inverse of a Z matrix is non-negative

Question

My question is : Suppose $A\in R^{n\times n}$ is a Z matrix (https://en.wikipedia.org/wiki/Z-matrix_(mathematics)), then $A^{-1}\geq 0$ (the inequality is element-wise) iff A is an M matrix (https://en.wikipedia.org/wiki/M-matrix). (This is given as an equivalent criterion in wikipedia).
The proof for inverse of an M matrix is non-negative is given here:How to prove that an M-matrix is inverse-non-negative?. I am trying to prove the converse part by showing that $A$ can be written as $sI-B$, where $B=(b_{ij})$ with $b_{ij}\geq 0 \forall 1\leq i,j\leq n$ where $s$ is at least as large as the maximum of the moduli of the eigenvalues of $B$, and $I$ is an identity matrix ($s\geq \rho(B)$).
So far, I have tried this : Write $A=|\lambda_i|I-A_1$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$ and $\rho(A)=|\lambda_i|$ for some $1\leq i\leq n$. Since, $A_1=|\lambda_i|I-A$, thus, $\rho(A_1)\leq |\lambda_i|$.
I am not sure if my proof is correct. A hint or reference material will be really helpful!

Answer 1

A key fact that we use here is that for $P,Q \geq 0$, we have $PQ \geq 0$.

Suppose that $A$ is a $Z$-matrix and that $A^{-1}$ is positive. Let $s$ be equal to the greatest diagonal entry of $A$. Note that this diagonal entry must be positive: otherwise, we see that $A \leq 0$ and $A^{-1} \geq 0$ which implies that $0 \geq AA^{-1} = I$, which is false.

We can write $A = s I - B$, where $B$ is a non-negative matrix. It remains to be seen that $s \geq \rho(B)$. So, suppose for contradiction that $\rho(B) > s$. Because $B$ is non-negative, by the Perron-Frobenius theorem there is a non-zero vector $x \geq 0$ for which $B x = \rho(B) x$. It follows that $$ Ax = (sI - B)x = sx - \rho(B) x = (s - \rho(B))x. $$ Because $s - \rho(B) < 0$, we see that $A x \leq 0$. Because $A^{-1} \geq 0$, it follows that $x = A^{-1}[A x] \leq 0$, which contradicts our premise that $x$ was non-zero with $x \geq 0$.