594 ig r 7810 does the result of exercise 775 extend

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Unformatted text preview: was sure it was on a firm mathematical foundation. Weierstrass once said that “a mathematician who is not also something of a poet will never be a perfect mathematician.” Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 590 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Jordan Form n×n It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] For every A ∈ C with distinct eigenvalues σ (A) = {λ1 , λ2 , . . . , λs } , there is a nonsingular matrix P such that ⎛ J(λ ) 0 1 ⎜ 0 J(λ2 ) P−1 AP = J = ⎜ . . ⎝. . . . 0 0 ··· ··· .. . ⎞ 0 0 . . . ⎟ ⎟. ⎠ (7.8.4) D E · · · J(λs ) • J has one Jordan segment J(λj ) for each eigenvalue λj ∈ σ (A) . • Each segment J(λj ) is made up of tj = dim N (A − λj I) Jordan blocks J (λj ) as described below. ⎛ J (λ ) 0 · · · 1j ⎜ 0 J2(λj ) · · · J(λj )=⎜ . . .. ⎝. . . . . 0 0 T H ⎛ ⎞ 0 0 . . . ⎜ ⎟ ⎟ with J (λj ) = ⎜ ⎜ ⎠ ⎝ 1 .. . λj IG R · · · Jtj(λj ) ⎞ .. . .. . ⎟ ⎟ ⎟. 1⎠ λj • The largest Jordan block in J(λj ) is kj × kj , where kj = index (λj ). • The number of i × i Jordan blocks in J(λj ) is given by Y P νi (λj ) = ri−1 (λj ) − 2ri (λj ) + ri+1 (λj ) with ri (λj ) = rank (A − λj I)i . • Matrix J in (7.8.4) is called the Jordan form for A. The structure of this form is unique in the sense that the number of Jordan segments in J as well as the number and sizes of the Jordan blocks in each segment is uniquely determined by the entries in A. Furthermore, every matrix similar to A has the same Jordan structure—i.e., A, B ∈ C n×n are similar if and only if A and B have the same Jordan structure. The matrix P is not unique—see p. 594. O C Example 7.8.1 ⎛ 5 ⎜2 ⎜ ⎜0 Problem: Find the Jordan form for A = ⎜ ⎜ −8 ⎝ 0 −8 Copyright c 2000 SIAM 4 3 −1 −8 0 −8 0 1...
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This document was uploaded on 03/06/2014 for the course MA 5623 at City University of Hong Kong.

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