4.3: 11, 12, 14
11. Show that if A and B are similar matrices then det(A) = det(B).
Proof. Suppose A and B are similar matrices. Then B = S −1 AS for some
nonsingular matrix S. Since S is nonsingular, we know that det(S) 6= 0 and
1
det(S −1 ) = det(S)
. Thus
det(B)
= det(S −1 AS)
= det(S −1 ) det(A) det(S)
1
=
det(A) det(S)
det(S)
1
=
det(S) det(A)
det(S)
= det(A),
and we have shown that det(A) = det(B).
12. Let A and B be similar matrices. Show that
(a) AT and B T are similar.
(b) Ak and B k are similar for each positive integer k.
Proof. Since A and B are similar, there is some nonsingular matrix S such that
B = S −1 AS.
(a) Recall if S is invertible then S T is also invertible, and (S T )−1 = (S −1 )T
(see Exercise 1.3#17). Thus
B T = (S −1 AS)T = S T AT (S −1 )T = S T AT (S T )−1
gives us that B T is similar to AT .
(b) Note that
k times
z
}|
{
B k = (S −1 AS)k = (S −1 AS)(S −1 AS) · · · (S −1 AS) = S −1 Ak S.
Since S is invertible, B k is similar to Ak .
14. Let A and B be similar matrices and let λ be any scalar. Show that
(a) A − λI and B − λI are similar.
(b) det(A − λI) = det(B − λI).
Proof. Since A and B are similar, there is some nonsingular matrix S such that
B = S −1 AS.
(a) Note that S −1 IS = S −1 S = I, so
B−λI = S −1 AS−λS −1 IS = S −1 AS−S −1 λIS = S −1 (AS−λIS) = S −1 (A−λI)S
and thus we see that B − λI is similar to A − λI.
(b) From part (a), we have that A − λI and B − λI are similar. By Exercise
11, we know therefore that det(A − λI) = det(B − λI).
1