Suppose $p, q \gt 0$ and $p+q = 1$. Suppose the PTM $P$ of a Markov chain is \begin{bmatrix} p &q \\ 0 & 1 \end{bmatrix} Is the stationary distribution unique for such PTM? If so, how do I prove that converges in the long run regardless of initial state? Also, how do I prove that the long run probabilities of each state must converge to the stationary distribution? If it's not unique how many such distributions are there?
After solving the equation $\pi =\pi P$ where $\pi $ is the stationary distribution $\array{[x & y]}$ where $x+y = 1$.
I found that $x = 0$ thus $y = 1$.
But I'm not sure if this is unique. If it is, I don't know how to prove it converges in the long run regardless of the initial state or whether the long run probabilities converge to the stationary distribution.
The basic result here is based on the Perron-Frobenius theorem. It says that if a Markov chain on a finite state space is irreducible and aperiodic then it has a unique stationary distribution which is approached in the long run, regardless of the initial distribution.
If the Markov chain is just aperiodic then it still approaches some stationary distribution in the long run but this distribution may depend on the initial distribution. (The exact character of the dependence can be quantified; ask in the comments if you'd like me to explain how.)
If the Markov chain is just irreducible then there is still a unique stationary distribution but it may fail to be approached in the long run, depending on the initial distribution.
It turns out that your chain falls into the second case, so that only thinking about the previous three results, there could be multiple stationary distributions, but one of them will be approached no matter what distribution you start with. However, there is actually just one, and the "probabilistic" reason for that is that you will eventually hit state $2$ if you start from state $1$, after which you will stay there forever. Thus the only stationary distribution is the one concentrated on state $2$, as you computed.
In your particular case this could have been computed using pure linear algebra: your matrix is triangular and the diagonal elements are $1$ and $p$. So these are the eigenvalues. This and the fact $0<p<1$ is enough to get the conclusion by diagonalization.