If $P$ is the transition matrix of a reversible Markov chain, why are its eigenvalues real?

If the MC is reversible, then $\pi(x)P(x,y) = \pi(y)P(y,x)$ for some distribution $\pi$ and for all states $x,y$. I see that if $\pi$ is the uniform distribution, then $P$ is symmetric and thus has real eigenvalues. But what if $P$ is not symmetric?

(Typed after the answer of Fnacool was already accepted, only a complement that may make the same argument "human" / structural.)

The usual argument considers the Hilbert spacce $H=L^2(\pi)$, and the operator $P$ (well, same letter, sorry) on $H$ given by $$ (Pf)(x)=\sum_{y\in\Omega}P(x,y)f(y)\ . $$ It is selfadjoint, $$ \begin{aligned} \langle Pf, g\rangle &= \sum_{x}\pi(x)\; (Pf)(x)\;\bar g(x)\\ &= \sum_{x,y}\pi(x)\; P(x,y)\;f(y)\;\bar g(x)\\ &= \sum_{x,y}\pi(y)\; P(y,x)\;f(y)\;\bar g(x)\\ &= \sum_{x,y}\pi(y) \;f(y)\;\overline{P(y,x)\; g(x)}\\ &= \sum_{y}\pi(y)\; f(y)\;\overline {Pg(y)}\\ &= \langle f, Pg\rangle \end{aligned} $$ so the operator $P$ is selfadjoint (and a contraction). Its eigenvalues are thus real and contained in $[-1,1]$.

You're pretty close. Here's what's missing.

We will assume further that $P$ is irreducible so that up to a multiplicative constant: $\pi$ is unique and strictly positive.

Let $D= \mbox{diag} (\sqrt{\pi(1)},\dots, \sqrt{\pi(n)})$. Let $Q = D P D^{-1}$. Observe that

$$ Q_{i,j} = (D P D^{-1})_{i,j} = \sqrt{\pi(i)} p_{i,j} \frac{1}{\sqrt{\pi(j)}}.$$

By assuming

$$(*)\quad \pi(i) p_{i,j} = \pi(j) p_{j,i},$$ we have

\begin{align*} Q_{j,i} &\overset{\mbox{def}}{=} \sqrt{\pi(j)} p_{j,i} \frac{1}{\sqrt{\pi(i)}} \\ & =\frac{1}{\sqrt{\pi(j)}} \pi (j) p_{j,i} \frac{1}{\sqrt{\pi(i)}}\\ & \overset{(*)}{=} \frac{1}{\sqrt{\pi(j)}} \pi(i) p_{i,j} \frac{1}{\sqrt{\pi(i)}} \\ & = \sqrt{\pi(i)} p_{i,j} \frac{1}{\sqrt{\pi(j)}}\\ & = Q_{i,j}. \end{align*}

Therefore $Q$ is symmetric. As a result, all its eigenvalues are real and it is diagonalizable. Since $P$ and $Q$ are similar, the same holds for $P$.