Given irreducible markov chain with $P=P^n$, show that all rows of $P$ are equal

Question

So I'm given that $P=P^n$ for $n=2, 3, …$, and I've deduced that for any distribution $\pi$, it must be true that $\pi P$ is a stationary distribution. But I'm stuck at this point, I've tried reducing it to a linear algebra problem by multiplying out the matrices in attempt to show that $p_{1,j}=p_{i,j}$ for all $j$, but this isn't leading me anywhere at all and the notation is getting messy. I'm wondering if I'm missing something simple? Any help would be great.

Answer 1

If the Markov chain is irreducible and aperiodic, then there is a unique stationary distribution, Additionally, in this case $P^k$ converges to a rank-one matrix in which each row is the stationary distribution $\pi$.

(This can be proved by using the fact that the Largest Eigenvalue of a Stochastic Matrix is $1$ and using perron-frobenius theorem(Perron projection)).

Now using that fact that $P=P^n$ for all natural number $n$ we trivially get the result as distinct constant sequences cannot converge to the same thing.

Answer 2

Notice that for an irreducible positive matrix, the spectral radius is a simple eigenvalue. In your case, $1$ is an eigenvalue with multiplicity 1.

$P^2=P$ is all you need, since it means that $P$ is a projection. A projection has only eigenvalues 0 and 1 and it is diagonalizable (you can prove it by noticing that the minimal polynomial of $P$ divides $x^2-x$).

As a consequence, $P$ is diagonalizable, has $n-1$ eigenvalues 0, and an eigenvalue $1$. In particular, it has rank 1, so all rows are linearly dependent (meaning that they are multiple of a fixed vector $v$).

To conclude, you can notice that the sum of the elements in each row is 1, so they must be equal.