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Unformatted text preview: n. Since these λj ’s are all distinct (cos θ is a strictly decreasing function of θ on (0, π ), and a = 0 = c), A must be diagonalizable—recall (7.2.6). Finally, the k k k th component of any eigenvector associated with λj satisﬁes xk = αr1 + βr2 with α + β = 0, so xk = α Copyright c 2000 SIAM c a k /2 eiπjk/(n+1) − e−iπjk/(n+1) = 2iα c a k /2 sin j kπ n+1 . Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 516 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Setting α = 1/2i yields a particular eigenvector associated with λj as It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] ⎛ (c/a)1/2 sin (1jπ/(n + 1)) ⎞ ⎜ (c/a)2/2 sin (2jπ/(n + 1)) ⎟ ⎜ ⎟ 3/2 xj = ⎜ (c/a) sin (3jπ/(n + 1)) ⎟ . ⎜ ⎟ . ⎝ ⎠ . . n/2 (c/a) sin (njπ/(n + 1)) Because the λj ’s are distinct, {x1 , x2 , . . . , xn } is a complete linearly independent set—recall (7.2.3)—so P = x1 | x2 | · · · | xn diagonalizes A. D E It’s often the case that a right-hand and left-hand eigenvector for some eigenvalue is known. Rather than starting from scratch to ﬁnd additional eigenpairs, the known information can be used to reduce or “deﬂate” the problem to a smaller one as described in the following example. T H Example 7.2.6 IG R Deﬂation. Suppose that right-hand and left-hand eigenvectors x and y∗ for an eigenvalue λ of A ∈ n×n are already known, so Ax = λx and y∗ A = λy∗ . Furthermore, suppose y∗ x = 0 —such eigenvectors are guaranteed to exist if λ is simple or if A is diagonalizable (Exercises 7.2.23 and 7.2.22). Problem: Use x and y∗ to deﬂate the size of the remaining eigenvalue problem. Y P Solution: Scale x and y∗ so that y∗ x = 1, and construct Xn×n−1 so that its columns are an orthonormal basis for y⊥ . An easy way of doing this is to build ˜ ˜ a reﬂector R = y | X having y = y/ y 2 as its ﬁrst column as described...
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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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