**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
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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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