# 2 diagonalization by similarity transformations

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Unformatted text preview: vectors associated with a simple eigenvalue λ ∈ σ (A) , then Y P G = xy∗ /y∗ x (7.2.12) is the projector onto N (A − λI) along R (A − λI). In the context of the spectral theorem (p. 517), this means that G is the spectral projector associated with λ. O C Proof. It’s not diﬃcult to prove y∗ x = 0 (Exercise 7.2.23), and it’s clear that G is a projector because G2 = x(y∗ x)y∗ /(y∗ x)2 = G. Now determine R (G). The image of any z is Gz = αx with α = y∗ z/y∗ x, so R (G) ⊆ span {x} = N (A − λI) and dim R (G) = 1 = dim N (A − λI). Thus R (G) = N (A − λI). To ﬁnd N (G), recall N (G) = R (I − G) (see (5.9.11), p. 386), and observe that y∗ (A − λI) = 0 =⇒ y∗ (I − G) = 0, so ⊥ ⊥ ⊥ R (A − λI) ⊆ R (I − G) = N (G) =⇒N (G) ⊆ R (A − λI) (Exercise 5.11.5). But dim N (G) = n − dim R (G) = n − 1 = n − dim N (A − λI) = dim R (A − λI), so N (G) = R (A − λI). Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.2 Diagonalization by Similarity Transformations http://www.amazon.com/exec/obidos/ASIN/0898714540 519 Example 7.2.7 Problem: Determine the spectral projectors for A = 1 8 −8 −4 −11 8 −4 −8 5 . It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] Solution: This is the diagonalizable matrix from Example 7.2.1 (p. 507). Since there are two distinct eigenvalues, λ1 = 1 and λ2 = −3, there are two spectral projectors, G1 = the projector onto N (A − 1I) along R (A − 1I), G2 = the projector onto N (A + 3I) along R (A + 3I). D E There are several diﬀerent ways to ﬁnd these projectors. 1. Compute bases for the necessary nullspaces and ranges, and use (5.9.12). T 2. Compute Gi = Xi Yi as described in (7.2.11). The required computations are essentially the same as those needed above. Since much of the work has already been done in Example 7.2.1, let’s complete the arithmetic. We have ⎛ ⎛ ⎞ ⎞ 1 −1 −1 T...
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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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