How many stationary distributions does a time homogeneous Markov chain have? I've been given the following definition:

For a THMC with one step transition matrix $\mathbf{P}$, the row vector $\mathbf{\pi}$ with elements $(\pi_{i})_{i \in S}$ (where $S$ is the state space) is a stationary distribution iff $\mathbf{\pi \; P} = \mathbf{\pi}$

However, I also know that many THMCs will have multiple stationary distributions. 
This leaves me with the following questions:


*

*How can you tell how many stationary distributions a THMC has?

*How can I show that a THMC has only one stationary distribution?

*The equation $\mathbf{\pi \; P} = \mathbf{\pi}$ looks like it should only have one solution, so how is it possible to have multiple stationary distributions?

 A: Stationary Distributions:
Let $\mathbf{P}$ be the transition probability matrix of a homogeneous Markov chain $\{X_n, n \geq 0\}$. If there exists  a probability vector $\mathbf{\pi}$ such that $$\mathbf{\pi} \mathbf{P} = \mathbf{\pi} \:\:\:\:\:\:\: (1)$$
then $\mathbf{\pi}$ is called a stationary distribution for the Markov chain.
Equation $(1)$ indicates that a stationary distribution $\mathbf{\pi}$ is a (left) eigenvector of $\mathbf{P}$ with eigenvalue $1$.  Note that any nonzero multiple of $\mathbf{\pi}$ is also an eigenvector of $\mathbf{P}$. But the stationary distribution $\mathbf{\pi}$ is fixed by being a probability vector;
that is, its components sum to unity.
Limiting Distributions:
A Markov chain is called regular if there is a finite positive integer $m$ such that after $m$ time-steps, every state has a nonzero chance of being occupied, no matter what the initial state. Let $A > 0$ denote that every element $a_{ij}$ of $A$ satisfies the condition $a_{ij} > 0$. Then, for a regular Markov chain with transition probability matrix $\mathbf{P}$, there exists an $m > 0$ such that $\mathbf{P}^m > 0$.  For a regular homogeneous
Markov chain we have the following theorem:

Thm:
Let $\{X_n, n \geq 0\}$ be a regular homogeneous finite-state Markov chain
with transition matrix $\mathbf{P}$.
Then
$$\lim \limits_{n \to \infty} \mathbf{P}^n = \mathbf{\hat{P}}$$
where $\mathbf{\hat{P}}$ is a matrix whose rows are identical and equal to the stationary distribution $\mathbf{\pi}$ for the Markov chain defined by $(1)$. This is a sufficient condition for $\mathbf{\pi}$ to be unique.

A: As mentioned in JKnecht's answer, a Markov chain being regular is just a sufficient condition for a unique stationary distribution. A necessary and sufficient condition is that there's only one closed communicating class.
Having one closed communicating class is not the same as being irreducible, for example
$$P=\left[\begin{matrix}0.5 & 0.5\\
     0 & 1\end{matrix}\right], \pi=\left[\begin{matrix}0 & 1\end{matrix}\right],\pi P=\pi$$
where $P$ is reducible and has one closed communicating class (consisting of only one state), and the unique stationary distribution is $\pi$.
See also this answer https://math.stackexchange.com/a/252244/291503, this answer http://qr.ae/TUTSij, and this tutorial http://wwwf.imperial.ac.uk/~ejm/M3S4/NOTES3.pdf 
