# Iterative Solution of Large Linear Systems by David M. Young

By David M. Young

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**Extra resources for Iterative Solution of Large Linear Systems**

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6 there exists a nonsingular matrix L such that || A \\L < 1. Since || An \\L < || A \\nL it follows that || An \\L -> 0. 1 it follows that An —>0. , converges to A. If convergence holds, then we write A = A{1) + A™ + A™ + . . 4. The matrix I — B is nonsingular and the series I + B + B2 + . . converges if and only if S(B) < 1. 6) 36 2 / MATRIX PRELIMINARIES Proof. Let S{n) = I + B + B2 + · · + Bn. 2 it follows that || Bn+1 || -> 0. 6) holds. If on the other hand the series I + B + B2 + · · · converges, then one can easily show that Bn —► 0.

Evidently P can be joined to Q by a path consisting of line segments connecting properly adjacent mesh points. Somewhere along this path there must be a seg ment joining a point P' of S to a point P" of T. 17). Hence aitj φ 0 where i e S and y e T. This contradiction shows that T is empty and that A is irreducible. Next we show that A has weak diagonal dominance. 11). Moreover, since R is bounded, there exists points of Rh which are properly adjacent to one or more points of Sh. 13). Thus some of the di do not appear in the matrix A.

3. If A is an irreducible matrix with weak diagonal dom inance, then det A φ 0 and none of the diagonal elements of A vanishes. Proof. If i V = l , then αλ1 > 0 by weak diagonal dominance and det A φ 0. Suppose that N > 1 and aiti = 0 for some /. By the weak diagonal dominance we have aifj = 0 for all j . But this contradicts the irreducibility. For if S = {i} and T = W — {i}> we have aitj = 0 for all i eS,jeT. 4, if det A = 0, then there exists a solution w 9ε 0 of the homogeneous system Au = 0.