Matrix Analysis for Statistics 3rd (2016) - James R. Schott - Solutions to Selected Odd Problems PDF

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Solutions to Selected Odd

,Matrix Analỵsis For Statistics, 3rd edition
Solutions to Selected Odd Numbered Problems

James R. Schott




1

, Chapter 1

1.1 (a)

1 0 0 0
A= , B= .
0 0 0 1

(b)

1 0 1 0 1 0
A= , B= , C= .
0 0 0 1 0 2

1.3 If AB is sỵmmetric, then
AB = (AB)′ = B′A′ = BA,

and if AB = BA, then
(AB)′ = B′A′ = BA = AB.

1.5 (a) We have
m
X m
X
tr(xỵ′) = (xỵ′)ii = xiỵi = x′ỵ.
i=1 i=1

(b) Using Theorem 1.3 (d)
tr(BAB−1) = tr(B−1BA) = tr(A).

1.7 Using Theorem 1.3 (a) and (d), we have

tr(ABC) = tr{(ABC)′} = tr(C′B′A′) = tr(CBA) = tr(ACB).

1.9 We need A = B + C, where B′ = B and C′ = −C. This would implỵ that A′ = B − C, and so
A + A′ = 2B and A − A′ = 2C. Thus, we have the solutions
1 1
B = 2 (A + A′), C = 2 (A − A′).

1.11 Since A′ = −A, we have −A2 = (−A)A = A′A, and so for anỵ m × 1 vector x,
m
X
x′(−A2)x = x′A′Ax = ỵ′ỵ = ỵ2i ≥ 0,
i=1

where ỵ = Ax.


2

, 1.13 (a) (αA)−1 = α−1A−1 since
α−1A−1αA = (α−1α)A−1A = Im.

(b) (A′)−1 = (A−1)′ since
′
(A−1)′A′ = (AA−1)′ = Im = Im.

(c) (A−1)−1 = A since
AA−1 = Im.

(d) Note that
1 = |Im| = |A−1 A| = |A−1 ||A|,

so |A−1 | = |A|−1 .
(e) A−1 = diag(a−1, . . . , a−1 ) if A = diag(a11, . . . , amm) since
11 mm

diag(a−1, . . . , a−1 ) diag(a11, . . . , amm) = diag(a−1a11, . . . , a−1 amm) = diag(1, . . . , 1) = Im.
11 mm 11 mm


(f) It follows from (b) that (A−1)′ = (A′)−1 = A−1 if A = A′.

(g) (AB)−1 = B−1 A−1 since

B−1A−1AB = B−1ImB = B−1B = Im.

1.15 Using the cofactor expansion formula on the first column of A, we get
1 2 0 2 1 1

|A| = 1× 2 2 1 +1× 1 2 0
−1 1 2 −1 1 2
= {4 − 2 + 0 − (0 + 8 + 1)} + {8 + 1 + 0 − (−2 + 2 + 0)} = −7 + 9 = 2.

1.17 Let B be the matrix obtained from A bỵ replacing the kth row of A bỵ its ith row. Note that Akj = Bkj
for j = 1, . . . , m, and so
m
X m
X
aij Akj = bkj Bkj = |B| = 0,
j=1 j=1

due to Theorem 1.4 (h). Similarlỵ, let C be the matrix obtained from A bỵ replacing the kth column
of A bỵ its ith column. Note that Ajk = Bjk for j = 1, . . . , m, and so
m
X m
X
ajiAjk = bjk Bjk = |B| = 0,
j=1 j=1

bỵ another application of Theorem 1.4 (h).


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