Assume to have an $N \times N$ stochastic matrix $W$, where $\sum_j w_{ij} = 1$ and $w_{ij}$ is a generic element on row $i$ column $j$ of the matrix $W$. Moreover you have the following two $N \times N$ matrices:
\begin{equation} \renewcommand*{\arraystretch} A = \begin{pmatrix} 1-\alpha_{1} & 1-\alpha_{1} & 1-\alpha_{1} & ... & 1-\alpha_{1}\\ 1-\alpha_{2} & & & & \vdots \\ 1-\alpha_{3} & & & & \vdots \\ \vdots & & & & \vdots \\ 1-\alpha_{N} & ... & ... & ... & 1-\alpha_{N} \end{pmatrix} \;. \end{equation}
and \begin{equation} \renewcommand*{\arraystretch} B = \begin{pmatrix} \alpha_{1} & \alpha_{2} & \alpha_{3} & ... & \alpha_{N}\\ \alpha_{1} & & & & \vdots \\ \alpha_{1} & & & & \vdots \\ \vdots & & & & \vdots \\ \alpha_{1} & ... & ... & ... & \alpha_{N} \end{pmatrix} \;. \end{equation}
where $0<\alpha_i<1 \; \forall i$. I need to prove that the matrix \begin{equation} C = B \circ (I - A \circ W)^{-1} \end{equation} where $\circ$ is the hadamard product, is also a stochastic matrix.
Do you have hint? I am getting crazy over this! Thank you very much!
J.
expand the inverse as, $$(I-A \circ W)^{-1} = \sum_{k=0}^{\infty} (A\circ W)^k$$ I think that the series is guaranteed to converge for those matrices. Then notice that $$B=1-A^T$$ Then it can be proved, with some little manipulations, that $$\sum_j \left[(A\circ W)^{k+1}\right] _{ij} - \left[A^T\circ(A\circ W)^{k}\right] _{ij} =0 $$ In fact one has $$\sum_j \left[(A\circ W)^{k+1}\right] _{ij} =\sum_j\sum_m \left[(A\circ W)^{k}\right] _{im} \left[A\circ W\right] _{mj} =\sum_m \left[(A\circ W)^{k}\right] _{im} (1-\alpha_m)$$ Once you do that it is easy to conclude the proof noticing that the only relevant term of the series is the one with $k=0$, therefore you have $$ \sum_j C_{ij} = \sum_j [1\circ I]_{ij} =\sum_j I_{ij} =1 \qquad \forall i$$