1

0sinnπx sinmπxdx =

Z 1

0

12

hcos(n+m)πx− cos(n−m)πx

idx = 0.

3.2.27. Any nonzero constant multiple of x2 − 13 .

3.2.28. p(x) = a“

(e− 1)x− 1”

+ b“x2 − (e− 2)x

”for any a, b ∈ R.

3.2.29. 1 is orthogonal to x, cosπx, sinπx; x is orthogonal to 1, cosπx; cosπx is orthogonal to1, x, sinπx; sinπx is orthogonal to 1, cosπx; ex is not orthogonal to any of the others.

3.2.30. Example: 1 and x− 23 .

3.2.31. (a) θ = cos−1 5√84≈ 0.99376 radians; (b) v ·w = 5 < 9.165 ≈

√84 = ‖v ‖ ‖w ‖,

‖v + w ‖ =√

30 ≈ 5.477 < 6.191 ≈√

14 +√

6 = ‖v ‖+ ‖w ‖; (c)“− 7

3 t,− 13 t, t

”T.

3.2.32.(a) ‖v1 + v2 ‖ = 4 ≤ 2

√5 = ‖v1 ‖+ ‖v2 ‖;

(b) ‖v1 + v2 ‖ =√

2 ≤ 2√

2 = ‖v1 ‖+ ‖v2 ‖;(c) ‖v1 + v2 ‖ =

√14 ≤

√2 +√

12 = ‖v1 ‖+ ‖v2 ‖;(d) ‖v1 + v2 ‖ =

√3 ≤√

3 +√

6 = ‖v1 ‖+ ‖v2 ‖;(e) ‖v1 + v2 ‖ =

√8 ≤√

10 +√

6 = ‖v1 ‖+ ‖v2 ‖.3.2.33.

(a) ‖v1 + v2 ‖ =√

5 ≤√

5 +√

10 = ‖v1 ‖+ ‖v2 ‖;(b) ‖v1 + v2 ‖ =

√6 ≤ 3 +

√19 = ‖v1 ‖+ ‖v2 ‖;

(c) ‖v1 + v2 ‖ =√

12 ≤√

13 +√

43 = ‖v1 ‖+ ‖v2 ‖.3.2.34.

(a) ‖ f + g ‖ =q

116 + 1

2 e2 ≈ 2.35114 ≤ 2.36467 ≈

q13 +

q12e

2 − 1 = ‖ f ‖+ ‖ g ‖;(b) ‖ f + g ‖ =

q23 + 1

2 e2 + 4e−1 − 1

2 e−2 ≈ 2.40105 ≤

81

≤ 2.72093 ≈q

23 +

q12 (e2 − e−2) = ‖ f ‖+ ‖ g ‖;

(c) ‖ f + g ‖ =√

2 + e− 5e−1 ≈ 1.69673 ≤ 1.71159 ≈√

2− 5e−1 +√e− 1 = ‖ f ‖+ ‖ g ‖.

3.2.35.(a) ‖ f + g ‖ =

q13345 ≈ 1.71917 ≤ 2.71917 ≈ 1 +

q2845 = ‖ f ‖+ ‖ g ‖;

(b) ‖ f + g ‖ =q

73 π ≈ 2.70747 ≤ 2.79578 ≈ √π +

q13 π = ‖ f ‖+ ‖ g ‖.

3.2.36.(a) 〈 1 , x 〉 = 0, so θ = 1

2 π. Yes, they are orthogonal.

(b) Note ‖ 1 ‖ =√

2, ‖x ‖ =q

23 , so 〈 1 , x 〉 = 0 <

q43 = ‖ 1 ‖ ‖x ‖, and

‖ 1 + x ‖ =q

83 ≈ 1.63299 < 2.23071 ≈

√2 +

q23 = ‖ 1 ‖+ ‖x ‖.

(c) 〈 1 , p 〉 = 2a+ 23 c = 0 and 〈x , p 〉 = 2

3 b = 0 if and only if p(x) = c“x2 − 1

3

”.

3.2.37.

(a)

˛˛˛

Z 1

0f(x) g(x) ex dx

˛˛˛ ≤

sZ 1

0f(x)2 ex dx

sZ 1

0g(x)2 ex dx ,

sZ 1

0

hf(x) + g(x)

i2ex dx ≤

sZ 1

0f(x)2 ex dx +

sZ 1

0g(x)2 ex dx ;

(b) 〈 f , g 〉 = 12 (e2 − 1) = 3.1945 ≤ 3.3063 =

√e− 1

q13 (e3 − 1) = ‖ f ‖ ‖ g ‖,

‖ f + g ‖ =q

13 e

3 + e2 + e− 73 = 3.8038 ≤ 3.8331 =

√e− 1 +

q13 (e3 − 1) = ‖ f ‖+ ‖ g ‖;

(c) cos θ =

√3

2

e2 − 1q

(e− 1)(e3 − 1)= .9662, so θ = .2607.

3.2.38.

(a)

˛˛˛

Z 1

0

hf(x)g(x) + f ′(x)g′(x)

idx

˛˛˛ ≤

sZ 1

0

hf(x)2 + f ′(x)2

idx

sZ 1

0

hg(x)2 + g′(x)2

idx;

sZ 1

0

h[f(x) + g(x)]2 + [f ′(x) + g′(x)]2

idx ≤

sZ 1

0

hf(x)2 + f ′(x)2

idx +

rhg(x)2 + g′(x)2

i.

(b) 〈 f , g 〉 = e− 1 ≈ 1.7183 ≤ 2.5277 ≈ 1 ·√e2 − 1 = ‖ f ‖ ‖ g ‖;

‖ f + g ‖ =√e2 + 2e− 2 ≈ 3.2903 ≤ 3.5277 ≈ 1 +

√e2 − 1 = ‖ f ‖+ ‖ g ‖.

(c) cos θ =

se− 1

e+ 1≈ .6798, so θ ≈ .8233.

3.2.39. Using the triangle inequality, ‖v ‖ = ‖ (v −w) + w ‖ ≤ ‖v −w ‖+ ‖w ‖.Therefore, ‖v −w ‖ ≥ ‖v ‖ − ‖w ‖. Switching v and w proves‖v −w ‖ ≥ ‖w ‖ − ‖v ‖, and the result is the combination of bothinequalities. In the figure, the inequality states that the length ofthe side v−w of the triangle opposite the origin is at least as largeas the difference between the other two lengths.

v −wv

w

3.2.40. True. By the triangle inequality,‖w ‖ = ‖ (−v) + (w + v) ‖ ≤ ‖−v ‖+ ‖v + w ‖ = ‖v ‖+ ‖v + w ‖.

♥ 3.2.41.(a) This follows immediately by identifying R

∞ with the space of all functions f : N → R

where N = 1, 2, 3, . . . are the natural numbers. Or, one can tediously verify all thevector space axioms.

82

(b) If x,y ∈ `2, then, by the triangle inequality on Rn,

nX

k=1

(xk + yk)2 ≤0B@

vuuutnX

k=1

x2k +

vuuutnX

k=1

y2k

1CA

2

≤0B@

vuuut∞X

k=1

x2k +

vuuut∞X

k=1

y2k

1CA

2

<∞,

and hence, in the limit as n→∞, the series of non-negative terms is also bounded:∞X

k=1

(xk + yk)2 <∞, proving that x + y ∈ `2.

(c) (1, 0, 0, 0, . . . ), (1, 12 ,

14 ,

18 , . . . ) and (1, 1

2 ,13 ,

14 , . . . ) are in `2, while

(1, 1, 1, 1, . . . ) and (1, 1√2, 1√

3, 1√

4, . . . ) are not.

(d) True. convergence of∞X

k=1

x2k requires x2

k → 0 as k →∞ and hence xk → 0.

(e) False – see last example in part (b).

(f )∞X

k=1

x2k =

∞X

k=1

α2k is a geometric series which converges if and only if |α | < 1.

(g) Using the integral test,∞X

k=1

x2k =

∞X

k=1

k2α converges if and only if 2α < −1, so α < − 12 .

(h) First, we need to prove that it is well-defined, which we do by proving that the series is

absolutely convergent. If x,y ∈ `2, then, by the Cauchy–Schwarz inequality on Rn for

the vectors ( |x1 |, . . . , |xn | )T , ( | y1 |, . . . , | yn | )

T ,

nX

k=1

|xk yk | ≤vuuut

nX

k=1

x2k

vuuutnX

k=1

y2k ≤

vuuut∞X

k=1

x2k

vuuut∞X

k=1

y2k <∞,

and hence, letting n → ∞, we conclude thatnX

k=1

|xk yk | < ∞. Bilinearity, symmetry

and positivity are now straightforward to verify.

(i)∞X

k=1

|xk yk | ≤vuuut

∞X

k=1

x2k

vuuut∞X

k=1

y2k ,

vuuut∞X

k=1

(xk + yk)2 ≤vuuut

∞X

k=1

x2k +

vuuut∞X

k=1

y2k .

3.3.1. ‖v + w ‖1 = 2 ≤ 2 = 1 + 1 = ‖v ‖1 + ‖w ‖1;‖v + w ‖2 =

√2 ≤ 2 = 1 + 1 = ‖v ‖2 + ‖w ‖2;

‖v + w ‖3 = 3√

2 ≤ 2 = 1 + 1 = ‖v ‖3 + ‖w ‖3;‖v + w ‖∞ = 1 ≤ 2 = 1 + 1 = ‖v ‖∞ + ‖w ‖∞.

3.3.2.(a) ‖v + w ‖1 = 6 ≤ 6 = 3 + 3 = ‖v ‖1 + ‖w ‖1;‖v + w ‖2 = 3

√2 ≤ 2

√5 =√

5 +√

5 = ‖v ‖2 + ‖w ‖2;‖v + w ‖3 = 3

√54 ≤ 2 3

√9 = 3√

9 + 3√

9 = ‖v ‖3 + ‖w ‖3;‖v + w ‖∞ = 3 ≤ 4 = 2 + 2 = ‖v ‖∞ + ‖w ‖∞.

(b) ‖v + w ‖1 = 2 ≤ 4 = 2 + 2 = ‖v ‖1 + ‖w ‖1;‖v + w ‖2 =

√2 ≤ 2

√2 =√

2 +√

2 = ‖v ‖2 + ‖w ‖2;‖v + w ‖3 = 3

√2 ≤ 2 3

√2 = 3√

2 + 3√

2 = ‖v ‖3 + ‖w ‖3;‖v + w ‖∞ = 1 ≤ 2 = 1 + 1 = ‖v ‖∞ + ‖w ‖∞.

(c) ‖v + w ‖1 = 10 ≤ 10 = 4 + 6 = ‖v ‖1 + ‖w ‖1;‖v + w ‖2 =

√34 ≤

√6 +√

14 = ‖v ‖2 + ‖w ‖2;

83

‖v + w ‖3 = 3√

118 ≤ 3√

10 + 3√

36 = ‖v ‖3 + ‖w ‖3;‖v + w ‖∞ = 4 ≤ 5 = 2 + 3 = ‖v ‖∞ + ‖w ‖∞.

3.3.3.(a) ‖u− v ‖1 = 5, ‖u−w ‖1 = 6, ‖v −w ‖1 = 7, so u,v are closest.

(b) ‖u− v ‖2 =√

13, ‖u−w ‖2 =√

12, ‖v −w ‖2 =√

21, so u,w are closest.(c) ‖u− v ‖∞ = 3, ‖u−w ‖∞ = 2, ‖v −w ‖∞ = 4, so u,w are closest.

3.3.4. (a) ‖ f ‖∞ = 23 , ‖ g ‖∞ = 1

4 ; (b) ‖ f + g ‖∞ = 23 ≤ 2

3 + 14 = ‖ f ‖∞ + ‖ g ‖∞.

3.3.5. (a) ‖ f ‖1 = 518 , ‖ g ‖1 = 1

6 ; (b) ‖ f + g ‖1 = 49√

3≤ 5

18 + 16 = ‖ f ‖1 + ‖ g ‖1.

3.3.6. (a) ‖ f − g ‖1 = 12 = .5, ‖ f − h ‖1 = 1− 2

π = .36338, ‖ g − h ‖1 = 12− 1

π = .18169, so g, h

are closest. (b) ‖ f − g ‖2 =q

13 = .57735, ‖ f − h ‖2 =

q32 − 4

π = .47619, ‖ g − h ‖2 =q56 − 2

π = .44352, so g, h are closest. (c) ‖ f − g ‖∞ = 1, ‖ f − h ‖∞ = 1, ‖ g − h ‖∞ = 1,so they are equidistant.

3.3.7.(a) ‖ f + g ‖1 = 3

4 = .75 ≤ 1.3125 ≈ 1 + 516 = ‖ f ‖1 + ‖ g ‖1;

(b) ‖ f + g ‖2 =q

3148 ≈ .8036 ≤ 1.3819 ≈ 1 +

q748 = ‖ f ‖2 + ‖ g ‖2;

(c) ‖ f + g ‖3 =3√

394 ≈ .8478 ≤ 1.4310 ≈ 1 +

3√

418 = ‖ f ‖3 + ‖ g ‖3;

(d) ‖ f + g ‖∞ = 54 = 1.25 ≤ 1.75 = 1 + 3

4 = ‖ f ‖∞ + ‖ g ‖∞.

3.3.8.(a) ‖ f + g ‖1 = e− e−1 ≈ 2.3504 ≤ 2.3504 ≈ (e− 1) + (1− e−1) = ‖ f ‖1 + ‖ g ‖1;(b) ‖ f + g ‖2 =

q12 e

2 + 2− 12 e

−2 ≈ 2.3721

≤ 2.4448 ≈q

12 e

2 − 12 +

q12 − 1

2 e−2 = ‖ f ‖2 + ‖ g ‖2;

(c) ‖ f + g ‖3 = 3

q23 e

3 + 3e− 3e−1 − 23 e

−3 ≈ 2.3945

≤ 2.5346 ≈ 3

q13 e

3 − 13 + 3

q13 − 1

3 e−3 = ‖ f ‖3 + ‖ g ‖3;

(d) ‖ f + g ‖∞ = e+ e−1 ≈ 3.08616 ≤ 3.71828 ≈ e+ 1 = ‖ f ‖∞ + ‖ g ‖∞.

3.3.9. Positivity : since both summands are non-negative, ‖x ‖ ≥ 0. Moreover, ‖x ‖ = 0 if and

only if x = 0 = x− y, and so x = (x, y )T = 0.

Homogeneity : ‖ cx ‖ = | cx |+ 2 | cx− cy | = | c |“|x |+ 2 |x− y |

”= | c | ‖v ‖.

Triangle inequality : ‖x + v ‖ = |x+ v |+ 2 |x+ v − y − w |≤“|x |+ | v |

”+ 2

“|x− y |+ | v − w |

”= ‖x ‖+ ‖v ‖.

3.3.10.(a) Comes from weighted inner product 〈v ,w 〉 = 2v1w1 + 3v2w2.

(b) Comes from inner product 〈v ,w 〉 = 2v1w1 − 12 v1w2 − 1

2 v2w1 + 2v2w2; positivity

follows because 〈v ,v 〉 = 2(v1 − 14 v2)

2 + 158 v

22 .

(c) Clearly positive; ‖ cv ‖ = 2 | cv1 |+ | cv2 | = | c |“

2 | v1 |+ | v2 |”

= | c | ‖v ‖;‖v + w ‖ = 2 | v1 + w1 |+ | v2 + w2 | ≤ 2 | v1 |+ | v2 |+ 2 |w1 |+ |w2 | = ‖v ‖+ ‖w ‖.

(d) Clearly positive; ‖ cv ‖ = maxn

2 | cv1 |, | cv2 |o

= | c | maxn

2 | v1 |, | v2 |o

= | c | ‖v ‖;‖v + w ‖ = max

n2 | v1 + w1 |, | v2 + w2 |

o≤ max

n2 | v1 |+ 2 |w1 |, | v2 |+ |w2 |

o

≤ maxn

2 | v1 |, | v2 |o

+ maxn

2 |w1 |, |w2 |o

= ‖v ‖+ ‖w ‖.(e) Clearly non-negative and equals zero if and only if v1 − v2 = 0 = v1 + v2, so v = 0;

‖ cv ‖ = maxn| cv1 − cv2 |, | cv1 + cv2 |

o= | c | max

n| v1 − v2 |, | v1 + v2 |

o= | c | ‖v ‖;

84

‖v + w ‖ = maxn| v1 + w1 − v2 − w2 |, | v1 + w1 + v2 + w2 |

o

≤ maxn| v1 − v2 |+ |w1 − w2 |, | v1 + v2 |+ |w1 + w2 |

o

≤ maxn| v1 − v2 |, | v1 + v2 |

o+ max

n|w1 − w2 |, |w1 + w2 |

o= ‖v ‖+ ‖w ‖.

(f ) Clearly non-negative and equals zero if and only if v1 − v2 = 0 = v1 + v2, so v = 0;

‖ cv ‖ = | cv1 − cv2 |+ | cv1 + cv2 | = | c |“| v1 − v2 |+ | v1 + v2 |

”= | c | ‖v ‖;

‖v + w ‖ = | v1 + w1 − v2 − w2 |+ | v1 + w1 + v2 + w2 |≤ | v1 − v2 |+ | v1 + v2 |+ |w1 − w2 |+ |w1 + w2 | = ‖v ‖+ ‖w ‖.

3.3.11. Parts (a), (c) and (e) define norms. (b) doesn’t since, for instance, ‖ ( 1,−1, 0 )T ‖ = 0.

(d) doesn’t since, for instance, ‖ ( 1,−1, 1 )T ‖ = 0.

3.3.12. Clearly if v = 0, then w = 0 since only the zero vector has norm 0. If v 6= 0, thenw = cv, and ‖w ‖ = | c | ‖v ‖ = ‖v ‖ if and only if | c | = 1.

3.3.13. True for an inner product norm, but false in general. For example,‖ e1 + e2 ‖1 = 2 = ‖ e1 ‖1 + ‖ e2 ‖1.

3.3.14. If x = ( 1, 0 )T ,y = ( 0, 1 )T , say, then ‖x + y ‖2∞ + ‖x− y ‖2∞ = 1 + 1 = 2 6= 4 =

2“‖x ‖2∞ + ‖y ‖2∞

”, which contradicts the identity in Exercise 3.1.12.

3.3.15. No — neither result satisfies the bilinearity property.

For example, if v = ( 1, 0 )T ,w = ( 1, 1 )T , then

〈 2v ,w 〉 = 14

“‖ 2v + w ‖21 − ‖ 2v −w ‖21

”= 3 6= 2 〈v ,w 〉 = 1

2

“‖v + w ‖21 − ‖v −w ‖21

”= 4,

〈 2v ,w 〉 = 14

“‖ 2v + w ‖2∞ − ‖ 2v −w ‖2∞

”= 2 6= 2 〈v ,w 〉 = 1

2

“‖v + w ‖2∞ − ‖v −w ‖2∞

”= 3

2 .

♦ 3.3.16. Let m = ‖v ‖∞ = max| v1 |, . . . , | v1 |. Then ‖v ‖p = m

0@

nX

i=1

| vi |m

!p1A

1/p

. Now if

| vi | < m then

| vi |m

!p

→ 0 as p→∞. Therefore, ‖v ‖p ∼ mk1/p → m as p→∞, where

1 ≤ k is the number of entries in v with | vi | = m.

♦ 3.3.17.

(a) ‖ f + g ‖1 =Z b

a| f(x) + g(x) | dx ≤

Z b

a

h| f(x) |+ | g(x) |

idx

=Z b

a| f(x) | dx+

Z b

a| g(x) | dx = ‖ f ‖1 + ‖ g ‖1.

(b) ‖ f + g ‖∞ = max | f(x) + g(x) | ≤ max | f(x) |+ | g(x) |≤ max | f(x) |+ max | g(x) | = ‖ f ‖∞ + ‖ g ‖∞.

♦ 3.3.18.(a) Positivity follows since the integrand is non-negative; further,

‖ cf ‖1,w =Z b

a| cf(x) | w(x) dx = | c |