# inverse M matrices structure

I am trying to figure out the structure of M matrix whose inverse has a special form: Let $$A$$ be an inverse M matrix (inverse M matrices are those matrices whose inverse is an M matrix, https://en.wikipedia.org/wiki/M-matrix) such that each row sum of the matrix is a fixed constant (greater than 1). Each diagonal entry is strictly greater than all off-diagonal entries of that row such that $$xa_{ii}\leq a_{ij}\leq ya_{ii} \forall i\neq j$$, where $$0. I am trying to show that $$A^{-1}$$ will have positive diagonal entries less than 1.
So far, I could not find any reference on why such a result must hold true, but I could not get a counter-example either (I tried numerical examples). Intuitively, I think it will be true due to the special structure of the matrix, the off-diagonal entries of $$A$$ are close to each other, so the inverse will have the above structure to compensate for that. Any idea or suggestion will be really helpful.

• What do you mean by "Each diagonal entry is strictly greater than all off-diagonal entries of that row such that $xa_{ii} \leq a_{ij} \leq ya_{ii}$ ..."? Do you mean those as 2 different conditions because the one doesnt follow from the other. Consider the identity matrix...
– jack
Oct 26, 2020 at 5:28
• the 2nd condition implies that the diagonal entry is strictly greater than the off-diagonal entries of the corresponding row. Oct 26, 2020 at 5:39
• I am also looking for any other conditions which might make the result true. Even a suggestion would be helpful. Oct 26, 2020 at 5:40

1. Let $$A= \begin{pmatrix}1.1 & 0.89\\0.89 & 1.1 \end{pmatrix}$$, then $$A^{-1}=\begin{pmatrix} 2.6322 &-2.1297\\-2.1297 & 2.6322 \end{pmatrix}$$.
Here $$x=0.8, y=0.9$$, so, $$(0.8)(1.1)\leq 0.89\leq (0.9)(1.1)$$.
2. Let $$A= \begin{pmatrix}5.1 & 4.9\\4.9 & 5.1 \end{pmatrix}$$, then $$A^{-1}=\begin{pmatrix} 2.55 &-2.45\\-2.45 & 2.55 \end{pmatrix}$$.
Here $$x=0.94, y=0.98$$, so, $$(0.94)(5.1)\leq 4.9\leq (0.98)(5.1)$$.