We define the matrix $Z_n = (z_{i,j}) \in \mathcal{M}_n(\mathbb{R})$ by $z_{i,j} = \begin{cases} 1 & \text{if } 1 \leqslant i < n \text{ and } j = i+1 \\ 1 & \text{if } (i,j) \in \{(n-1,1),(n,2)\} \\ 0 & \text{in all other cases} \end{cases}$
Show that $Z_n^{n^2-2n+2} = 2^{n-1} Z_n$ and recover the fact that $Z_n$ is not primitive.