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Unformatted text preview: S1 0
S1
0
−1
Let S =
. Then S =
−
0 S2
0 S2 1
and
S −1 A0
0B S= −
S1 1 AS1
0
−1
0
S2 BS2 which is a diagonal matrix, so C is diagonalizable.
Conversely, assume that C is diagonalizable. Then ∃S ∈ Mn+m such that S is invertible and
D = S −1 CS is a diagonal matrix. Write S = [s1 s2 . . . sn+m ] with each sj ∈ Cn+m . Then,
because SD = CS , each sj is an eigenvector for C . Each sj may then be written as
xj
sj =
with xj ∈ Cn and yj ∈ Cm
yj
Then, the block form of C and the fact that Csj = λj sj implies that Axj= λj j and Byj = λj yj .
x
X
If we let X = [x1 x2 . . . xn+m ] and Y = [y1 y2 . . . yn+m ] then S =
. The matrix S is
Y
invertible, so rank (S ) = n + m. By the dimensions of X and Y , we also have rank (X ) ≤ n
and rank (Y ) ≤ m. When looking at row rank, we see that
n + m = rank (S ) ≤ rank (X ) + rank (Y ) ≤ n + m
so rank (X ) + rank (Y ) = n + m. This can only occur if rank (X ) = n and rank (Y ) = m.
Thus X contains n linearly independent columns, each of which is an eigenvector for A, and Y
contains m linearly independent columns, each of which is an eigenvector for B . Thus, we have
bases of eigenvectors of A and B , and we conclude that A and B are diagonalizable.
1.5.32 Theorem. Let A, B ∈ Mn be diagonalizable. Then AB = BA if and only if A and B
are simultaneously diagonalizable.
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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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