Show that there does not exist a unique stationary distribution.

Question

So I was doing some self study and came across a proposition in one of my chemical engineering course's prescribed textbooks. I can't quite get the proof out. It's to do with a particle moving through a medium such that when it makes contact with to either of two plates $L$ units apart (i.e. one at $0$ and one at $L$), it remains there.

Consider that the movement of a single particle follows a random walk which can be described by a Markov chain with states $[0, L]$ where $P(X_n = -1) = p_{-1}$,$P(X_n = 0) = p_{0}$ and $P(X_n = 1) = p_{1}$ with $p_{-1} + p_{0} + p_{1} = 1$. Show that if states $0$ and $L$ are completely absorbing, then there does not exist a stationary distribution. Hint: Start by considering ${\pi} = \pi P$

This makes sense intuitively since we have two recurrent classes $\{0\}$ and $\{L\}$ and one transient class $\{1, 2, ..., L - 2, L - 1\}$. However, once I try and expand ${\pi} = \pi P$, I don't know how to proceed next. Ideally I'd like a few more hints rather than an answer.

Answer 1

$(1,0,0,...,0)$ and $(0,0,...0,1)$ are two invariant distributions so uniqueness fails.

Answer 2

Write your equation as $\mathbf\pi(P-I)=0$. The first and last rows of $P-I$ are zero, so $(1,0,\dots,0)$ and $(0,\dots,0,1)$ are obvious independent solutions of the equation.