**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
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Chapter 7
Eigenvalues and Eigenvectors
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Please report violations to [email protected] 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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