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A complex infinity-norm defect bounds the inverse norm

Statement

Let E∈Cn×n, n≥1, and 0≤q<1 satisfy ∥E∥∞,C≤q. Then I−E is invertible and

∥(I−E)−1∥∞,C≤11−q.

Facts & Assumptions

Given: E,q,n as in the statement.

[F1]

The complex infinity matrix norm is submultiplicative (The complex infinity matrix norm is compatible and submultiplicative).

[F2]

Its vector norm is the maximum coordinate modulus and its matrix norm is the maximum absolute row sum; in particular ∥I∥∞,C=1 and ∣Yij∣≤∥Y∥∞,C≤nmax⁡i,j∣Yij∣ (The complex induced infinity matrix norm).

[F3]

An injective endomorphism of a finite-dimensional vector space is invertible (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

Proof

1.1givenF1F3algebra

If (I−E)x=0, then x=Ex. The vector compatibility bound in [F1] gives ∥x∥∞≤q∥x∥∞, so x=0 because q<1. Thus I−E is injective, and [F3] makes it invertible; put T=(I−E)−1.

2.1givenF1F2step 1.1algebra∎

Define SN=∑k=0NEk. By induction and [F1], ∥Ek∥∞,C≤qk. The row-sum definition [F2] gives ∥I∥∞,C=1 and the triangle inequality, so ∥SN∥∞,C≤(1−qN+1)/(1−q). The identity (I−E)SN=I−EN+1, multiplied on the left by T from step 1.1, gives SN=T−TEN+1 and hence T=SN+TEN+1. By [F2] and submultiplicativity [F1], (1−qN+1)∥T∥∞,C≤∥SN∥∞,C≤(1−qN+1)/(1−q). Since qN+1<1, division proves the asserted bound.

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