Stationary Distributions of a irreducible Markov chain

I was trying to get all Stationary Distributions of the following Markov chain. Intuitively, I would say there are two resulting from splitting op the irreducible Markov chain into two reducible ones. However, I feel this is not mathematically correct. How else would I be able to find all stationary distributions? $$\begin{bmatrix} \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{4} & \frac{3}{4} \\ 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} \\ \end{bmatrix}$$

• Looking at the limit of $\mathbb P^n$. Nov 27 '19 at 14:13
• @Xi'an It is true that this will yield the stationary distributions of the Markov chain, but it is not trivial to compute $P^n$. Dec 19 '19 at 2:17

Conditioned on $$X_0\in\{0,1\}$$ we have from solving \begin{align} \pi_0 &= \frac13\pi_0 + \frac12\pi_1\\ \pi_1 &= \frac23\pi_0 + \frac12\pi_1\\ \pi_0 + \pi_1 &= 1 \end{align} $$\pi_0 = \frac37$$, $$\pi_1=\frac 47$$.
Conditioned on $$X_0\in\{3,4\}$$ we have by solving a similar system of equations $$\pi_3 = \frac4{13}$$, $$\pi_4=\frac9{13}$$.
Conditioned on $$X_0=2$$ we have $$\tilde\pi_i = \frac12\pi_i$$ for $$i\in\{0,1,3,4\}$$.
So we have three stationary distributions: those obtained by conditioning on $$X_0\in\{0,1\}$$ and $$X_0\in\{3,4\}$$, and the one obtained by conditioning on $$X_0=2$$: $$\tilde\pi = \left( \begin{array}{ccccc} \frac{3}{14} & \frac{2}{7} & 0 & \frac{2}{13} & \frac{9}{26} \\ \end{array} \right).$$