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Unformatted text preview: = 1, we have x∗ Dx ≤ λi , and βi = min dim V =n−i+1 ˜ ˜ max x∗ Bx ≤ max x∗ Bx = max x∈V x 2 =1 x∈T x 2 =1 x∈T x 2 =1 ˜ x∗ Dx + x∗ Ex ˜ ˜ ≤ max x∗ Dx + max x∗ Ex ≤ λi + max x∗ Ex = λi + n x∈T x 2 =1 Copyright c 2000 SIAM x∈T x 2 =1 x∈C x 2 =1 1. Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 552 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Note: Because E often represents an error, only E (or an estimate thereof) is known. But for every matrix norm, | j | ≤ E for each j (Example 7.1.4, p. 497). Since the j ’s are real, − E ≤ j ≤ E , so (7.5.6) guarantees that It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu λi − E ≤ βi ≤ λi + E . (7.5.7) In other words, • the eigenvalues of a hermitian matrix A are perfectly conditioned because a hermitian perturbation E changes no eigenvalue of A by more than E . D E It’s interesting to compare (7.5.7) with the Bauer–Fike bound of Example 7.3.2 (p. 528). When A is hermitian, (7.3.10) reduces to minλi ∈σ(A) |β − λi | ≤ E because P can be made unitary, so, for induced matrix norms, κ(P) = 1. The two results diﬀer in that Bauer–Fike does not assume E and B are hermitian. T H Example 7.5.3 Interlaced Eigenvalues. For a hermitian matrix A ∈ C n×n with eigenvalues λ1 ≥ λ2 ≥ · · · ≥ λn , and for c ∈ C n×1 , let B be the bordered matrix B= A c∗ c α IG R with eigenvalues n+1×n+1 β1 ≥ β2 ≥ · · · ≥ βn ≥ βn+1 . Problem: Explain why the eigenvalues of A interlace with those of B in that Y P β1 ≥ λ1 ≥ β2 ≥ λ2 ≥ · · · ≥ βn ≥ λn ≥ βn+1 . (7.5.8) Solution: To see that βi ≥ λi ≥ βi+1 for 1 ≤ i ≤ n, let U be a unitary matrix such that UT AU = D = diag (λ1 , λ2 , . . . , λn ) . Since V = U 0 is 01 also unitary, the eigenvalues of B agree with those of O C ˜ B = V...
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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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