# notice that x 0 x 1 x k 1 counts the

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Unformatted text preview: er Markowic Ostrowski, who made several contributions to the analysis of classical iterative methods. The “M” is short for “Minkowski” (p. 278). Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.10 Diﬀerence Equations, Limits, and Summability http://www.amazon.com/exec/obidos/ASIN/0898714540 627 It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] the form A = rI − B, where B ≥ 0 and r > ρ (B) , then (7.10.14) guarantees that A−1 exists and A−1 ≥ 0, and it’s clear that aij ≤ 0 for each i = j, so A must be an M-matrix. Proof of (7.10.26). If A is an M-matrix, then, by (7.10.25), A = rI − B, where r > ρ (B) . This means that if λA ∈ σ (A) , then λA = r − λB for some λB ∈ σ (B) . If λB = α + iβ, then r > ρ (B) ≥ |λB | = α2 + β 2 ≥ |α| ≥ α implies that Re (λA ) = r − α ≥ 0. Now suppose that A is any matrix such that aij ≤ 0 for all i = j and Re (λA ) > 0 for all λA ∈ σ (A) . This means that there is a real number γ such that the circle centered at γ and having radius equal to γ contains σ (A)—see Figure 7.10.1. Let r be any real number such that r > max{2γ, maxi |aii |}, and set B = rI − A. It’s apparent that B ≥ 0, and, as can be seen from Figure 7.10.1, the distance |r − λA | between r and every point in σ (A) is less than r. D E T H iy x IG R σ(A) r γ Y P Figure 7.10.1 All eigenvalues of B look like λB = r − λA , and |λB | = |r − λA | < r, so ρ (B) < r. Since A = rI − B is nonsingular (because 0 ∈ σ (A) ) with B ≥ 0 / and r > ρ (B) , it follows from (7.10.14) in Example 7.10.3 (p. 620) that A−1 ≥ 0, and thus A is an M-matrix. O C Proof of (7.10.27). If Ak×k is the principal submatrix lying on the intersection of rows and columns i1 , . . . , ik in an M-matrix A = rI − B, where B ≥ 0 and r > ρ (B) , then A = rI − B, where B ≥ 0 is the corresponding principal submatrix of B. Let P be a permutation matrix such that PT BP = B Y X Z , or B = P B Y X Z PT , and let C = P B0 00 PT . Clearly, 0 ≤ C ≤ B, so, by (7.10...
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