We can always expand \( \boldsymbol{w}(0) \) in terms of the right eigenvectors \( \boldsymbol{v} \) of \( \boldsymbol{W} \) as
$$ \begin{equation*} \boldsymbol{w}(0) = \sum_i\alpha_i\boldsymbol{v}_i, \end{equation*} $$resulting in
$$ \begin{equation*} \boldsymbol{w}(t) = \boldsymbol{W}^t\boldsymbol{w}(0)=\boldsymbol{W}^t\sum_i\alpha_i\boldsymbol{v}_i= \sum_i\lambda_i^t\alpha_i\boldsymbol{v}_i, \end{equation*} $$with \( \lambda_i \) the \( i^{\mathrm{th}} \) eigenvalue corresponding to the eigenvector \( \boldsymbol{v}_i \).
If we assume that \( \lambda_0 \) is the largest eigenvector we see that in the limit \( t\rightarrow \infty \), \( \boldsymbol{w}(t) \) becomes proportional to the corresponding eigenvector \( \boldsymbol{v}_0 \). This is our steady state or final distribution.