For a matrix $A$, does convergence of a subsequence $A^{k_n}\to0$ imply convergence of the sequence of powers $A^n\to0$?

Question

In particular, I'd like to know whether $A^{n(n-1)/2}\to0$ implies $A^n\to0$.

I thought of using Gauss' eureka theorem to write $A^n=A^{k_n(k_n-1)/2}A^{l_n(l_n-1)/2}A^{m_n(m_n-1)/2}\to0$. Is there a simpler way?

For the general case, we have $(\det A)^{k_n}\to0$ which implies $(\det A)^n\to0$ (both equivalent to $|\det A|<1$), but this isn't helpful; for example a projection has $\det P=0$ but $P^n=P\not\to0$.

If the jumps $k_{n+1}-k_n$ are bounded by $\Delta$, so every $n$ can be written as $n=k_{m_n}+r_n$ with remainder $0\leq r_n<\Delta$, then (using the Frobenius norm) $\lVert A^n\rVert\leq\lVert A^{k_{m_n}}\rVert\lVert A\rVert^{r_n}\leq\lVert A^{k_{m_n}}\rVert\max(1,\lVert A\rVert^\Delta)\to0$, and thus $A^n\to0$.

Answer 1

We can assume that we have a sub multiplicative norm.

Choose $k$ such that $\|A^k\| < 1$.

Then $\|A^{m+nk}\| \le (\max_{i=0,...,k-1}) \|A\|^i \|A^k\|^n$.