Lecture #4 Notes

0 1 i m 1 0 0 2 i m 2 0 0 0 0

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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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