**Unformatted text preview: **ation:
⎛
⎞
1
−4 −4
A = ⎝ 8 −11 −8 ⎠ .
−8
8
5 Y
P Solution: Determine whether or not A has a complete set of three linearly
independent eigenvectors. The characteristic equation—perhaps computed by
using (7.1.5)—is
λ3 + 5λ2 + 3λ − 9 = (λ − 1)(λ + 3)2 = 0. O
C Therefore, λ = 1 is a simple eigenvalue, and λ = −3 is repeated twice (we
say its algebraic multiplicity is 2). Bases for the eigenspaces N (A − 1I) and
N (A + 3I) are determined in the usual way to be
⎧⎛
⎧⎛ ⎞ ⎛ ⎞⎫
⎞⎫
1⎬
1⎬
⎨
⎨1
N (A − 1I) = span ⎝ 2 ⎠
and N (A + 3I) = span ⎝ 1 ⎠ , ⎝ 0 ⎠ ,
⎩
⎭
⎩
⎭
−2
0
1
and it’s easy to check that when combined these three eigenvectors constitute a
linearly independent set. Consequently, A must be diagonalizable. To explicitly
exhibit the similarity transformation that diagonalizes A, set
⎛
⎞
⎛
⎞
111
1
0
0
P = ⎝ 2 1 0 ⎠ , and verify P−1 AP = ⎝ 0 −3
0 ⎠ = D.
−2 0 1
0
0 −3 Copyright c 2000 SIAM Buy online from SIAM
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508
Chapter 7
Eigenvalues and Eigenvectors
http://www.amazon.com/exec/obidos/ASIN/0898714540
Since not all square matrices are diagonalizable, it’s natural to inquire about
the next best thing—i.e., can every square matrix be triangularized by similarity?
This time the answer is yes, but before explaining why, we need to make the
following observation. It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] Similarity Preserves Eigenvalues
Row reductions don’t preserve eigenvalues (try a simple example). However, similar matrices have the same characteristic polynomial, so they
have the same eigenvalues with the same multiplicities. Caution! Similar matrices need not have the same eigenvectors—see Exercise 7.2.3. D
E Proof. Use the product rule for determinants in conjunction with the fact that
det P−1 = 1/det (P) (Exercise 6.1.6) to write T
H det (A − λI) = det P−1 BP − λI = det P−1 (B − λI)P = det...

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