Given the discrete time markov chain
P= $\begin{pmatrix}0.6&0.4&0&0\\ 0.2&0.8&0&0\\ 0&0&0.6&0.4\\ 0&0&0.4&0.6\end{pmatrix}$
How do I find $\lim _{n\to \:\infty }P^n$ without calculating $P^n$?
I know it has something to do with the fact that this markov chain is two isolated ergotic chains, but I don't know how to find the limiting distribution without finding $P^n$
The eigenvalues of this matrix are $1,0.4,1,0.2$ and you can find a matrix $U$ such that $$P=U^{-1}\begin{pmatrix}1&0&0&0\\ 0&0.4&0&0\\ 0&0&1&0\\ 0&0&0&0.2\end{pmatrix}U.$$
Then $P^n$ tends to $$U^{-1}\begin{pmatrix}1&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&0&0\end{pmatrix}U.$$
Since $P$ is block-diagonal you can indeed split this problem into subproblems that involve the individual $2\times2$ blocks. Now, there’s a theorem that the rows of powers of the transition matrix of an ergodic chain tend toward the stationary distribution as $n\to\infty$, so you just need to find the stochastic left eigenvector of $1$ for each of the $2\times2$ blocks. The column sums of the lower-right block are equal, so its stationary distribution is $\left(\frac12,\frac12\right)$. With a little bit of work you should be able to find the stationary distribution for the other block.
If you don’t happen to know this fundamental theorem, as described here you can decompose each block into the form $\lambda_1P_1+\lambda_2P_2$. Since $P_1^2=P_1$, $P_2^2=P_2$ and $P_1P_2=P_2P_1=0$, $(\lambda_1P_1+\lambda_2P_2)^n=\lambda_1^nP_1+\lambda_2^nP_2$. When $\lambda_1=1$ and $\lvert\lambda_2\rvert\lt1$, as is the case here, we therefore have $\lim_{n\to\infty}(P_1+\lambda_2P_2)^n=P_1$. Taking the lower-left block as before, examining the trace gives $\lambda_2=0.6+0.6-1=0.2$, and so $$P_1=\frac1{1-0.2}\begin{bmatrix}0.6-0.2 & 0.4 \\ 0.4 & 0.6-0.2 \end{bmatrix} = \begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix},$$ which agrees with the eigenvector method used above. I’ll leave working out the other block to you.
