Irreducible Markov chain with $P^2=P$, then all entries are equal

Question

For a matrix $P$, suppose $P^2=P$ and all entries of $P$ are positive, and $\sum_j P_{ij}=1$. Prove $P_{ij}=P_{ii}$ for all $i,j$. The original problem is from stochastic processes. I reduced it into this linear algebra problem, and wonder if there is a result in linear algebra about it.
The original problem is exercise 2.22 of Sidney Resnick's book "Adventures in Stochastic Processes". It says suppose an irreducible Markov chain, not neccessarily finite many states, has the property that $P^2 = P$. Show that $p_{ij} = p_{jj}$ for all $i, j\in S$.

Answer 1

You have misunderstood the original problem; note that "$p_{ij} = p_{ii}$" is different from "$p_{ij} = p_{jj}$."

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Because $P^2 = P$, each column $P_{(i)}$ of $P$ is an eigenvector of $P$, associated with the eigenvalue $1$. By the Perron-Frobenius Theorem, all $P_{(j)}$ for $1 \leq j \leq n$ must be positive multiples of $P_{(1)}$, denoted as $$ P_{(j)} = \alpha_j\cdot P_{(1)} $$ where $\alpha_j > 0$ and $\alpha_1 = 1$. We have $$ \sum_{j} P_{ij} = \sum_{j} \alpha_jP_{i1} = P_{i1}\sum_{j}\alpha_j = 1 $$ for all $i$. It is hence not hard to conclude that all elements of $P_{(1)}$ are the same.