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Strict column dominance gives an entrywise nonnegative inverse

Statement

Let M=(mij)∈Rm×m for a finite m≥1. Suppose mjj>0, mij≤0 for i≠j, and

mjj>∑i≠j∣mij∣(j=1,…,m).

Then M is invertible and every entry of M−1 is nonnegative. If b∈Rm has nonnegative coordinates, then M−1b≥0; if every coordinate of b is positive, then every coordinate of M−1b is positive.

Facts & Assumptions

Given: A finite real matrix satisfying the displayed positive-diagonal, nonpositive-off-diagonal, and strict column-dominance hypotheses.

[F1]

The hypotheses give cij:=−mij/mjj≥0 for i≠j and cjj=0.

Proof

1.1givenF1algebraconstruct

Let D=diag⁡(m11,…,mmm) and C=I−MD−1. Then D is invertible, C≥0, and for each column j, rj:=∑iCij=∑i≠j∣mij∣/mjj<1. Since there are finitely many columns, r:=max⁡jrj<1.

2.1step 1.1algebrainduction

For every k≥0, Ck≥0 and ∑i(Ck)ij≤rk for every column j: it holds at k=0, and multiplying by C gives ∑i(Ck+1)ij=∑ℓ(∑iCiℓ)(Ck)ℓj≤r∑ℓ(Ck)ℓj. Thus each entry series in S:=∑k=0∞Ck converges and is nonnegative.

3.1step 1.1step 2.1algebra

The finite geometric identity (I−C)∑k=0NCk=∑k=0NCk(I−C)=I−CN+1 holds by cancellation of adjacent powers. Step 2.1 gives CN+1→0 entrywise, so (I−C)S=S(I−C)=I. Since M=(I−C)D, M−1=D−1S≥0, proving invertibility and inverse nonnegativity.

4.1step 3.1algebra∎

If b≥0, then M−1b≥0 by step 3.1. If b>0, then S≥I entrywise, so D−1Sb≥D−1b>0 coordinatewise. This proves both vector conclusions.

Used by

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