Aperiodicity of diagonally dominant Markov Matrices. I would greatly appreciate any further thoughts on the following proposition:
"Given $P \in \mathbb{R}^{n\times n}$ a diagonally dominant Markov matrix; it follows that $P$ will always converge to a steady state."
I am not too familiar with the Jordan normal form, though I suspect it can shed some light on the proposition. I have only been able to prove that -1 cannot be an eigenvalue and suspect a similar device can be used to prove it cannot be complex:
$$|p_{ii} - z| \leq \sum_{j = 1\\j\neq i}^{n}|p_{ij}|\\
|p_{ii} + 1| \leq \sum_{j = 1\\j\neq i}^{n}|p_{ij}|\\
|p_{ii} + 1| \leq |p_{ii}|$$
The last inequality being absurd. I brought up the Jordan form as it can clarify the existence of $P^{\infty}$, an equivalent statement.
 A: In general, any Markov chain whose transition matrix has positive diagonal entries must be aperiodic. It is easiest to see this through the definition of a period: for each state $i = 1,\dots,n$, we find that $\Pr(X_1 = i|X_0 = i) = p_{ii} > 0$. Thus, the period $\gcd\{n>0:\Pr(X_n = i|X_0 = i)>0\}$ of each state $i$ must be equal to $1$.
Alternatively, we could take an eigenvalue-based approach. If $P = I$, then $P$ has no complex eigenvalues, so clearly $P$ is aperiodic. Suppose then that that $P \neq I$. Let $\alpha = \min_{i=1,\dots,n} p_{ii}>0$; because $P \neq I$, we have $\alpha < 1$. Let $Q$ denote the transition matrix $Q = \frac 1{1-\alpha}(P - \alpha I)$. Notably, $P$ can be seen as a "lazy" version of the Markov chain described by $Q$. That is, we have $P = (1-\alpha)Q + \alpha I$.
By the Perron Frobenius theorem, each eigenvalue $\lambda$ of $Q$ satisfies $|\lambda| \leq 1$. Now, each eigenvalue of $P$ is of the form $(1 - \alpha)\lambda + \alpha$ for some eigenvalue $\lambda$ of $Q$. By the triangle inequality, we have
$$
\left|(1-\alpha)\lambda + \alpha \right| \overset{!}\leq (1-\alpha)|\lambda| + \alpha \leq (1-\alpha) + \alpha = 1,
$$
where the first inequality is an equality if and only if $\lambda$ is a non-negative real number. Conclude that if $P$ has an eigenvalue $\mu = (1-\alpha)\lambda + \alpha$ with $|\mu| = 1$, then it must hold that $\mu= 1$.
