TITLE: How to prove that an M-matrix is inverse-non-negative?
QUESTION [8 upvotes]: Wikipedia says that 

The inverse of any non-singular M-matrix is a non-negative matrix."

To be more precise, if $A$ is an M-matrix,
then the entries of the inverse of $A$ are all non-negative,
i.e. $A^{-1} \geq 0$.
How do I prove this result?

REPLY [13 votes]: Let $A$ be a real matrix with entries $a_{ij}$ be an $M$ matrix.  We have:

$a_{ij} \leq 0$ when $i \neq j$
the eigenvalues of $A$ have positive real part

Let $s$ be equal to the greatest diagonal entry of $A$ (which must be positive! otherwise, $A$ would have a negative eigenvalue).  We can rewrite $A$ as
$A = sI - B$,
where $B$ is a non-negative matrix.
By the Perron-Frobenius theorem, $B$ must have a positive eigenvalue equal to $\rho(B)$.  
Furthermore, for any eigenvalue $\lambda$ of $B$, $s - \lambda$ is an eigenvalue of $A$.  Thus, $s - \rho(B)$ is an eigenvalue of $B$.  Since the eigenvalues of $A$ all have positive real part, we note that
$$
\text{Re}(s - \rho(B)) > 0 \implies \rho(B) < s
$$
So, we have $\rho(B) < s$.  Denote $B' = \frac 1s B$.
We note that $\rho(B') = \rho(B)/|s| < 1$.
Let $A' = \frac 1s A = I - B'$.  Since $\rho(B') < 1$, we can show that the infinite series $\sum_n (B')^n$ converges.  More importantly, we note that (defining the zeroeth power to be the identity),
$$
A' \sum_{n=0}^\infty (B')^n = (I - B') \sum_{n=0}^\infty (B')^n
= \left(\sum_{n=0}^\infty (B')^n \right) - \left(\sum_{n=1}^\infty (B')^n \right) = (B')^0 = I
$$
Thus, we have
$$
A^{-1} = (sA')^{-1} = \frac 1s(A')^{-1} = \frac 1s \sum_{n=0}^\infty (B')^n.
$$
Since $A^{-1}$ is a positive multiple of the sum of non-negative matrices, it must be non-negative.