8.20 HINTS AND ANSWERS
8.5 Use the property of the determinant of a matrix product.
8.7 (d)S=
(
0 −tan(θ/2)
tan(θ/2) 0)
.
(e) Note that (I+K)(I−K)=I−K^2 =(I−K)(I+K).
8.9 (b) 32iA.
8.11 a=bcosγ+ccosβ, and cyclic permutations;a^2 =b^2 +c^2 − 2 bccosα, and cyclic
permutations.
8.13 (a) 2−^1 /^2 (0011)T, 6 −^1 /^2 (2 0 −11)T,
39 −^1 /^2 (− 16 −11)T, 13 −^1 /^2 (212−2)T.
(b) 5−^1 /^2 (1200)T,(345)−^1 /^2 (14 −7100)T,
(18 285)−^1 /^2 (−56 28 98 69)T.
8.15 Cdoes not commute with the others;A,BandDhave (1 −2)Tand (2 1)Tas
common eigenvectors.
8.17 ForA:(1 0−1)T,(1 α 1 1)T,(1 α 2 1)T.
ForB:(111)T,(β 1 γ 1 −β 1 −γ 1 )T,(β 2 γ 2 −β 2 −γ 2 )T.
Theαi,βiandγiare arbitrary.
Simultaneous and orthogonal: (1 0 −1)T,(111)T,(1 −21)T.
8.19 αj=(v·ej∗)/(λj−μ), whereλjis the eigenvalue corresponding toej.
(a) x=(213)T.
(b) Sinceμis equal to one ofA’s eigenvaluesλj, the equation only has a solution
ifv·ej∗= 0; (i) no solution; (ii)x=(1 1 3/2)T.8.21 U= (10)−^1 /^2 (1, 3 i;3i,1), Λ = (1,0; 0,11).
8.23 J=(2y^2 − 4 y+4)/(y^2 + 2), with stationary values aty=±
√
2 and corresponding
eigenvalues 2∓√
- From the trace property ofA, the third eigenvalue equals 2.
8.25 Ellipse;θ=π/4,a=
√
22;θ=3π/4,b=√
10.
8.27 The direction of the eigenvector having the unrepeated eigenvalue is
(1, 1 ,−1)/
√
3.
8.29 (a) A=SA′S†,whereSis the matrix whose columns are the eigenvectors of the
matrixAto be constructed, andA′=diag(λ, μ, ν).
(b)A=(λ+2μ+3ν, 2 λ− 2 μ, λ+2μ− 3 ν;2λ− 2 μ, 4 λ+2μ, 2 λ− 2 μ;
λ+2μ− 3 ν, 2 λ− 2 μ, λ+2μ+3ν).
(c)^13 (1, 5 ,−2; 5, 4 ,5;− 2 , 5 ,1).
8.31 The null space is spanned by (2 0 1 0)Tand (1 −201)T.
8.33 x=3,y=1,z=2.
8.35 First show thatAis singular.η=1,x=1+2z,y=− 3 z;η=2,x=2z,
y=1− 3 z.
8.37 L=(1, 0 ,0;^13 , 1 ,0;^23 , 3 ,1),U=(3, 6 ,9; 0,− 2 ,2; 0, 0 ,4).
(i)x=(−112)T. (ii)x=(−322)T.
8.39 Ais not positive definite, asL 33 is calculated to be
√
−6.
B=LLT, where the non-zero elements ofLare
L 11 =√
5 ,L 31 =
√
3 / 5 ,L 22 =
√
3 ,L 33 =
√
12 /5.
8.41
A†A=
(
21
12
)
,U=
1
√
6
− 1
√
3
√
2
20
√
2
− 1 −
√
3
√
2
,V=√^1
2
(
11
1 − 1
)
.
8.43 The singular values are 12
√
6 , 0 , 18
√
3 and the calculated best solution isx=
1. 71 ,y=− 1. 94 ,z=− 1 .71. The null space is the linex=z, y= 0 and the general
SVD solution isx=1.71 +λ, y=− 1. 94 ,z=− 1 .71 +λ.