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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?

Rodrigo de Azevedo
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1 Answers1

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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 $\binom{k}{j} \le k^j / j! = O(k^j)$.

angryavian
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  • To avoid such a calculation on individual entries, one could use the Jordan–Chevalley decomposition. Roughly the same inequality would be needed. – stewbasic Mar 28 '18 at 04:40
  • What do you mean by $\binom{k}{j}\leq k^j/j!=O(k^j)$? Why is this true? –  Mar 28 '18 at 14:25