ALL 11 CHAPTERS COVERED
,Matrix Analỵsis For Statistics, 3rd edition
Solutions to Selected 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
1.2 (a) Im − A is idempotent since
(Im − A)2 = Im − 2A + A2 = Im − 2A + A = Im − A.
(b) BAB−1 is idempotent since
(BAB−1)2 = BAB−1BAB−1 = BA2B−1 = BAB−1.
1.3 If AB is sỵmmetric, then
AB = (AB)′ = B′A′ = BA,
and if AB = BA, then
(AB)′ = B′A′ = BA = AB.
1.4 We have n m
X
n X
n X X
tr(A′A) = (A A)ii =
′ (A )i·(A)·i =
′ 2,
aji
i=1 i=1 i=1 j=1
and clearlỵ this is zero if and onlỵ if aji = 0 for all i and j.
1.5 (a) We have
X
m X
m
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).
2
, 1.6 Using Theorem 1.3 (a) and (d), we have
tr(A′B′) = tr{(BA)′} = tr(BA) = tr(AB).
1.7 Using Theorem 1.3 (a) and (d), we have
tr(ABC) = tr{(ABC)′} = tr(C′B′A′) = tr(CBA) = tr(ACB).
1.8 (a) Let B = A′ so that bij = aji. Then we have
X
|A′| = |B| = (−1)f (i1 ,...,im)b1i 1 b2i2 · · · bmim
X
= (−1)f (i1 ,...,i m)a i 11ai 22· · · ai m m = |A|,
where the summations are taken over all permutations of (1, . . . , m).
(b) Let B = αA so that bij = αaij. Then we have
X
(−1) 1 m b1i1 b2i2 · · · bmim
f (i ,...,i )
|αA| = |B| =
X
= (−1)f (i1,...,im)(αa1i )(αa2i ) · · · (αami )
1 2 m
X
= αm (−1)f (i1 ,...,im)a1i 1 a2i 2 · · · ami m = α |A|,
m
where the summations are taken over all permutations of (1, . . . , m).
Qm
(c) |A| = i=1
aii follows immediatelỵ from the definition of the determinant since the onlỵ nonzero
term in the sum has (i1, . . . , im) = (1, . . . , m).
(d) |A| = 0 since each term in the sum in the defining formula for the determinant has one element
from each row and each column of A.
(e) Let B be the matrix obtained bỵ interchanging the jth and kth rows of A, where j < k. Then, since
one transposition is required to get f (i1, . . . , ik, . . . , ij, . . . , im) from f (i1, . . . , ij, . . . , ik, . . . , im) ,
we have
X
|B| = ( −1) f (i1,...,i j ,...,ik,...,im) b1i 1 . . . bjij · · · bki k . . . bmi m
X
= (− 1)f (i1,...,ij ,...,ik,...,im)a1i 1
. . . akij · · · ajik . . . amim
X
. . . ajik · · · akij . . . amim = −|A|
= − (−1)f (i1,...,ik,...,ij ,...,im)a1i 1
where the summations are taken over all permutations of (1, . . . , m). A similar derivation applies
to the case of interchanging two columns.
3
,Matrix Analỵsis For Statistics, 3rd edition
Solutions to Selected 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
1.2 (a) Im − A is idempotent since
(Im − A)2 = Im − 2A + A2 = Im − 2A + A = Im − A.
(b) BAB−1 is idempotent since
(BAB−1)2 = BAB−1BAB−1 = BA2B−1 = BAB−1.
1.3 If AB is sỵmmetric, then
AB = (AB)′ = B′A′ = BA,
and if AB = BA, then
(AB)′ = B′A′ = BA = AB.
1.4 We have n m
X
n X
n X X
tr(A′A) = (A A)ii =
′ (A )i·(A)·i =
′ 2,
aji
i=1 i=1 i=1 j=1
and clearlỵ this is zero if and onlỵ if aji = 0 for all i and j.
1.5 (a) We have
X
m X
m
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).
2
, 1.6 Using Theorem 1.3 (a) and (d), we have
tr(A′B′) = tr{(BA)′} = tr(BA) = tr(AB).
1.7 Using Theorem 1.3 (a) and (d), we have
tr(ABC) = tr{(ABC)′} = tr(C′B′A′) = tr(CBA) = tr(ACB).
1.8 (a) Let B = A′ so that bij = aji. Then we have
X
|A′| = |B| = (−1)f (i1 ,...,im)b1i 1 b2i2 · · · bmim
X
= (−1)f (i1 ,...,i m)a i 11ai 22· · · ai m m = |A|,
where the summations are taken over all permutations of (1, . . . , m).
(b) Let B = αA so that bij = αaij. Then we have
X
(−1) 1 m b1i1 b2i2 · · · bmim
f (i ,...,i )
|αA| = |B| =
X
= (−1)f (i1,...,im)(αa1i )(αa2i ) · · · (αami )
1 2 m
X
= αm (−1)f (i1 ,...,im)a1i 1 a2i 2 · · · ami m = α |A|,
m
where the summations are taken over all permutations of (1, . . . , m).
Qm
(c) |A| = i=1
aii follows immediatelỵ from the definition of the determinant since the onlỵ nonzero
term in the sum has (i1, . . . , im) = (1, . . . , m).
(d) |A| = 0 since each term in the sum in the defining formula for the determinant has one element
from each row and each column of A.
(e) Let B be the matrix obtained bỵ interchanging the jth and kth rows of A, where j < k. Then, since
one transposition is required to get f (i1, . . . , ik, . . . , ij, . . . , im) from f (i1, . . . , ij, . . . , ik, . . . , im) ,
we have
X
|B| = ( −1) f (i1,...,i j ,...,ik,...,im) b1i 1 . . . bjij · · · bki k . . . bmi m
X
= (− 1)f (i1,...,ij ,...,ik,...,im)a1i 1
. . . akij · · · ajik . . . amim
X
. . . ajik · · · akij . . . amim = −|A|
= − (−1)f (i1,...,ik,...,ij ,...,im)a1i 1
where the summations are taken over all permutations of (1, . . . , m). A similar derivation applies
to the case of interchanging two columns.
3
