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.


1 Answer 1


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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.