Stationary distribution and the period of chain brand - on example Let's define two stats Markov chain:
$P=\left( \begin{array}{ccc}
0 & 1 \\
1/2 &1/2  \end{array} \right)$
Then, stationary distribution is:
$$sP=s$$
$$\left(0s_1+\frac12s_2, s_1+\frac12s_2\right)=\left(\frac12s_2, s_1+\frac12 s_2 \right) = (s_1, s_2)$$
And, $\frac12 s_2+ s_1+\frac12s_2 = 1$
So have exactly one solution,  $s_1=\frac13, s_2=\frac23$.
Our chain has exactly one stationary distribution.  
Is it ok computed ?
My mainly question is-  how to compute period for state $1$ and state $2$  ?
From definition period is $o(s) =\gcd\left\{n : p_{ss}(n) > 0\right\}$.
 A: Since the Markov chain is irreducible and aperiodic with a finite state space, we may also compute the stationary distribution $s$ by computing $\lim_{n\to\infty} P^n$ and taking any of its rows. Since $P^n$ is a stochastic matrix (i.e. the rows sum to $1$), it is clear that $(1,1)$ is an eigenvector with associated eigenvalue $1$. The sum of the eigenvalues of a matrix is equal to its trace (the sum of the diagonal elements), so the other eigenvalue is $-\frac12$. Solving $\left(P-\left(\frac12\right)I\right)x=0$ yields $(2,-1)$ as a solution. So we may write $P=ADA^{-1}$ where
$$A = \begin{pmatrix}1&2\\1&-1\end{pmatrix},\quad D = \begin{pmatrix}1&0\\0&-\frac12\end{pmatrix}. $$
By induction we see that $P^n = AD^nA^{-1}$, and since $A^{-1}=\frac13 A$ we have
$$P^n = \frac13\begin{pmatrix}1&2\\1&-1\end{pmatrix} \begin{pmatrix}1&0\\0&\left(-\frac12\right)^n\end{pmatrix}\begin{pmatrix}1&2\\1&-1\end{pmatrix}=\frac13\begin{pmatrix}1 - \left(\frac12\right)^{n-1}& 2\left(1-\left(\frac12\right)^n\right) \\ 1- \left(\frac12\right)^n & 2\left(1+\left(\frac12\right)^n\right)\end{pmatrix}. $$
It follows that 
$$\lim_{n\to\infty} P^n = \begin{pmatrix}\frac13&\frac23\\\frac13&\frac23, \end{pmatrix} $$
and so the stationary distribution is given by $$s = \left(\frac13,\frac23\right).$$
