Unformatted text preview: s lemma, we deduce that each block Cj is diagonalizable.
Thus, for each j, ∃Tj ∈ Mmj such that each Tj is invertible and Tj−1 Cj Tj is diagonal. −1 T1 0 . . . 0
T1
0 ...
0
.
. −
.
.
.
. 0 T2 0 T2 1
−1
Let T = . . Then T = . ..
..
.
.
. 0
.
.
.
0
0 . . . 0 Tr
0
...
0 Tr−1
and T −1 S −1 BST = T −1 which is a diagonal matrix. Also CT = −
T 1 1 C1 T 1 0 0
.
.
. −
T 2 1 C2 T 2 0 ... T −1 S −1 AST = T −1 DT −1
T 1 λ1 I m 1 T 1
0 −1
0
T 2 λ2 I m 2 T 2 =
.
. .
0
... λ1 I m 1
0
...
0
. .
λ2 I m 2
.
0
= .
..
.
.
.
0
0
...
0 λr I m r
= D. ...
..
0 ...
.. 0 . . 0
.
.
.
0
Tr−1 Cr Tr 0
.
.
.
0
Tr−1 λr Imr Tr Thus, T −1 S −1 AST and T −1 S −1 BST are both diagonal matrices, so A and B are simultaneously
diagonalizable by ST ∈ Mn .
4 We want to make this equivalence more general.
1.5.33 Remark. Let A ∈ Mn and consider polynomials
p(A) = p0 I + p1 A + · · · + pr Ar and
q (A) = q0 I + q1 A + · · · + qs As with r, s ∈ N.
Then p(A)q (A)...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.
 Fall '12
 BernhardBodmann
 Algebra, Matrices

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