Limit of Jordan Form and the largest eigenvalue As an intermediate step of my homework solution, I want to show that
$$\lim_{k\to\infty} J^k=0$$
if $\max |\lambda_i|<1$, where $J$ is the Jordan form of a matrix $A$ and $\lambda_i$ denotes the $i$-th eigenvalue of $A$.
Assume this is true. I try to prove it entry by entry. This is really messy. 
Is this statement indeed true? is there any shortcut to prove it?

Going entry-by-entry is enough.
You can compute a general form of $J_i^k$ (where $J_i$ is the $i$th block)
to see that the entries that are $j$ steps above the main diagonal equal $\binom{k}{j} \lambda^{k-j}$, which tends to zero as $k \to \infty$ since $|\lambda| < 1$ and $\binom{k}{j} \le  k^j / j! = O(k^j)$.
