# Cyclic Markov Chain

Say I have 3 states: $$A,B,C$$, where the transition probabilities are $$P_{B\leftarrow A}=P_{C\leftarrow B}=P_{A\leftarrow C}=p$$ and $$P_{i\leftarrow i}=1-p$$ for $$i=A,B,C$$. All other transition probabilities are zero. The transition matrix is $$P=\begin{pmatrix} 1-p & 0 & p\\ p & 1-p & 0\\ 0 & p & 1-p\\ \end{pmatrix}.$$

Here, $$0\leq p\leq 1$$.

I find that the stationary distribution for $$P$$ is$$\rho^*=\begin{pmatrix} 1/3\\ 1/3\\ 1/3\\ \end{pmatrix}.$$

If $$p=0$$, then any initial distribution vector $$\rho_0$$ is a stationary distribution; if $$p=1$$, however, then the components of any initial $$\rho_0\neq \rho^*$$ will keep hopping/cycling positions. In other words, if I start in say, state $$A$$ ($$\rho_0=(1,0,0)^T$$), then the system will never reach a time-independent steady state.

For any $$p$$ in between 0 and 1, it seems that I will always converge to $$\rho^*$$.

Two Questions:

1) Can whether or not a state is steady be determined by the eigenvalues of $$P$$, e.g. whether they are complex or not?

(Eigenvalues of $$P$$: $$\lambda_1=1, \lambda_2=\frac{1}{2}(2 - 3 p - i\sqrt{3}p), \lambda_3=\frac{1}{2}(2 - 3p + i\sqrt{3}p)$$).

2) Even though $$P$$ does not satisfy the detailed balance property, it stills converges to an equilibrium state (depending on the value of $$p$$). Why is this?