I am trying to figure out the structure of an M-matrix (https://en.wikipedia.org/wiki/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, 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 $0<x<y<1$. 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.


I got 2 counter-examples:

  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, $(0.94)(5.1)\leq 4.9\leq (0.98)(5.1)$.
    So, I guess intuition does not always work in research.

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