I have a question about the following Markov chain with state space S=$\{1,2,3,4,5,6\}$. \begin{equation} P= \begin{bmatrix} \frac{1}{3}&0&\frac{2}{3}&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&1&0&0&0&0\\ \frac{1}{2}&0&\frac{1}{2}&0&0&0\\ 0&\frac{1}{4}&\frac{1}{12}&\frac{1}{4}&\frac{1}{12}&\frac{1}{3} \end{bmatrix} \end{equation}
For this Markov chain I have to indicate if the limit distribution exists for each possible initial state $i\in S$. Then for the ones that do exist I need to determine the limit distribution.
From what I understand, the definition of the limit distribution is that $\pi^{(n)} \rightarrow \pi$ as $n\rightarrow \infty$, but I cannot figure out how to apply this to since I do not know the initial distribution.
At first you should look for certain properties of your Markov Chain like irreducibile or aperiodic.
Note that you can never leave the sets $\{1,3,5\}$ and $\{2,4\}$ after the chain has entered them. So if you start in one of these states you may consider the reduced chain, which only has the corresponding set as state space. On the other hand there is no way to enter the state $6$ from any other state and the chain is going to leave it with positive probability if you have entered it.
A better property to use then $\pi^{(n)}\to \pi$ as $n\to\infty$ is $\pi P = \pi$. This is an easy property to prove for stationary distributions using Chapman-Kolmogorov equations. For small matrices like yours this helps because then $\pi$ is a left eigenvector to the eigenvalue $1$ and normalized. For example $\tilde{\pi} = (0,1,0,1,0,0)$ is one left eigenvector. How do you get a stationary distribution $\pi$ from that? And from which initial states can it be attained (recall that if you start in $\{1,3,5\}$ you stay in $\{1,3,5\}$ and the same for $\{2,4\}$)?
Hint: There are more than one.
If you start in state $6$ there is no stationary distribution in the sense that $\pi^{(n)}\to\pi$ as $n\to\infty$ because the convergence depends on the state you enter when leaving $6$.