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For the system of equations, changed the condition $\forall j$ to $\forall i$
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If you write $A\nu=\nu$ as a system of equations it can be written as $$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall j. $$$$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall i=1, ..., n. $$ 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 $\nu_i=\alpha_i(1-\alpha_i)[\sum_j\alpha_j(1-\alpha_j)]^{-1}$.

If you write $A\nu=\nu$ as a system of equations it can be written as $$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall j. $$ 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 $\nu_i=\alpha_i(1-\alpha_i)[\sum_j\alpha_j(1-\alpha_j)]^{-1}$.

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 $\nu_i=\alpha_i(1-\alpha_i)[\sum_j\alpha_j(1-\alpha_j)]^{-1}$.

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If you write $A\nu=\nu$ as a system of equations it can be written as $$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall j. $$ 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 $\nu_i=\alpha_i(1-\alpha_i)[\sum_j\alpha_j(1-\alpha_j)]^{-1}$.