This preview shows page 1. Sign up to view the full content.
Unformatted text preview: oof. We have already shown that if A and B are simultaneously diagonalizable then AB = BA.
All that remains to show is the converse. Assume that AB = BA. Because A is diagonalizable,
∃S ∈ Mn such that S is invertible and D = S −1 AS is diagonal. We may multiply S by an
invertible matrix to permute elements on the diagonal of D, so we may assume without loss of
generality that λ1 I m 1
0
...
0
. .
λ2 I m 2
.
0
D= . with λj = λk ∀j = k
..
.
.
.
0
0
...
0 λr I m r
3 where Im is the m × m identity and mj is the multiplicity of λj . Because AB = BA, we have
(S −1 AS )(S −1 BS ) = S −1 ABS = S −1 BAS = (S −1 BS )(S −1 AS ). If we let C = (S −1 BS ), then the above gives that DC = CD. If we denote C = [ci,j ]n =1
i,j
and D = [di,j ]n =1 , then by the diagonal structure of D we have di,i ci,j = ci,j dj,j . Then
i,j
(di,i − dj,j )ci,j = 0 implies that if di,i = dj,j then ci,j = 0. By the block structure of D, this
implies that C is block diagonal with blocks of the same size. That is C1 0 . . . 0
. .
. 0 C2
C= . with Cj ∈ Mmj ∀j
..
.
. 0
.
0 . . . 0 Cr
Because B is diagonalizable, ∃R ∈ Mn such that R is invertible and R−1 BR is diagonal. Then
R−1 SCS −1 R = R−1 SS −1 BSS −1 R = R−1 BR so C is diagonalizable. From the previou...
View Full
Document
 Fall '12
 BernhardBodmann
 Algebra, Matrices

Click to edit the document details