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 $\lambda=1$.
Best Answer
Take a random walk on the integers where the jump distribution satisfies $$\mathbb{P}(\xi=-1)=\mathbb{P}(\xi=0)=\mathbb{P}(\xi=1)=1/3.$$ Allowing the walk to sit still with positive probability "kills" the periodicity.
This chain is null for the same reason as the simple, symmetric random walk. If the new chain were positive, it would have a unique invariant probability measure $\pi$. The only non-negative solution to $\pi=\pi P$ is a constant sequence $\pi(k)\equiv c$, which cannot be a probability measure.