I'm stuck with the following problem.
Let be $W\in \mathbb{R}^{n\times n}$ matrix, $n\geq 2$. $W$ is such that:
- $W_{ij}\geq 0$ for all $0\leq i,j\leq n$.
- $W_{ii}=0$ for all $0\leq i\leq n$.
- $W$ is symmetric.
- Let be $w_i=\sum_{j=1,\dots,n}W_{ij}$, then $0<w_i<1$.
Let be $M$ inverse matrix of $\mathbb{I}_n-W$, where $\mathbb{I}_n$ is the identity matrix of size $n$. Let be $W^{(-i)}$ a square matrix obtained by removing the $i^{th}$-column and $i^{th}$-row from $W$. So let be $M^{(-i)}$ inverse matrix of $\mathbb{I}_{n-1}-W^{(-i)}$. I've also to prove the following equation:
$$M_{nj}M_{nk}=M_{nn}(M_{jk}-M_{jk}^{(-n)})\text{,}$$ for all $k\ne n \ne j$.
Thanks to last property and Perron-Frobenius theorem I've proved that $M$ and $M^{(-i)}$ are well-defined. But I'm not able to prove the equation above.
Even an hint should be helpful.
Thank you in advance for your support.
Write $W$ in block form $$W = \begin{bmatrix} W^{(-n)} & v \\ v^T & 0\end{bmatrix}$$ where $v$ is a certain vector. Then $$I - W = \begin{bmatrix} I - W^{(-n)} & -v \\ -v^T & 1\end{bmatrix}.$$ We can use the inversion formula for block matrices: if $X = (I - W^{(-n)} - v v^T)^{-1}$, then $$M = \begin{bmatrix} X & Xv \\ v^T X & 1 + v^T X v\end{bmatrix}.$$ Now your formula reduces to the fact that $$X v v^T X = (1 + v^T X v)(X - (I - W^{(-n)})^{-1})$$ We note that $$(I - W^{(-n)})^{-1} = (X^{-1} + vv^T)^{-1} = (I + X vv^T )^{-1}X.$$ So it suffices to show that $$X v v^T X = (1 + v^T X v)(I - (I + X vv^T )^{-1})X.$$ Or $$X v v^T = (1 + v^T X v)(I - (I + X vv^T )^{-1}).$$ This is actually the formula for the inverse of rank-one pertubation. To derive it, note that the formula is equivalent to $$(I + X vv^T )^{-1} = I - (1 + v^T X v)^{-1} X v v^T.$$ Or $$(I + X vv^T )(I - (1 + v^T X v)^{-1} X v v^T) = I.$$ Or $$I + X vv^T - (1 + v^T X v)^{-1} X v v^T - (1 + v^T X v)^{-1} X vv^TX v v^T = I.$$ Note that $$X vv^TX v v^T = (v^TX v) X v v^T$$ and simplifying the above gives the proof.