# Bounding the minimum entry of an inverse matrix

Suppose $A$ is an $n\times n$ stochastic matrix, that is, entrywise nonnegative and row sums are all $1$. If $A$ is invertible, is it true that the minimum diagonal entry of $A^{-1}$ is no larger than $1$?

Small matrices support this claim, but for larger ones, I don't know how to (dis)prove it.

Edited I forgot to add the condition that the diagonal entries of $A$ are all zero.

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## 2 Answers

Here's a counterexample: $$A = \frac{1}{5} \left[ \matrix{ 0 & 1 & 3 & 1 \cr 2 & 0 & 1 & 2 \cr 3 & 1 & 0 & 1 \cr 0 & 3 & 2 & 0}\right], \quad A^{-1}= \frac{1}{3} \left[ \matrix{ 7 & -9 & 11 & -6 \cr -8 & 6 & -4 & 9 \cr 12 & -9 & 6 & -6 \cr -13 & 21 & -14 &9} \right].$$

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Many thanks for your example. – Betrand Jun 20 '12 at 22:16

Try $$A = \pmatrix{1/2 & 1/2\cr 1/2 - s & 1/2 + s\cr},\ A^{-1} = \pmatrix{1 + 1/(2s) & -1/(2s)\cr 1 - 1/(2s) & 1/(2s)\cr}$$ where the minimum diagonal entry of $A^{-1}$ goes to $+\infty$ as $s \to 0+$.

EDIT: with your added condition, try $$A = \pmatrix{ 0&0&1/2&1/2\cr s &0&1/2-2s&1/2+s\cr 1/2&1/2&0&0\cr 1/2-2s &1/2+s&s&0\cr}$$ $$A^{-1} = \pmatrix{ 1/(8{s})+1/4&-1/(8{s})&3/(8{s})+3/4&-3/(8{s})\cr -1/(8{s})-1/4&1/(8{ s})&5/4-3/(8{s})&3/(8{s})\cr 3/(8{s})+3/4&-3/(8{s})&1/(8{s})+1/4&-1/(8{s}) \cr5/4-3/(8{s})&3/(8{s})&-1/(8{s})-1/ 4&1/(8{s})\cr}$$

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+1: I forgot to add the key condition, otherwise I know it is false. – Betrand Jun 20 '12 at 20:59
Very nice, your example shows "it is unbounded". – Betrand Jun 21 '12 at 15:26