Hints and answers

MATRICES AND VECTOR SPACES
8.40
Find the equation satisfied by the squares of the singular values of the matrix
associated with the following over-determined set of equations:
2x + 3y + z
x−y−z
2x + y
2y + z
8.41
8.42
8.43
=0
=1
=0
= −2.
Show that one of the singular values is close to zero. Determine the two larger
singular values by an appropriate iteration process and the smallest one by
indirect calculation.
Find the SVD of


0
−1
1 ,
A= 1
−1
0
√
showing that the singular values are 3 and 1.
Find the SVD form of the matrix


22 28 −22
−2 −19
 1
.
A=
19 −2 −1 
−6 12
6
Use it to determine the best solution x of the equation Ax = b when (i) b =
(6 − 39 15 18)T , (ii) b = (9 − 42 15 15)T , showing√that (i) has an exact
solution, but that the best solution to (ii) has a residual of 18.
Four experimental measurements of particular combinations of three physical
variables, x, y and z, gave the following inconsistent results:
13x + 22y − 13z
10x − 8y − 10z
10x − 8y − 10z
9x − 18y − 9z
= 4,
= 44,
= 47,
= 72.
Find the SVD best values for x, y and z. Identify the null space of A and hence
obtain the general SVD solution.
8.20 Hints and answers
8.1
(a) False. ON , the N × N null matrix,
is not
non-singular.
1 0
0 0
(b) False. Consider the sum of
and
.
0 0
0 1
(c) True.
(d) True.
2
(e) False. Consider bn = an + an for which N
n=0 |bn | = 4 = 1, or note that there
is no zero vector with unit norm.
(f) True.
(g) False. Consider the two series defined by
a0 = 12 ,
8.3
an = 2(− 21 )n
for
n ≥ 1;
bn = −(− 12 )n
for n ≥ 0.
The series that is the sum of {an } and {bn } does not have alternating signs
and so closure does not hold.
(a) x = a, b or c; (b) x = −1; the equation is linear in x.
314
8.20 HINTS AND ANSWERS
8.5
8.7
8.9
8.11
8.13
8.15
8.17
8.19
Use the property of the determinant
of a matrix product.
0
− tan(θ/2)
.
tan(θ/2)
0
2
(e) Note that (I + K)(I − K) = I − K = (I − K)(I + K).
(b) 32iA.
a = b cos γ + c cos β, and cyclic permutations; a2 = b2 + c2 − 2bc cos α, and cyclic
permutations.
(a) 2−1/2 (0 0 1 1)T , 6−1/2 (2 0 − 1 1)T ,
39−1/2 (−1 6 − 1 1)T , 13−1/2 (2 1 2 − 2)T .
(b) 5−1/2 (1 2 0 0)T , (345)−1/2 (14 − 7 10 0)T ,
(18 285)−1/2 (−56 28 98 69)T .
C does not commute with the others; A, B and D have (1 − 2)T and (2 1)T as
common eigenvectors.
For A : (1 0 − 1)T , (1 α1 1)T , (1 α2 1)T .
For B : (1 1 1)T , (β1 γ1 − β1 − γ1 )T , (β2 γ2 − β2 − γ2 )T .
The αi , βi and γi are arbitrary.
Simultaneous and orthogonal: (1 0 − 1)T , (1 1 1)T , (1 − 2 1)T .
αj = (v · ej∗ )/(λj − µ), where λj is the eigenvalue corresponding to ej .
(d) S =
(a) x = (2 1 3)T .
(b) Since µ is equal to one of A’s eigenvalues λj , the equation only has a solution
if v · ej∗ = 0; (i) no solution; (ii) x = (1 1 3/2)T .
8.21
8.23
8.25
8.27
8.29
8.31
8.33
8.35
8.37
8.39
8.41
U = (10)−1/2 (1, 3i; 3i, 1), Λ = (1, 0; 0, 11).
√
2
+ 2), with stationary values at y = ± 2 and corresponding
J = (2y 2 − 4y + 4)/(y
√
eigenvalues 2 ∓ 2. From
√ of A, the third eigenvalue equals 2.
√ the trace property
Ellipse; θ = π/4, a = 22; θ = 3π/4, b = 10.
The direction
of the eigenvector having the unrepeated eigenvalue is
√
(1, 1, −1)/ 3.
(a) A = SA S† , where S is the matrix whose columns are the eigenvectors of the
matrix A to be constructed, and A = diag (λ, µ, ν).
(b) A = (λ + 2µ + 3ν, 2λ − 2µ, λ + 2µ − 3ν; 2λ − 2µ, 4λ + 2µ, 2λ − 2µ;
λ + 2µ − 3ν, 2λ − 2µ, λ + 2µ + 3ν).
(c) 13 (1, 5, −2; 5, 4, 5; −2, 5, 1).
The null space is spanned by (2 0 1 0)T and (1 − 2 0 1)T .
x = 3, y = 1, z = 2.
First show that A is singular. η = 1, x = 1 + 2z, y = −3z; η = 2, x = 2z,
y = 1 − 3z.
L = (1, 0, 0; 13 , 1, 0; 23 , 3, 1), U = (3, 6, 9; 0, −2, 2; 0, 0, 4).
(i) x = (−1 1 2)T . (ii) x = (−3 2 2)T .
√
A is not positive definite, as L33 is calculated to be −6.
B = LL√T , where the
of L are
non-zero elements
√
L11 = 5, L31 = 3/5, L22 = 3, L33 = 12/5.
√ 
√

3
2
−1
√
1
1
1
1
.
, U= √  2
A† A =
0
2 , V = √
√
√
1
−1
6
2
−1 − 3
2
√
√
The singular values are 12 6, 0, 18 3 and the calculated best solution is x =
1.71, y = −1.94, z = −1.71. The null space is the line x = z, y = 0 and the general
SVD solution is x = 1.71 + λ, y = −1.94, z = −1.71 + λ.
8.43
2
1
1
2
315