71044 t xi et x ex y p o c in other words the limiting

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Unformatted text preview: .13) on p. 619, ρ(B) = ρ (C) ≤ ρ (B) < r. Consequently, (7.10.25) insures that A is an M-matrix. Proof of (7.10.28). If A is an M-matrix, then det (A) > 0 because the eigenvalues of a real matrix appear in complex conjugate pairs, so (7.10.26) and (7.1.8), Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 628 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 n It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu p. 494, guarantee that det (A) = i=1 λi > 0. It follows that each principal minor is positive because each submatrix of an M-matrix is again an M-matrix. Now prove that if An×n is a matrix such that aij ≤ 0 for i = j and each principal minor is positive, then A must be an M-matrix. Proceed by induction on n. For n = 1, the assumption of positive principal minors implies that A = [ρ] with ρ > 0, so A−1 = 1/ρ > 0. Suppose the result is true for n = k, and consider the LU factorization A(k+1)×(k+1) = Ak×k T d c I = α T 0 dA −1 A 0 1 c α − d A−1 c = LU. D E T We know that A is nonsingular (det (A) is a principal minor) and α > 0 (it’s a 1 × 1 principal minor), and the induction hypothesis insures that A−1 ≥ 0. Combining these facts with c ≤ 0 and dT ≤ 0 produces ⎛ −A−1 c −1 A−1 = U−1 L−1 ⎜A ⎜ =⎜ ⎝ 0 α− dT A−1 c T H ⎞ ⎟ ⎟ ⎟ ⎠ I 0 −dT A−1 1 IG R 1 α − dT A−1 c ≥ 0, and thus the induction argument is completed. Y P Proof of (7.10.29). If A = M − N is an M-matrix, and if M−1 ≥ 0 and N ≥ 0, then the iteration matrix H = M−1 N is clearly nonnegative. Furthermore, −1 (I − H) − I = (I − H)−1 H = A−1 N ≥ 0 =⇒ (I − H)−1 ≥ I ≥ 0, O C so (7.10.14) in Example 7.10.3 (p. 620) insures that ρ (H) < 1. Convergence of Jacobi’s method is a special case because the Jacobi splitting is A = D − N, where D = diag (a11 , a...
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