Eigenvector of a nonnegative matrix in closed form

Question

Consider $n\times 1$ vector $\alpha = (\alpha_{1}, ..., \alpha_{n})$, where $0<\alpha_{i}<1$, and $\sum_{i=1}^{n}\alpha_i = 1$. Construct the $n\times n$ zero-diagonal matrix $A$ with $(i,j)$-th entry

$$A_{ij} = \begin{cases}\frac{\alpha_{i}}{1-\alpha_{j}} & \text{for} \; i\neq j,\\ 0 & \text{otherwise.}\end{cases}$$

I would like to compute (whenever exist) the elements of vector $v = (v_{1}, ..., v_{n})$ as functions of $(\alpha_{1}, ..., \alpha_{n})$, where $Av = v$, $0<v_{i}<1$, and $\sum_{i=1}^{n}v_{i}=1$. In other words, $v$ is the (normalized) eigenvector corresponding to eigenvalue 1. Any idea or suggestions will be appreciated.

Answer 1

If you write $A\nu=\nu$ as a system of equations it can be written as $$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall i=1, ..., n. $$ Then the change of variable $x_i=\frac{\nu_i}{1-\alpha_i}$. This gives the simple system $x_i=\alpha_i\sum_jx_j$, $i=1,\dots,n$, which has a solution $x_i=k\alpha_i$, where $k$ is a normalising constant. Thus $\nu_i=\alpha_i(1-\alpha_i)[\sum_j\alpha_j(1-\alpha_j)]^{-1}$.