Markov Chain Eigenvalues Proof Suppose $M$ is a reversible Markov chain with respect to $\pi$, on the finite set $X$. Note, operator M defined by $Mg(x)=\sum_y M(x,y)g(y)$, where $g$ is a real-valued function. Show that if M is not periodic, then $-1$ is not an eigenvalue of M.
Proof:
All we have to show is that $-1$ is not an eigenvalue of $M$. We do this by contradiction.
Note that the definition of $M$ being aperiodic means for every $x,y \in X$ that $gcd(\{n: M^{n}(x,y)>0\})=1$
Now, suppose by contradiction that $f$ is an eigenvector of $M$ with eigenvalue $-1$. Thus $Mf=-f$. Choose $x^*$ satisfying $|f(x^{*})| \geq |f(y)|$ for all $y\in X$.
We have then that $f(x^{*})=-Mf(x^{*})=-\sum_y M(x^{*},y)f(y)$.
This implies that $f(y)=-f(x^{*})$ for all $y$ with $M(x^{*},y)>0$. Thus all values of $f$ one step from $x^{*}$ are equal to $-f(x^{*})$.
So apparently, all of the steps are correct and do not need further justification. However, apparently I am missing a step now to finish the proof and show that the period of the state $x^{*}$ is greater than 1. Specifically, I need to show that the period of $x^{*}$ is even. That is, I need to show that the $gcd(\{n:M^{n}(x^{*},x^{*})>0\})=2k$ for some positive integer $k$ (period is even).
Thus, I am just not sure how to complete the proof. 
How does $f(y)=-f(x^{*})$ for all $y$ with $M(x^{*},y)>0$ imply that the period of $x^{*}$ is even, $gcd(\{n:M^{n}(x^{*},x^{*})>0\})=2k$ for some positive integer $k$? (really, I just have to prove the period is $\geq 2$, so that it is not aperiodic). Any help to help finish this last step to complete the proof would be much appreciated! Thank you!
 A: A quick observation: since $f$ is an eigenvector (by definition non-zero), we must have $f(x^*) \neq 0$.
I'll now extend your argument a bit: for any integer $k \geq 1$, we have 
$$
f(x^*) = (-1)^k M^k f(x^*) = (-1)^k\sum_{y}M^k(x^*,y) f(y)
$$
so we must have $f(y) = (-1)^k f(x^*)$ for all $y$ such that there is a walk of length $k$ from $x^*$ to $y$ of non-zero probability.
Now, suppose (for the purpose of contradiction) that $x^*$ does not have an even period.  That is, we suppose that for some odd number $2n+1$, we have $M^{2n+1}(x^*,x^*) > 0$.  So, there exists an odd cycle at $x^*$ with non-zero probability: that is, there is a sequence $ x_0,x_1,\dots,x_{2n},x_{2n+1}$ with $x_0=x_{2n+1} = x^*$ such that $M(x_i,x_{i+1}) > 0$ for $i = 0,1,\dots,2n$.
Consider $x_1$.  Since $M(x^*,x_1) > 0$, we must have $f(x_1) = -f(x^*)$.  On the other hand, the existence of our cycle means that 
$$
M^{2n}(x_1,x^*) \geq M(x_1,x_2) M(x_2,x_3) \cdots M(x_{2n},x^*)> 0
$$ 
which is to say that $M^{2n}(x_1,x^*) > 0$. The fact that $M$ is reversible means that $M^{2n}(x^*,x_1) > 0$. By the earlier argument, this means that $f(x_1) = (-1)^{2n}f(x^*) = f(x^*)$.  That is, we have concluded that
$$
f(x_1) = f(x^*) = -f(x^*)
$$
which is impossible since $f(x^*)$ is non-zero.  The conclusion follows.
