Let $A$ be an $m\times m$ matrix with entries $0,1$. We say that $A$ is irreducible if for all $0\leq l,j\leq m-1$ there exists $n$ s.t. $A^n(l,j)>0$. Similarly we say that $A$ is aperiodic if there exists $n$ s.t. for all $0\leq l,j\leq m-1$ $A^n(l,j)>0$
I mean it is clear that aperiodicity $\Rightarrow $ irreducibility. But is it true that in the finite dimensional case also the other implication is true? Because I mean if I have finitely many $l,j$ and I assume that for every $l,j$ there exists $n_{l,j}$ s.t. $A^{n_{l,j}}(l,j)>0$, can't I then define $n:=\max_{l,j} (n_{l,j})$ and then say that for all $l,j$, $A^n(l,j)>0$?
Observe that $A^n(\ell,j)=A^{n_{\ell,j}}(\ell,j)$ is not necessarily true if $n\neq n_{\ell,j}$, so no clever definition of $n$ is going to work as easily as that!
Consider the matrix:
$$A:=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$
$A^2=\mathrm{Id}_2$ so the sequence $A^1,A^2,A^3,\cdots$ is very periodic - so this matrix is certainly not aperiodic, not in the natural sense of the word nor in the sense of the definition in your post. Yet, $A$ is irreducible. So irreducible does not imply aperiodic.