# Periodic markov chain - finding initial conditions causing convergence to steady state?

Let's say I have the Markov chain $\begin{bmatrix} 0 && 1 \\ 1 && 0 \end{bmatrix}$. From my understanding this is periodic and thus its convergence depends on the initial conditions. However, I have two questions about this:

1) How am I able to compute that it's periodic (and its period) from the transition matrix? I found the following, but I'm not sure how to interpret this for a transition matrix.$$d( x) = \{\gcd{n \in \mathbb{N}+ : P^n ( x, x) > 0}\}$$

2) Am I able to compute the vectors which will cause this periodic Markov chain to converge to a steady state? I think for this matrix the only vector $v_0$ which does this is $\begin{bmatrix} \frac{1}{2} \\ \frac{1}{2}\end{bmatrix}$.

Note that for $$A= \begin{bmatrix} 0 && 1 \\ 1 && 0 \end{bmatrix}$$ we have $$A^2=I$$ thus it is periodic with period $2$.
The eigenvalues of $A$ are the roots of its characteristic polynomial, $$P(\lambda ) = \lambda ^2 -1$$ namely $\lambda =\pm 1$ with corresponding eigenvectors of $\begin {bmatrix} 1\\1\end {bmatrix}$ and $\begin {bmatrix} 1\\-1\end {bmatrix}$
The eigenvalue $\lambda =1$ gives us the equilibrium state which is $\begin {bmatrix} 1\\1\end {bmatrix}$ Upon scaling we get $V=\begin{bmatrix} {1/2}\\{1/2} \end{bmatrix}$
• The eigenvector associated with eigenvalue of $1$ is the equilibrium state. For the first question, if the matrix is periodic, the period could be found by the smallest positive $n$ for which $A^n=I$ – Mohammad Riazi-Kermani Apr 25 '18 at 8:19