Are irreducible, positiv-definite Markov chains aperiodic? If $M$ is the transition matrix of a discrete Markov chain, and $M$ is both irreducible, symmetric and positiv-definite, is the resulting Markov chain necessarily aperiodic? 
In my intuition, periodicity would correspond to an $-1$-eigenvalue of $M$, but I don't know if that is true or how to formalize it.
 A: I think so... Here's my thought.
Let $M$ be  an irreducible, symmetric and positive-definite $n\times n$ stochastic matrix, with spectrum $\sigma(M)=\{\lambda_1, \lambda_2,\ldots, \lambda_n\}$.


*

*Since $M$ is stochastic, we have that $\lambda_1=1$.

*Since $M$ is symmetric we have that $M$ is diagonalizable.

*Since $M$ is positive definite, we have that all eigenvalues of $M$ are positive.


Thus, there exists an invertible matrix $U$ s.t.:
$$M= U \cdot J\cdot U^{-1},$$
where $J=\begin{bmatrix} 1  &&&\\ & \lambda_2 & \\ &&\ddots &\\ &&&\lambda_n\end{bmatrix}$ is the Jordan normal form of $M$.
By Perron - Frobenius theorem for non - negative, irreducible matrices, we have that eigenvalue $\lambda_1=1$, which happens to be the spectral radius of $M$ has algebraic multiplicity $1$ and for all other eigenvalues we have $$|\lambda_i|=\lambda_i<1, \, i=2, \ldots, n.$$
Thus, 
$$\begin{array}[t]{l}
M^k=U\cdot \begin{bmatrix} 1^k & && \\ & \lambda_2^k  && \\ &&\ddots & \\ &&& \lambda_n^k\end{bmatrix}\cdot U^{-1}\\\\
 \lim_{k\to\infty}M^k=U\cdot \begin{bmatrix} 1 & && \\ & 0  && \\ &&\ddots & \\ &&&0\end{bmatrix}\cdot U^{-1} \end{array}$$
must be a stochastic matrix, since $M^k$ is a stochastic matrix for every $k\in \mathbb N$.
Now it is easy to prove that 
$$ U \cdot \begin{bmatrix} 1 & && \\ & 0  && \\ &&\ddots & \\ &&&0\end{bmatrix}\cdot U^{-1}=\begin{bmatrix} \pi_1 & \pi_2 & \cdots & \pi_n\\
\pi_1 & \pi_2 & \cdots & \pi_n\\
\vdots & \vdots & \ddots & \vdots\\
\pi_1 & \pi_2 & \cdots & \pi_n\end{bmatrix}=\mathbf{\Pi}$$
plus $\displaystyle \sum_{i=1}^n \pi_i =1$.
Thus, $M$ must be aperiodic, since $\lim_{k\to\infty}M^k$ exists and equals to  a stochastic matrix with identical rows. 
Note: We can prove that $\mathbf{\Pi}$ has all its elements strictly positive, since $$\pi = \begin{bmatrix} \pi_1 & \cdots & \pi_n\end{bmatrix}$$ is a left eigenvector which corresponds to Perron-Frobenius eigenvalue $\lambda=1$.
A: In addition to your  comment:
Let $P$ be a stochastic, irreducible matrix. Since $P$ is irreducible it will be either aperiodic or periodic.
$P$ aperiodic$\iff\displaystyle\lim_{k\to\infty}P^k$ exists.
$P$  periodic  matrix  with period $d>1$ $\iff \displaystyle\lim_{k\to\infty}P^k$ does not exist.
We can prove the latter using cyclic subclasses.
A: It follows immediately from the definition of positive definite matrix that  $M_{i,i} = (e_i, M e_i)>0$ for all $i$. That is, all diagonal elements of $M$ are strictly positive. This implies aperiodicity of the resulting chain.  
