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I had a quick question regarding a result that I see a lot regarding M-matrix properties, but whose proof keeps eluding me. It is stated as follows:

If A is an M-matrix, i.e. its inverse is non-negative element-wise, then the diagonal elements of A are positive. To be more specific, I would like to use inverse-positivity to prove that the first diagonal element of A is positive.

Any quick and easy ways of doing this (preferably without any theorems and the like)?

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    $\begingroup$ Your definition of M-matrix is different from the usual one, in which a real square matrix is called an M-matrix if its off-diagonal entries are non-positive and all of its eigenvalues have positive real parts. The inverses of $\pmatrix{0&1\\ 1&0}$ and $\pmatrix{-3&0&3\\ 0&2&-1\\ 3&-1&-1}$ are nonnegative, but they are not considered M-matrices in the usual definition and none of them has a positive diagonal. $\endgroup$
    – user1551
    Oct 29 '19 at 15:15
  • $\begingroup$ @user1551 but according to wikipedia the definition of a nonnegative inverse is equivalent with the usual definition of an M-matrix? Or is it that the listed result there is faulty after all? $\endgroup$ Oct 29 '19 at 20:17
  • $\begingroup$ In the Wikipedia entry, the equivalent conditions are listed under the assumption that $A$ is a real invertible Z-matrix. So, an M-matrix is first a Z-matrix. $\endgroup$
    – user1551
    Oct 29 '19 at 20:39
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The way to prove $A_{i,i}$ is positive is as follows: We know 2 things because $A$ is an M-matrix:

  • $A^{-1}_{i,j} \geq 0 \quad \forall i,j$
  • $A_{i,j} \leq 0 \quad \forall i\neq j$

Then $0 \leq 1 = (A*A^{-1})_{1,1} = A_{1,1}*\underbrace{A^{-1}_{1,1}}_{\geq 0} + \underbrace{\underbrace{A_{1,2}}_{\leq 0}*\underbrace{A^{-1}_{2,1}}_{\geq 0}}_{\leq 0} +\dots + \underbrace{\underbrace{A_{1,n}}_{\leq 0}*\underbrace{A^{-1}_{n,1}}_{\geq 0}}_{\leq 0} = A_{1,1}*\underbrace{A^{-1}_{1,1}}_{\geq 0} - |c| $ where some $c\in \mathbb{R}$.

So, in order for this summation to be at least positive, you need $A_{1,1}*A^{-1}_{1,1} \geq 0$, implying $A_{1,1}\geq 0$

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