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.