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Let $A\equiv\left\{ a_{ij}\right\} $ be a strictly substochastic matrix (i.e., $a_{ij}\geq0,\forall i,j$ and $\sum_{j}a_{ij}<1,\forall i$) and let $$M\equiv(I-A)^{-1}.$$

Next, let $D$ be a nonnegative diagonal matrix with diagonal elements being weakly less than one, and $$W\equiv\left(I-AD\right)^{-1}.$$

I want to show that $\forall i,j$ we have $$ w_{ij}\left(\frac{\sum_{k}w_{ki}-1}{\sum_{k}w_{kj}}\right)\leq m_{ij}\left(\frac{\sum_{k}m_{ki}-1}{\sum_{r}m_{kj}}\right). $$

Note that $M,W$ are inverse M-matrices so all their entries are weakly positive. Also, $w_{ii},m_{ii} \geq 1$ and $w_{ij} \leq m_{ij}$ from the Newmann series expression for $W$ and $M$. One implication is that the column sums of $W$ and $M$ in the inequality are weakly greater than one.

The result is obvious for $D = I$ and for $D = 0$, as well as for $i=j$ since in that case the inequality is $$ w_{ii}\left(1-\frac{1}{\sum_{k}w_{ki}}\right)\leq m_{ii}\left(1-\frac{1}{\sum_{r}m_{ki}}\right), $$ which follows from the fact that $w_{ij} \leq m_{ij}$ for all $i,j$.

Any suggestions for proving this apparent result would be greatly appreciated.

UPDATE: I can show this for the case in which $A$ is a one-rank matrix, so that $a_{ij}=a_{i}b_{j}$ or $A=b^{T}a$, with $a,b$ column vectors, with $\sum_{k}a_{k}b_{k}<1$, while $A\mathrm{diag}\left(\mu\right)=b^{T}\mathrm{diag}\left(\mu\right)a$. The Sherman-Morrison formula implies that $$ W=I+\frac{A}{1-\sum a_{i}b_{i}\mu_{i}} $$ and hence $$ \sum_{r}w_{ri}=1+\frac{b_{i}\mu_{i}\sum_{r}a_{r}}{1-\sum_{i}a_{i}b_{i}\mu_{i}} $$ After some rearranging, our inequality is then $$ \left(\delta_{ij}+\frac{a_{i}b_{j}\mu_{j}}{1-\sum_{k}a_{k}b_{k}\mu_{k}}\right)\left(\mu_{i}-\mu_{i}\sum_{k}a_{k}b_{k}+\mu_{i}b_{j}\sum_{r}a_{r}\right)\leq\left(\delta_{ij}+\frac{a_{i}b_{j}}{1-\sum_{k}a_{k}b_{k}}\right)\left(1-\sum_{k}a_{k}b_{k}\mu_{k}+b_{j}\mu_{j}\sum_{r}a_{r}\right), $$ where $\delta_{ij}=1$ if $i=j$ and zero otherwise. Consider two cases: (i) $i=j$ and (ii) $i\neq j$. In case (i) we can show that $$ \mu_{i}-\mu_{i}\sum_{k}a_{k}b_{k}+\mu_{i}b_{j}\sum_{r}a_{r}\leq1-\sum_{k}a_{k}b_{k}\mu_{k}+b_{j}\mu_{j}\sum_{r}a_{r}, $$ which is of course enough to ensure the original inequality. Since $i=j$ then this inequality simplifies to $$ \sum_{k}a_{k}b_{k}\mu_{k}+\mu_{i}\left(1-\sum_{k}a_{k}b_{k}\right)\leq1. $$ Since $1-\sum_{k}a_{k}>0$ then it is enough to show that $$ \sum_{k}a_{k}b_{k}\mu_{k}+1-\sum_{k}a_{k}b_{k}\leq1, $$ which is equivalent to $$ \sum_{k}a_{k}b_{k}\mu_{k}\leq\sum_{k}a_{k}b_{k}, $$ which is true. In case (ii) $i\neq j$ then $\delta_{ij}=0$ and so we need to show that $$ \frac{\mu_{j}}{1-\sum_{k}a_{k}b_{k}\mu_{k}}\left(\frac{\mu_{i}}{1-\sum_{k}a_{k}b_{k}\mu_{k}+b_{j}\mu_{j}\sum_{r}a_{r}}\right)\leq\frac{1}{1-\sum_{k}a_{k}b_{k}}\left(\frac{1}{1-\sum_{k}a_{k}b_{k}+b_{j}\sum_{r}a_{r}}\right), $$ or $$ \mu_{i}\mu_{j}\left(1-\sum_{k}a_{k}b_{k}\right)^{2}+b_{j}\mu_{j}\sum_{r}a_{r}\left(\sum_{k}a_{k}b_{k}\mu_{k}+\mu_{i}\left(1-\sum_{k}a_{k}b_{k}\right)-1\right)\leq\left(1-\sum_{k}a_{k}b_{k}\mu_{k}\right)^{2} $$ Above we showed that $$ \sum_{k}a_{k}b_{k}\mu_{k}+\mu_{i}\left(1-\sum_{k}a_{k}b_{k}\right)-1\leq0, $$ and so since $$ \mu_{i}\mu_{j}\left(1-\sum_{k}a_{k}b_{k}\right)^{2}\leq\left(1-\sum_{k}a_{k}b_{k}\mu_{k}\right)^{2}, $$ then we are done.

Andres
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    Suppose every entry in $D$ is 1, except one. $AD$ is then a rank-1 modification of $A$, so you can try to analyze that. An arbitrary $D$ is just the successive application of such modifications to matrices satisfying your initial conditions, so if you manage to prove it for the aforementioned $D$, you're done. – cangrejo Aug 23 '21 at 22:37
  • @cangrejo I followed your recommendation by setting all elements of the diagonal of $D$ equal to 1 except the first one. In one special case that may be quite informative, I set $d_{11} = 0$ and after some algebra find that the inequality becomes $$\frac{m_{1i}/m_{11}+\sum_{k}m_{ki,11}-1}{m_{1j}/m_{11}+\sum_{k}m_{kj,11}}\leq\frac{m_{ij}}{m_{ij,11}}\left(\frac{\sum_{k}m_{ki}-1}{\sum_{k}m_{kj}}\right),$$ where $m_{ij,11}$is the $ij$ element of the inverse of $I-A$ after removing the first row and first column. Any thoughts on how to proceed? – Andres Sep 29 '21 at 03:32
  • Note that the numerator is positive since $m_{ii,11} > 1$. Also note that $m_{ij} \geq m_{ij,11}$ thanks to https://math.stackexchange.com/a/4247032/165163 – Andres Sep 29 '21 at 03:34
  • One way to proceed is to use blockwise inversion of matrix $I-A$ (with lower right block being $(I-A)$ with first row and column removed) to relate $m_{ij,11}$ to $m_{ij}$. – Andres Sep 29 '21 at 04:02

1 Answers1

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Here is a counterexample to the inequality in question: \begin{equation} A=\begin{pmatrix}0.3102 & 0.0127 & 0.3517\\ 0 & 0.3547 & 0.1750\\ 0 & 0.2743 & 0.2941 \end{pmatrix}, \end{equation} \begin{equation} D=\begin{pmatrix}0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, \end{equation} \begin{equation} M=(I-A)^{-1}=\begin{pmatrix}1.4497 & 0.3751 & 0.8154\\ 0 & 1.7323 & 0.4295\\ 0 & 0.6731 & 1.5835 \end{pmatrix}, \end{equation} \begin{equation} W=\left(I-A D \right)^{-1}=\begin{pmatrix}1 & 0.2587 & 0.5624\\ 0 & 1.7323 & 0.4295\\ 0 & 0.6731 & 1.5835 \end{pmatrix}, \end{equation} For $i = 2$ and $j = 3$: \begin{equation} w_{ij}\left(\frac{\sum_{r}w_{ri}-1}{\sum_{r}w_{rj}}\right)=0.2775>0.2704=m_{ij}\left(\frac{\sum_{k}m_{ki}-1}{\sum_{k}m_{kj}}\right) \end{equation}

Konstantin
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