Let $A$ be an $K\times K$ strictly substochastic matrix (i.e., $\sum_{k=1}^{K}a_{sk}<1,\forall s$), let $W\equiv(I-A)^{-1},$ and let $Z_{i}\equiv\sum_{k=1}^{K}w_{ki}$ be the column sums of $W$. Note that since $W$ is an inverse $M$ matrix then all elements are nonnegative, $w_{ii}\geq w_{ki}$ for all $k$ (see https://math.stackexchange.com/a/4223105/165163), and the diagonal elements are weakly higher than one (from the Newmann expansion of $(I-A)^{-1}$, implying that $w_{ii},Z_{i}\geq1$ for all $i$. Consider the symmetric matrix $H=\left\{ h_{ij}\right\} $ with $$ h_{ij}=w_{j1}\left(w_{ij}-\delta_{ij}\right)\left(Z_{i}-1\right)Z_{1}+w_{i1}w_{ji}\left(Z_{j}-1\right)Z_{1}-w_{i1}w_{j1}\left(Z_{i}-1\right)\left(Z_{j}-1\right), $$ where $\delta_{ij}$ is the indicator function (i.e., $\delta_{ij}=1$ if $i=j$ and zero otherwise). Numerically I see that this matrix is positive definite. Any thoughts on how to prove this?
We can show that the diagonal elements of H are positive, since \begin{align*} h_{ii} & =w_{i1}\left(w_{ii}-1\right)\left(Z_{i}-1\right)Z_{1}+w_{i1}w_{ii}\left(Z_{i}-1\right)Z_{1}-w_{i1}w_{i1}\left(Z_{i}-1\right)\left(Z_{i}-1\right)\\ & =w_{i1}\left(Z_{i}-1\right)\left[\left(w_{ii}-1\right)Z_{1}+\sum_{k}\left(w_{k1}w_{ii}-w_{ki}w_{i1}\right)+w_{i1}\right]. \end{align*} We know that $Z_{i},w_{ii}\geq1$, and we also know that (see Property of elements of inverse matrix?) $$ w_{k1}w_{ii}=w_{ki}w_{i1}+\frac{m_{k1,ii}}{\det (I-A)}, $$ where $m_{k1,ii}$ is the $\left(k,1\right)$ element of the inverse of $\left(I-A\right)_{[ii]},$ where $B_{[ii]}$ means matrix $B$ after removing row and column $i$. Since $\left(I-A\right)_{[ii]}$ is also an M matrix, then all elements of its inverse are nonnegative, and so $w_{k1}w_{ii} \geq w_{ki}w_{i1}$.
Unfortunately, the matrix is not diagonally dominant, so that approach to show that $H$ is PD does not work.
