Lecture #4 Notes

# Then because sd cs each sj is an eigenvector for c

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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. Pr...
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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.

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