Lecture #4 Notes

0 2 i m 2 0 d with j k j k 0 0 0

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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...
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