How to prove that an M-matrix is inverse-non-negative? 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?
 A: 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.
A: Take home messages:
If $B$ is small ($\rho(B)<1$) then the inverse of $I-B$ is the sum of the convergent series $I + B + B^2 + \cdots$.
Moreover, if $B$ has all its entries $\ge 0$, then so does $B^k$ for all $k$. If the above series is convergent the same for its sum.
So it's all about in cases of interest that $B$ is small. For instance, if we use some norm on the space and $\|B\|<1$ for that norm.
Note: it may well be that $B$ has all entries positive, $I-B$ is invertible, yet $(I-B)^{-1}$ does not have entries positive. Note that the invertibility condition is $1 \ne \sigma(B)$ ($1$ is not in the spectrum of $B$), which is weaker than $\rho(B) <1$. The problem is that we cannot use the Neumann series to get the inverse of $I-B$. In fact, for any matrix $B$ the Neumann series converges if and only if $\rho(B)<1$.
