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}

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 $X_0=2$: 
$$
\tilde\pi = \left(
\begin{array}{ccccc}
 \frac{3}{14} & \frac{2}{7} & 0 & \frac{2}{13} & \frac{9}{26} \\
\end{array}
\right).
$$
