Lecture #4 Notes

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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)...
View Full Document

## 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.

Ask a homework question - tutors are online