How to prove that irreducible aperiodic finite markov chain's transition matrix can only have one eigenvalue of modulus 1.
I know how to prove irreducible aperiodic finite markov chain's transition matrix can only have one eigenvalue of 1, given by the stationary probability distribution.
However, how to prove that there is no left eigenvector $u$ such that
$$u P = \alpha u$$ with P the transition matrix, $|\alpha| = 1$ and $\alpha \neq 1$.
My attempt:
$\alpha \neq 1$ and $|\alpha| = 1$
Case1: $\alpha$ is a $k$th root of unit for some integer $k$.
$\alpha^k = 1$, then $P^k$ has at least two different stationary states. It implies that $P$ has periodicity $> 1$, which is contradict to the assumption of aperiodic.
Case2: $\alpha^n = 1$ for some irrational number $n$.
However, I have no idea how to exclude this possibility.