The following theorem is well-known: $$ \lim_k A^k = 0 \text{ if and only if } \rho(A)<1 $$ (see wiki for context and proofs).

What if now $\rho(A)=1$ and $\lambda\neq -1$ for all $\lambda \in Spec(A)$. Then we also have convergence ? (not to $0$ but $A$ converges)

up vote 2 down vote accepted

No. This already fails when all eigenvalues of $A$ are ones. E.g. $A^k=\pmatrix{1&k\\ 0&1}$ doesn't converge when $A=\pmatrix{1&1\\ 0&1}$.

When $|\lambda|=1$ but $\lambda\ne1$, we don't even have convergence of $\lambda^k$.

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