Show $A^k$ doesn't converge to $0$ Let $A \in \mathbb C^{n \times n}$. Let $r=\{\max \lvert \lambda \rvert \text{ such that }\lambda \in \mathbb C \text { is an eigenvalue of } A\}$. If $r\geq 1$ show that $A^k$ doesn't converge to the zero matrix as $k\to \infty$.
Here is the proof I wrote :
$A=P^{-1}JP$ with $P$ inversible and $J$ in a Jordan normal form. Let $\lambda_m$ be the eigenvalue that has $\lvert \lambda_m \rvert=r\geq 1$.
We have that $A^k=P^{-1}J^kP$. Then I show by induction that $J^k$ has a coefficient $A_{ij}=a\lambda_m^k+b$ where $a, b\in \mathbb C^{n}$ (they do not matter). Now since that $\lvert \lambda_m \rvert \geq 1$ then $A_{ij}$ diverges when $n$ goes to $+\infty$ that means that $A^k$ does not converge to the zero matrix.
Is it correct ?

This is almost correct. You don’t need to use the normal form. You have an eigenvector $v_m$, so $A^k v_m = \lambda^k_m v_m$. In particular, there’s no problem with multiplication of Jordan matrices. This is the weakest spot in your proof.
