**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
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7.10 Diﬀerence Equations, Limits, and Summability
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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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