# Markov chain which is also a projection

Let $P$ be the transition matrix of a Markov chain, and assume that $P^2=P$. One immediate conclusion is that $P=P^\infty$.

Furthermore, assume that there is a state $i$ such as each state $j$ (including $j=i$), is accessible from $i$: $i\rightarrow j$.

I am fairly sure that this means that $P$ is irreducible (and thus, has only one stationary distribution, and all the columns of its matrix are equal), but I can't find a quick argument why this is true.

I think I can get a proof using the coefficients, to show that if there is a state $j$ such as $j\not\rightarrow i$, then $P^\infty\neq P$, because the probability of being in state $j$ will "grow over time", but I was wondering whether there was a more direct argument, perhaps making use of a standard result.

• You defined irreducibility in your second sentence. To say that something is irreducible does not need proof. – tintinthong Sep 4 '17 at 12:35
• perhaps take a look at this link. stats.stackexchange.com/questions/126412/… – tintinthong Sep 4 '17 at 12:37
• @tintinthong I only assume one direction: there exists a state $i$ with transitions to all other states. I'm looking for an easy way to show that the reverse is also true (= all states $j$ have a transition to $i$) if the Markov chain is a projection. – Ted Sep 4 '17 at 12:40
• Ah, that link is interesting. The result I'm looking for seems to be a consequence of Theorem 1.16 (page 48) of this book. This seems to use some heavy artillery to prove it, and I'm curious to know whether there is a more elementary proof, but this'll be enough to cite this. Thanks! – Ted Sep 4 '17 at 13:01
• It was just a matter of terminology; a Markov chain is a process, a transition matrix gives the probabilities of the possible state-transitions in that process. I think it's fine now. – Glen_b Sep 4 '17 at 23:14