I stumbled across an odd fact while thinking about Markov chains, and I have an odd proof for it.
Claim: Let $P$ be the transition matrix of a finite irreducible aperiodic Markov chain, and assume that $P$'s left eigenvectors are orthogonal. Then $P$ is doubly stochastic.
Is this a special case of some well known result? In any case, here's my proof.
Proof: Let $n$ denote the number of states. Since $P$ is the transition matrix of a finite irreducible aperiodic Markov chain, there is a unique stationary distribution $\pi$ and for any initial distribution $\mu$ there holds $\mu P^t \rightarrow \pi$ when $t \rightarrow \infty$. Let $\mu$ be an arbitrary distribution, let $t \geq 0$, and consider the inner product $\langle \mu P^t,\pi \rangle$.
Let $\{v_i\}$ be an orthonormal basis of left eigenvectors for $P$ whose corresponding eigenvalues are $\{\lambda_i\}$ with $v_1 = \pi/\|\pi\|_2$ and $\lambda_1 = 1$, and write $\mu = \sum_{i=1}^n{\alpha_i v_i}$. Then $$ \langle \mu P^t,\pi \rangle = \langle \sum_{i=1}^n{\alpha_i v_i P^t},\pi \rangle = \sum_{i=1}^n{\alpha_i \lambda_i^t \langle v_i,\pi\rangle} = \alpha_1 \langle v_1,\pi \rangle = \alpha_1 \|\pi\|_2. $$
So $(\langle \mu P^t,\pi \rangle)_{t=0}^{\infty}$ is a contant series. On the other hand, $\mu P^t \rightarrow \pi$ so $\langle \mu P^t,\pi \rangle \rightarrow \langle \pi,\pi \rangle$.
Now, if a constant series converges to a value, then all of the elements of the series are equal to the value. Therefore $\langle \mu,\pi \rangle=\langle \pi,\pi \rangle$.
This holds for any distribution $\mu$. In particular consider indicator vectors $e_i$: We have that for any $i,j \in [n]$, $$ \langle e_i,\pi\rangle = \langle \pi,\pi\rangle=\langle e_j,\pi\rangle \Rightarrow \pi_i = \pi_j. $$ So $\pi$ is uniform. Thus, $\bar{1}\cdot P = \bar{1}$, so $P$'s columns all sum to 1 and $P$ is doubly stochastic. $\square$
This is a very strange proof - I've never seen a proof that shows that two values are equal because the first belongs to a constant series that converges to the second value.
Can anyone think of a direct, simple proof for this claim? Is this claim a special case of some well known result in linear algebra?
