We can always expand \( \hat{w}(0) \) in terms of the right eigenvectors \( \hat{v} \) of \( \hat{W} \) as $$ \begin{equation*} \hat{w}(0) = \sum_i\alpha_i\hat{v}_i, \end{equation*} $$ resulting in $$ \begin{equation*} \hat{w}(t) = \hat{W}^t\hat{w}(0)=\hat{W}^t\sum_i\alpha_i\hat{v}_i= \sum_i\lambda_i^t\alpha_i\hat{v}_i, \end{equation*} $$ with \( \lambda_i \) the \( i^{\mathrm{th}} \) eigenvalue corresponding to the eigenvector \( \hat{v}_i \).

If we assume that \( \lambda_0 \) is the largest eigenvector we see that in the limit \( t\rightarrow \infty \), \( \hat{w}(t) \) becomes proportional to the corresponding eigenvector \( \hat{v}_0 \). This is our steady state or final distribution.