Markov chain - Stationnary distribution - Unique

Question

Consider the following respective infinitesimal generators of Markov chains in continuous time:

\begin{equation} A=\begin{bmatrix} -4 & 1 & 3 \\ 3 & -5 & 2 \\\ 0 & 3 & -3 \end{bmatrix}. \end{equation}

\begin{equation} B=\begin{bmatrix} -3 & 2 & 1 \\ 0 & 0 & 0 \\\ 0 & 0 & 0 \end{bmatrix}. \end{equation}

Decide for each of the considered Markov chains, if it admits a stationary distribution, whether it is unique.

I did a course in stochastic processes it long ago, and I can't find the information I want on the Internet. Does someone could give me the theorem that would allow me to conclude to the existence and uniqueness of a stationary distritibution for $A$ and $B$.

Answer 1

A continuous Markov chain $X(t)$ with the state space $S$ and the generator matrix $G$ has a probability distribution $\pi$ on $S$ that is a stationary distribution for $X(t)$ if and only if it satisfies $\pi G=0$.

And it is unique if the Markov chain is irreducible, i.e. if its state space is a single communicating class; in other words, if it is possible to get to any state from any state.

For the general non-irreducible case, you can partition the state space into classes of recurrent states and a class of transient states. Then each recurrent class is associated with an invariant distribution. All invariant distributions are linear combinations of those invariant distributions associated with recurrent classes. Thus if there are at least two recurrent classes, then there are infinitely many invariant distributions

Answer 2

The stationary distribution of a continuous-time Markov chain with finitely many states is unique iff the Markov chain has only one recurrent class of states, i.e. there is only one set $S$ of states such that it is possible to go from any state $i$ in $S$ to any other state $j$ in $S$ by allowed transitions (corresponding to positive entries of the generator), but it is not possible to leave the set $S$ by allowed transitions.