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.



Eigenvector of a nonnegative matrix in closed form
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 $\nu\_i=\alpha\_i(1-\alpha\_i)[\sum\_j\alpha\_j(1-\alpha\_j)]^{-1}$.


