1028 if a is an m matrix then det a 0 because the

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Unformatted text preview: , there is a positive integer K such that for all k ≥ K , so Ak R Y 1/k Because this holds for each O C Ak /(ρ (A) + )k < 1 for k ≥ K . < ρ (A) + > 0, it follows that limk→∞ Ak P Example 7.10.2 1/k . for all k ≥ K , and thus < ρ (A) + ρ (A) ≤ Ak 1/k Ak = 0. k→∞ (ρ (A) + )k IG = 0 =⇒ (7.10.12) 1/k = ρ(A). For A ∈ C n×n let |A| denote the matrix having entries |aij |, and for matrices B, C ∈ n×n define B ≤ C to mean bij ≤ cij for each i and j. Problem: Prove that if |A| ≤ B, then ρ (A) ≤ ρ (|A|) ≤ ρ (B) . (7.10.13) Solution: The triangle inequality yields |Ak | ≤ |A|k for every positive integer k. Furthermore, |A| ≤ B implies that |A|k ≤ Bk . This with (7.10.12) produces Ak ∞ = |Ak | =⇒ =⇒ ∞ ≤ |A|k 1/k Ak ∞ lim k→∞ ∞ ≤ |A| 1/k Ak ∞ k ≤ Bk 1/k ∞ ∞ ≤ Bk ≤ lim ≤ |A| k k→∞ 1/k ∞ 1/k ∞ ≤ lim ≤ Bk k→∞ 1/k ∞ =⇒ ρ (A) ≤ ρ (|A|) ≤ ρ (B) . Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 620 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Example 7.10.3 Problem: Prove that if 0 ≤ Bn×n , then ρ (B) < r if and only if (rI − B)−1 exists and (rI − B)−1 ≥ 0. (7.10.14) Solution: If ρ (B) < r, then ρ(B/r) < 1, so (7.10.8)–(7.10.11) imply that B r It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] rI − B = r I − is nonsingular and (rI − B)−1 = 1 r ∞ k=0 B r k ≥ 0. To prove the converse, it’s convenient to adopt the following notation. For any P ∈ m×n , let |P| = |pij | denote the matrix of absolute values, and notice that the triangle inequality insures that |PQ| ≤ |P| |Q| for all conformable P and Q. Now assume that rI − B is nonsingular and (rI − B)−1 ≥ 0, and prove ρ (B) < r. Let (λ, x) be any eigenpair for B, and use B ≥ 0 together with (rI − B)−1 ≥ 0 to...
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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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