Finding a limit of a matrix raised to the $n$-th power

Question

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?

Answer 1

State $2$ is absorbing and the other three states are all transient (because they all lead to state $2$). Hence $\lim_{n \to \infty} p_{ij}^{n}=0$ if $j \neq 2$ and $\lim_{n \to \infty} p_{ij}^{n}=1$ if $j = 2$.

Answer 2

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 $$ P^\infty = \pmatrix{ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ } $$