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# Proof use induction on n the size of the matrix for n

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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 http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 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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