Let $m \in \mathbb{N}$ and let $M \in \{0,1\}^{m \times m}$. Let
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\begin{align*}
\Sigma_M := \{x \in \{1,\dots,m\}^{\mathbb{N}_0} : M_{x_k x_{k+1}} = 1 \text{ for every } k \in \mathbb{N}_0\}
\end{align*}
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be the one-sided topological Markov chain defined by $M$, and let $\sigma: \Sigma_M \to \Sigma_M$ be the left shift map. Assume every symbol is active, meaning that the one-symbol cylinder
is nonempty for every $i \in \{1,\dots,m\}$. Then $(\Sigma_M,\sigma)$ is topologically mixing if and only if $M$ is primitive, that is, for every $i,j \in \{1,\dots,m\}$ there exists $N_{ij} \in \mathbb{N}$ such that $(M^n)_{ij} > 0$ for every integer $n \geq N_{ij}$.