Mathematics Chapter 7 And represent either a reﬂection through the origin

with 0

@

3

23

2

1

21

A= 1 2

0 1 !0

@

1

21

2

1

21

A, 2

2!= 1 2

0 1 ! 1

0!+ 1

2!;

(d)T6◦T3[x] = 0

@

1

23

2

1

23

21

Ax+0

@

5

2

3

21

A,

with 0

@

1

23

2

1

23

21

A=0

@

1

21

2

1

21

A 1 2

0 1 !,0

@

5

2

3

21

A=0

@

1

21

2

1

21

A 1

2!+ 4

−3!;

7.3.3. (a) True. (b) True. (c) False: in general, squares are mapped to parallelograms. (d) False:

in general circles are mapped to ellipses. (e) True.

7.3.4. The triangle with vertices (−1,−6),(7,−2),(1,6).

7.3.5.

7.3.6.

(a)F[x] = Ax+bhas an inverse if and only if Ain nonsingular.

(b) Yes: F−1[x] = A−1x−A−1b.

7.3.8. It can be regarded as a subspace of the vector space of all functions from Rnto Rnand

so one only needs to prove closure. If F[x] = Ax+band G[x] = Cx+d, then (F+

7.3.10. (a), (b), (e) are isometries

7.3.11. (a) True. (b) True if nis even; false if nis odd, since det(−In) = (−1)n.

7.3.12. Write y=F[x] = Q(x−a) + awhere Q= 0−1

1 0 !represents a rotation through

7.3.14.

(a)F[x] = 0

B

@

1

√2−1

√2

1

√2

1

√2

1

C

Ax;G[x] = x+ 1

0!;

F◦G[x] = 0

B

@

1

√2−1

√2

1

√2

1

√2

1

C

Ax+0

B

@

1

√2

1

√2

1

C

A=0

B

@

1

√2−1

√2

1

√2

1

√2

1

C

A2

4x−0

@−1

2

√2+1

21

A3

5+0

@−1

2

√2+1

21

A

is counterclockwise rotation around the point 0

@

1

2

√2+1

21

Aby 45◦.

(b)F[x] = 0

@

√3

2−1

2

1

2

√3

2

1

A“x− 1

1!#+ 1

1!=0

@

√3

2−1

2

1

2

√3

2

1

Ax+0

@

3−√3

2

1−√3

2

1

A;

G[x] = 0−1

1 0 !“x− −2

1!#+ −2

1!= 0−1

1 0 !x+ −1

3!;

192

G◦F[x] = 0

@−1

2−√3

2

√3

2−1

2

1

Ax+0

@−3+√3

2

9−√3

2

1

A=0

@−1

2−√3

2

√3

2−1

2

1

A2

4x−0

@−1−√3

2

5−√3

2

1

A3

5+0

@−1−√3

2

5−√3

2

1

A

is counterclockwise rotation around the point 0

@−1−√3

2

5−√3

2

1

Aby 120◦.

(c)F[x] = 0 1

1 0 !x+ −1

1!;

♥7.3.15.

(a) If F[x] = Qx+aand G[x] = Rx+b, then G◦F[x] = RQ x+ (Ra+b) = Sx+cis

a isometry since S=QR, the product of two orthogonal matrices, is also an orthogonal

matrix.

(b)F[x] = x+aand G[x] = x+b, then G◦F[x] = x+ (a+b) = x+c.

♦7.3.16. (a) 1 0

0−1!x+ 2

0!= x+ 2

−y!, (b) 0 1

1 0 !x+0

B

@

3

√2

3

√2

1

C

A=0

B

@

y+3

√2

x+3

√2

1

C

A,

(c) 0−1

−1 0 ! x− 1

0!!+ 1

0!+ √2

−√2!= −y+ 1 + √2

−x+ 1 −√2!.

♦7.3.17.

(a)F[x] = R(x−a) + awhere R= 2u uT−I is the elementary reﬂection matrix corre-

sponding to the line in the direction of uthrough the origin.

(b)G[x] = R(x−a) + a+du, where R= 2u uT−I is the same reﬂection matrix.

193

xi=ai−a1∈Wand b=a1, and so all aibelong to the aﬃne subspace A.

(b) Let vi=ai−a0,wi=bi−b0, for i= 1, . . . , n. Then, by the assumption, vi·vi=

kvik2=kai−a0k2=kbi−b0k2=kwik2=wi·wifor all i= 1, . . . , n, while

♦7.3.19. First, kv+wk2=kvk2+ 2 hv,wi+kwk2,

kL[v+w]k2=kL[v] + L[w]k2=kL[v]k2+ 2 hL[v], L[w]i+kL[w]k2.

If Lis an isometry, kL[v+w]k=kv+wk,kL[v]k=kvk,kL[w]k=kwk.Thus,

equating the previous two formulas, we conclude that hL[v], L[w]i=hv,wi.

♦7.3.20. First, if Lis an isometry, and kuk= 1 then kL[u]k= 1, proving that L[u]∈S1.

Conversely, if Lpreserves the unit sphere, and 06=v∈V, then u=v/r ∈S1where

r=kvk, so kL[v]k=kL[rv]k=kr L[u]k=rkL[u]k=r=kvk, proving (7.37).

7.3.21.

(a) All aﬃne transformations F[x] = Qx+bwhere bis arbitrary and Qis a symmetry

of the unit square, and so a rotation by 0,90,180 or 270 degrees, or a reﬂection in the x

7.3.22. Same answer as previous exercise. Now the transformation must preserve the unit dia-

mond/octahedron, which has the same (linear) symmetries as the unit square/cube.

♥7.3.23.

(a)q(Hx) = (xcosh α+ysinh α)2−(xsinh α+ycosh α)2

7.4.1.

(a)L(x) = 3x; domain R; target R; right hand side −5; inhomogeneous.

194

(b)L(x, y, z) = x−y−z; domain R3; target R; right hand side 0; homogeneous.

(c)L(u, v, w) = u−2v

v−w!; domain R3; target R2; right hand side −3

−1!; inhomogeneous.

(d)L(p, q) = 3p−2q

(i)L[u] = 0

B

@

u′′(x) + x2u(x)

u(0)

u′(0)

1

C

A; domain C2(R); target C0(R)×R2; right hand side 0

B

@

3x

1

01

C

A;

inhomogeneous.

(j)L[u, v ] = u′(x)−v(x)

−2u(x) + v′(x)!; domain C1(R)×C1(R); target C0(R)×C0(R); right

hand side 0

0!; homogeneous.

(o)L[u, v ] = Z1

0u(y)dy −Z1

0y v(y)dy; domain C0(R)×C0(R); target R; right hand side

0; homogeneous.

(p)L[u] = ∂u

∂t + 2 ∂u

∂x ; domain C1(R2); target C0(R2); right hand side the constant

function 1; inhomogeneous.

195

7.4.4.

(a) Since ais constant, by the Fundamental Theorem of Calculus,

du

dt =d

dt a+Zt

0k(s)u(s)ds !=k(t)u(t). Moreover, u(0) = a+Z0

0k(s)u(s)ds =a.

(b) (i)u(t) = 2e−t, (ii )u(t) = et2−1, (iii )u(t) = 3eet−1.

7.4.5. True, since the equation can be written as L[x] + b=c, or L[x] = c−b.

7.4.6.

(a)u(x) = c1e2x+c2e−2x, dim = 2;

7.4.7.

7.4.8. (a) If y∈C2[a, b], then y′∈C1[a, b], y′′ ∈C0[a, b] and so L[y] = 3 y′′ −2y′−5y∈

C0[a, b]. Further, L[cy +dz ] = 3(cy +dz)′′−2(cy +dz)−5(cy +dz) = c(3 y′′ −2y′−5y)+

d(3z′′ −2z′−5z) = cL[y]+d L[z]. (b) ker Lis the span of the basic solutions e−x, e5x/3.

7.4.9.

(a)p(D) = D3+ 5D2+ 3D−9.

(b)ex, e−3x, xe−3x. The general solution is y(x) = c1ex+c2e−3x+c2xe−3x.

7.4.10. (a) Minimal order 2: u′′ +u′−6u= 0. (b) minimal order 2: u′′ +u′= 0.

(c) minimal order 2: u′′ −2u′+u= 0. (d) minimal order 3: u′′ −6u′′ + 11u′−6u= 0.

196

7.4.13.

(i) Using the chain rule, dv

dt =etdu

dx =xdu

dx ,d2v

dt2=e2td2u

dx2+etdu

dx =x2d2u

dx2+xdu

dx ,

and so v(t) solves ad2v

dt2+ (b−a)dv

dt +cv = 0.

(ii ) In all cases, u(x) = v(log x) gives the solutions in Exercise 7.4.11.

♦7.4.14.

(a)v(t) = c1er t +c2t er t, so u(x) = c1|x|r+c2|x|rlog |x|.

(b) (i)u(x) = c1x+c2xlog |x|, (ii )u(x) = c1+c2log |x|.

7.4.17. For u= log(x2+y2), we compute ∂2u

∂x2=2y2−2x2

(x2+y2)2=−∂2u

∂y2. Similarly, when v=

x

x2+y2, then ∂2v

∂x2=6xy2−2x3

(x2+y2)3=−∂2v

∂y2. Or simply notice that v=1

2

∂u

∂x , and so if

∂2u

∂x2+∂2u

∂x2= 0, then ∂2v

∂x2+∂2v

∂y2=1

2

∂

∂x 0

@∂2u

∂x2+∂2u

∂y21

A= 0.

7.4.18. u=c1+c2log r. The solutions form a two-dimensional vector space.

7.4.21.

(a) Basis: 1, x, y, z, x2−y2, x2−z2, xy, xz, y z; dimension = 9.

197

7.4.23.

(a) If x∈ker M, then L◦M[x] = L[M[x] ] = L[0] = 0and so x∈ker L.

(b) For example, if L= O, but M6= O then ker(L◦M) = {0} 6= ker M.

Other examples: L=M= 0 1

0 0 !, and L=M=D, the derivative function.

7.4.24. (a) Not in the range. (b)x=0

B

@

0

1

C

A+z0

B

@

−7

5

−6

5

1

C

A. (c)x=0

B

@

3

2

−3

4

1

C

A; kernel element is 0.

7.4.25. (a)x= 1, y =−3, unique; (b)x=−1

7+3

7z, y =4

7+2

7z, not unique; (c) no solution;

(d)u= 2, v =−1, w = 0, unique; (e)x= 2 + 4 w, y =−2w, z =−1−6w, not unique.

7.4.26.

7.4.27.

(a)u(x) = 1

4ex−1

4e4−3x,

7.4.28. (a) Unique solution: u(x) = x−πsin √2x

sin √2π; (b) no solution; (c) unique solution:

7.4.29. (a)u(x) = 1

2xlog x+c1x+c2

x, (b)u(x) = 1

2log x+3

4+c1x+c2x2,

(c)u(x) = 1 −3

8x+c1x5+c2

198

♦7.4.31.

(a) First, Y⊂rng Lsince every y∈Ycan be written as y=L[w] for some w∈W⊂U,

and so y∈rng L. If y1=L[w1] and y2=L[w2] are elements of Y, then so is cy1+

dy2=L[cw1+dw2] for any scalars c, d since cw1+dw2∈W, proving that Yis a

subspace.

(b) Suppose w1,…,wkform a basis for W, so dim W=k. Let y=L[w]∈Yfor w∈W.

We can write w=c1w1+···+ckwk, and so, by linearity, y=c1L[w1] + ··· +

ckL[wk]. Therefore, the kvectors y1=L[w1], . . . , yk=L[wk] span Y, and hence, by

Proposition 2.33, dim Y≤k.

♦7.4.34.

(a) If Lwere invertible, then the solution to L[x] = bwould be unique, namely x=L−1[y].

But according to Theorem 7.38, we can add in any element of the kernel to get another

solution, which would violate uniqueness.

7.4.35.

(a)u(x) = 1

2+2

5cos x+1

5sin x+c e−2x,

7.4.36. (a)u(x) = 5x+ 5 −7ex−1, (b)u(x) = c1(x+ 1) + c2ex.

7.4.37. u(x) = −7 cos √x−3 sin √x.

7.4.38. u′′ +xu = 2, u(0) = a, u(1) = b, for any a, b.

7.4.40. u(x, y) = 1

4(x2+y2) + 1

12 (x4+y4).

ordinary diﬀerential equation for w.

(b) (i)u(x) = c1ex+c2xex,(ii )u(x) = c1(x−1) + c2e−x,(iii )u(x) = c1e−x2

+c2xe−x2

,

(iv )u(x) = c1ex2/2+c2ex2/2Ze−x2

dx.

♦7.4.42. We use linearity to compute

L[u⋆] = L[c1u⋆

1+···+cku⋆

k] = c1L[u⋆

1] + ···+ckL[u⋆

k] = c1f1+··· +ckfk,

and hence u⋆is a particular solution to the diﬀerential equation (7.66). The second part of

the theorem then follows from Theorem 7.38.

7.4.43. An example: Let A= 1 2

i 2 i !. Then ker Aconsist of all vectors c −2

1!where c=

7.4.44.

(a)u(x) = c1cos 2x+c2sin 2x,

(b)u(x) = c1e−3xcos x+c2e−3xsin x,

7.4.45.

(a) Minimal order 2: u′′ + 2u′+ 10u= 0;

7.4.46.

(a)u(x) = c e ix=ccos x+ i csin x.

(b)u(x) = c1ex+c2e( i −1) x=“c1ex+c2e−xcos x”+ i e−xsin x,

(c)u(x) = c1e(1+ i ) x/√2+c2e−(1+ i ) x/√2

=hc1ex/√2cos x

√2+c2e−x/√2cos x

√2)i+ i hc1ex/√2sin x

√2−c2e−x/√2sin x

√2)i.

200

7.4.49. u(t, x) = ek2t+k x,u(t, x) = e−k2t+k x, where kis any real or complex number. When

k=a+ i bis complex, we obtain the four independent real solutions

e(a2−b2)t+a x cos(bx + 2abt), e(a2−b2)t+a x sin(bx + 2abt),

e(b2−a2)t+a x cos(bx −2abt), e(b2−a2)t+a x sin(bx −2abt).

7.4.50. (a), (c), (e) are conjugated. Note: Case (e) is all of C3.

♦7.4.51. If v= Re u=u+u

2, then v=u+u

2=u+u

2=v. Similarly, if w= Im u=u−u

2 i ,

then w=u−u

−2 i =u−u

2 i =w.

7.4.55. (a)L[u] = L[u] = 0. (b)u=““−3

2+1

2i”y+“−1

2−1

2i”z, y, z ”Twhere y, z ∈C

are the free variables, is the general solution to the first system, and so its complex conjugate

u=““−3

2−1

2i”y+“−1

2+1

2i”z, y, z ”T, where y, z ∈Care free variables, solves the conju-

gate system. (Note: Since y, z are free, they could be renamed y, z if desired.)

♦7.4.56. They are linearly independent if and only if uis not a complex scalar multiple of a real

solution. Indeed, if u= (a+ i b)vwhere vis a real solution, then x=av,y=bvare

201

75

71

715

71

7.5.3. (a)0

B

@

1−1 0

1 0 −1

0 1 2 1

C

A, (b)0

B

@

1−2 0

1

20−3

2

02

32

1

C

A, (c)0

B

@

01

43

2

1−3

2−4

011

49

2

1

C

A.

7.5.4. Domain (a), target (b): 0

B

@

1−2 0

1 0 −3

0 2 6 1

C

A; domain (a), target (c): 0

B

@

1−1−1

2 0 −2

1 4 5 1

C

A;

1−1 0

1

1−1−1

1

7.5.5. Domain (a), target (a): 1 0 −1

3 2 1 !; domain (a), target (b): 1 0 −3

3 4 3 !.

domain (a), target (c): 2 0 −2

8 8 4 !; domain (b), target (a): 0

@

1

20−1

2

12

31

A.

domain (b), target (b): 0

@

1

20−3

2

14

311

A; domain (b), target (c): 1 0 −1

8

38

34

3!.

domain (c), target (a): 0

@12

7−3

7

14

71

A; domain (c), target (b): 0

@14

7−9

7

18

73

71

A.

domain (c), target (c): 0

@

16

78

7−4

7

18

716

76

71

A.

7.5.6. Using the monomial basis 1, x, x2, the operator Dhas matrix representative

202

7.5.9. In all cases, L=L∗if and only if its matrix representative A, with respect to the stan-

dard basis, is symmetric. (a)A= −1 0

0−1!=AT, (b)A= 0 1

1 0 !=AT,

(c)A= 3 0

0 3 !=AT, (d)A=0

@

1

21

2

1

21

A=AT.

2a21, a13 =1

3a31,1

2a23 =1

3a32, (b)0

3 3 2 1

7.5.13.

(a) 2a12−a22 =−a11+2a21−a31,2a13−a23 =−a21+2 a31,−a13+2a23−a33 =−a22+2 a32,

(b)0

B

@

0 1 1

2 3 −6

5 3 −16 1

C

A.

7.5.14. True. hI [u],vi=hu,vi=hu,I [v]ifor all u,v∈U, and so, according to (7.74),

I∗= I .

♦7.5.17.

(a) Write the condition as hN[u],ui= 0 where N=K−Mis also self-adjoint. Then,

for any u,v∈U, we have 0 = hN[u+v],u+vi=hN[u],ui+hN[u],vi+

hN[v],ui+hN[v],vi= 2 hu, N[v]i,where we used the self-adjointness of Nto

203

7.5.18. (a) =⇒(b): Suppose hL[u], L[v]i=hu,vifor all u,v∈U, then

kL[u]k=qhL[u], L[u]i=qhu,ui=kuk.

(b) =⇒(c): Suppose kL[u]k=kukfor all u∈U. Then

7.5.19.

(a)hMa[u], v i=Zb

aMa[u(x)] v(x)dx =Zb

aa(x)u(x)v(x)dx =Zb

au(x)Ma[v(x)] dx =

hu , Ma[v]i, proving self-adjointness.

(b) Yes, by the same computation, hhMa[u], v ii =Zb

aa(x)u(x)v(x)w(x)dx =hhu , Ma[v]ii.

♥7.5.20.

(a) If AT=−A, then (Au)·v= (Au)Tv=uTATv=−uTAv=−u·Avfor all u,v∈Rn,

♦7.5.21. Define L:U→V1×V2by L[u] = (L1[u], L2[u]). Using the induced inner product

hhh(v1,v2),(w1,w2)iii =hhv1,w1ii1+hhv2,w2ii2on the Cartesian product V1×V2given

in Exercise 3.1.18, we find

1◦L1[u] + L∗

2◦L2[u] = K1[u] + K2[u] = K[u].

7.5.22. Minimizer: “1

5,−1

5”T; minimum value: −1

5.

7.5.23. Minimizer: “14

13 ,2

13 ,−3

13 ”T; minimum value: −31

26 .

7.5.24. Minimizer: “2

3,1

3”T; minimum value: −2.

7.5.25.

204

7.5.27. (a)1

3, (b)6

11 , (c)3

5.

♦7.5.28. Suppose L:U→Vis a linear map between inner product spaces with ker L6={0}and

adjoint map L∗:V→U. Let K=L∗◦L:U→Ube the associated positive semi-definite

operator. If f∈rng K, then any solution to the linear system K[u⋆] = fis a minimizer for

205