Finding a limit of a matrix raised to the $n$-th power There is a stochastic matrix given
$$\mathbb{P} = \begin{bmatrix} 0&\frac{1}{2}&\frac{1}{2}&0 \\ 0&1&0&0 \\ \frac{1}{2}&0&0&\frac{1}{2} \\ \frac{1}{3}&0&\frac{1}{3}&\frac{1}{3} \end{bmatrix}$$
I am to find $\mathbb{P}^n$. I know a theorem which says that if $\exists n$ $[p_{i,j}^n] >0$ then the limit exists and can be easily calculated (of course by $[p_{i,j}^n]$ I mean elements of $\mathbb{P}^n$).
In my case however $[p_{2,1}^n] = [p_{2,3}^n] = [p_{2,4}^n] =0, \forall n$. How can I solve this problem? Is there another theorem?

We have $P=SDS^{-1}$, where
$$
S=\pmatrix{               
-1 & -1 & \frac{9}{16} & 1 \\
0 & 0 & 0 & 1 \\
1 & 0 & \frac{15}{16} & 1 \\
0 & 1 & 1 & 1 \\
}
,\quad
D=\pmatrix{
-\frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & \frac{5}{6} & 0 \\
0 & 0 & 0 & 1 \\
}
$$
Therefore, $P^\infty = S D^\infty S^{-1}$.
Since
$$
D^\infty = \pmatrix{
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
}
$$
we have
$$
P^\infty = \pmatrix{
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
}
$$
