That is ig r y p 0 0 jtjj j1 0

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lues of A. For example, if the Jordan form for A is ⎛ O C ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ J=⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 4 ⎞ 10 41 4 4 0 ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 1 4 3 0 1 3 2 2 then we know that A9×9 has three distinct eigenvalues, namely σ (A) = {4, 3, 2}; alg mult (4) = 5, alg mult (3) = 2, and alg mult (2) = 2; geo mult (4) = 2, geo mult (3) = 1, and geo mult (2) = 2; Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.8 Jordan Form http://www.amazon.com/exec/obidos/ASIN/0898714540 593 index (4) = 3, index (3) = 2, and index (2) = 1; λ = 2 is a semisimple eigenvalue, so, while A is not diagonalizable, part of it is; i.e., the restriction A/ is a diagonalizable linear operator. N (A−2I) It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu Of course, if both P and J are known, then A can be completely reconstructed from (7.8.4), but the point being made here is that only J is needed to reveal the eigenstructure along with the other similarity invariants of A. Now that the structure of the Jordan form J is known, the structure of the similarity transformation P such that P−1 AP = J is easily revealed. Focus on a single p × p Jordan block J (λ) contained in the Jordan segment J(λ) associated with an eigenvalue λ, and let P = [ x1 x2 · · · xp ] be the portion of P = [ · · · | P | · · ·] that corresponds to the position of J (λ) in J. Notice that AP = PJ implies AP = P J (λ) or, equivalently, ⎛ ⎜ ⎜ A[ x1 x2 · · · xp ] = [ x1 x2 · · · xp ] ⎜ ⎝ λ D E T H 1 .. . ⎞ . .. IG R .. . ⎟ ⎟ ⎟, 1⎠ λ p×p so equating columns on both sides of this equation produces Ax1 = λx1 =⇒ x1 is an eigenvector =⇒ (A − λI) x1 = 0, Ax2 = x1 + λx2 =⇒ (A − λI) x2 = x1 =⇒ (A − λI) x2 = 0, =⇒ (A − λI) x3 = x2 . . . =⇒ (A − λI) x3 = 0, . . . =⇒ (A − λI) xp = xp−1 =⇒ (A − λI) xp = 0. Y P Ax3 = x2...
View Full Document

Ask a homework question - tutors are online