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Unformatted text preview: y algebra states that if the coefficients αi in λn + αn−1 λn−1 + · · · + α1 λ + α0 = 0 T H are integers, then every integer solution is a factor of α0 . For our problem, this means that if there exist integer eigenvalues, then they must be contained in the set S = {±1, ±2, ±3, ±6, ±9, ±18}. Evaluating p(λ) for each λ ∈ S reveals that p(3) = 0 and p(−2) = 0, so λ = 3 and λ = −2 are eigenvalues for A. To determine the other eigenvalue, deflate the problem by dividing IG R λ3 − 4λ2 − 3λ + 18 = λ2 − λ − 6 = (λ − 3)(λ + 2). λ−3 Thus the characteristic equation can be written in factored form as Y P (λ − 3)2 (λ + 2) = 0, so the spectrum of A is σ (A) = {3, −2} in which λ = 3 is repeated—we say that the algebraic multiplicity of λ = 3 is two. The eigenspaces are obtained as follows. For λ = 3, ⎧⎛ ⎛ ⎞ ⎞⎫ 1 0 1/2 ⎨ −1 ⎬ A − 3I −→ ⎝ 0 1 0 ⎠ =⇒ N (A − 3I) = span ⎝ 0 ⎠ . ⎩ ⎭ 00 0 2 O C For λ = −2, ⎛ 1 A + 2I −→ ⎝ 0 0 0 1 0 ⎧⎛ ⎞ ⎞⎫ 1 ⎨ −1 ⎬ −2 ⎠ =⇒ N (A + 2I) = span ⎝ 2 ⎠ . ⎩ ⎭ 0 1 Notice that although the algebraic multiplicity of λ = 3 is two, the dimension of the associated eigenspace is only one—we say that A is deficient in eigenvectors. As we will see later, deficient matrices pose significant difficulties. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.1 Elementary Properties of Eigensystems http://www.amazon.com/exec/obidos/ASIN/0898714540 497 It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu Example 7.1.3 Continuity of Eigenvalues. A classical result (requiring complex analysis) states that the roots of a polynomial vary continuously with the coefficients. Since the coefficients of the characteristic polynomial p(λ) of A can be expressed in terms of sums of principal minors, it follows that the coefficients of p(λ) vary continuously with the entr...
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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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