Element-wise ordering of the inverse of two M-matrices Assume $A$ and $B$ are two M-mtrices and $A \geq B$, where "$\geq$" is element-wise ordering. Can I show that $A^{-1} \leq B^{-1}$?
I think it holds if $D = A - B$ does not have all-zero rows/columns.
 A: Since $A$ and $B$ are $M$-matrices, their inverses $A^{-1}$ and $B^{-1}$ are entry-wise non-negative. In particular, 
$$
B^{-1} - A^{-1}
= A^{-1}(A-B)B^{-1}
$$
is the product of three entry-wise non-negative matrices, since $A\ge B$ by assumption. The product of non-negative matrices is non-negative, so $B^{-1} \ge A^{-1}$.
A: $A$ has a positive diagonal because it is a nonsingular M-matrix. As invertibility, entrywise order and inverse-positivity of matrices are all preserved under congruence by positive diagonal matrix, we may assume that all diagonal entries of $A$ are equal to $1$. With this additional assumption, we may write $A=I-S$, where $S$ is a nonnegative matrix with a zero diagonal and $\rho(S)<1$.
Now, as $B$ is a nonsingular M-matrix that is $\le A$, it can be written as $bI-T$ where $\rho(T)<b:=\max_ib_{ii}\le1$ and $T\ge S$ (this is where we need $A$ to have a constant diagonal: without this assumption, we cannot force the diagonal of $S$ to zero and in turn we cannot guarantee that the diagonal part of $S$ is dominated by the diagonal of $T$, even though it is assumed that $A\ge B$). But then $\frac1{b^{k+1}}T^k \ge T^k\ge S^k$ for every nonnegative integer $k$. Therefore
\begin{aligned}
A^{-1}=(I-S)^{-1}
&=I+S+S^2+\ldots\\
&\le\frac1b\left(I+\frac1bT+\frac1{b^2}T^2+\ldots\right)\\
&=\frac1b\left(I-\frac1bT\right)^{-1}=B^{-1}.
\end{aligned}
