# Steady state probabilities of divergent Markov matrices

I have a problem for which the Markov matrix turns out to be the following:

$$P = \begin{pmatrix} 0 & 0.5 & 0 & 0.5\\ 0.5 & 0 & 0.5 & 0 \\ 0 & 0.5 & 0 & 0.5\\ 0.5 & 0 & 0.5 & 0\\ \end{pmatrix}$$

This matrix has eigenvalues $$-1$$ and $$1$$. Hence, it does not converge to some matrix $$A$$ when raised to some power $$n$$. However, there exists a solution to $$Pv = v$$. This is basically, the eigenvector corresponding to the eigenvalue of $$1$$. Is $$v$$ the steady state probabilites?

This Markov chain is periodic with period $$2$$, meaning that for any initial state $$i$$, the limiting probability $$\lim_{n\to\infty}\mathbb P(X_n=j\mid X_0=i)$$ does not exist. However, $$\{X_n\}$$ is irreducible, positive recurrent, and has finitely many states, so a unique stationary distribution $$\pi$$ exists. Since $$P$$ is doubly stochastic (both the rows and the columns sum to one), it follows that $$\pi$$ is the uniform distribution over $$0,1,2,3$$, i.e. $$\pi_0=\pi_1=\pi_2=\pi_3=\frac14$$.
Now, because $$\{X_n\}$$ is periodic with period $$2$$, the limits of $$P^{2n}$$ and $$P^{2n+1}$$ as $$n\to\infty$$ exist. In particular, $$P^{2n} = \left( \begin{array}{cccc} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ \end{array} \right),\quad P^{2n+1} = P$$ for all $$n$$, so $$\lim_{n\to\infty} P^{2n} = \left( \begin{array}{cccc} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ \end{array} \right),\quad \lim_{n\to\infty}P^{2n+1} = P.$$