Convergence of $A^n v$ for a matrix $A$ and a vector $v$ In this question, Martin Sleziak shows that for a complex square matrix $A$, we have $$\lim A^n = 0 $$ if and only if the spectral radius of $A$ is less than $1$. I was wondering what happens if we are interested only in the convergence with respect to a given vector.
To be more specific, suppose now that all eigenvalues of $A$ have modulus $\ge 1$ and let v be some given vector such that $$\lim A^n v = 0. $$ Does it imply that $ v = 0 $?
I suspect the answer is yes, but I don't know how to deal with the case where two distinct eigenvalues of $A$ have the same modulus.
Edit: The comments encouraged me to add more details. Let $J$ be the Jordan form of $A$ and $P$ such that $A=PJP^{-1}$. Then
$$A^n =
P\begin{pmatrix}
J_1^n & \;     & \; \\
\;  & \ddots & \; \\ 
\;  & \;     & J_m^n
\end{pmatrix}P^{-1} $$
where $J_i$ is a Jordan block of size $k_i\times k_i$, so that
$$ J_i^n = \begin{pmatrix}
\lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \binom{n}{2}\lambda_i^{n-2} & \cdots & \cdots & \binom{n}{k_i-1}\lambda_i^{n-k_i+1} \\
 & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1} & \cdots & \cdots & \binom{n}{k_i-2}\lambda_i^{n-k_i+2} \\
 &  & \ddots & \ddots & \vdots & \vdots\\
 &  & & \ddots & \ddots & \vdots\\
 &  & &  & \lambda_i^n & \binom{n}{1}\lambda_i^{n-1}\\
 &  &  &  &  & \lambda_i^n
\end{pmatrix}. $$
Thus every entry of $A^n v$ has the form
$$\tag{1}\label{sum}\sum_{i=1}^m p_i(n)\lambda_i^n$$
where $p_i$ is a complex polynomial such that $\deg p_i< k_i$ for $i=1,\dots,m$.
So essentially what I'm asking is this:
Suppose $|\lambda_i|\ge1$ for $i=1,\dots,m$ and that \eqref{sum} converges to $0$ (when $n\to\infty$). Does it imply that the sequence \eqref{sum} is identically zero?
 A: By Schur triangulation, we may assume that $A$ is lower triangular. Hence all diagonal elements of $A$ (which are the eigenvalues of $A$) have moduli greater than $1$. Since $A$ is lower triangular, the first entry of $A^nv$ is $a_{11}^nv_1$. By assumption, $|a_{11}|\ge1$ and $\lim_{n\to\infty}a_{11}^nv_1=0$. Hence $v_1$ is necessarily zero. But then the second entry of $A^nv$ is $a_{22}^nv_2$. Using a similar argument to the above, we can show that $v_2=0,v_3=0,\ldots$ and so on. Hence $v=0$.
A: Yes: if $A$ has all eigenvalues of modulus at least one, and $\lim_{k\to\infty} A^kv = 0$, then $v=0$.

Let me first give a slick answer in the case that all eigenvalues of $A$ have modulus $r>1$. The spectral radius of $A^{-1}$ is $r^{-1} < 1$. For all $v$,
$$ |v| = |A^{-n} A^n v| \leq |A^{-n}| |A^n v|.$$
The spectral radius $\rho$ satisfies $\rho(B) = \lim_{k \to \infty} |B^k|^{1/k}$ for all matrix norms $|\cdot|$. Hence $\lim_{n\to\infty}|A^{-n}|^{1/n} = r^{-1}$.
Given $\epsilon>0$, there exists a natural number $N$ such that $|A^nv| \leq \epsilon$ and $|A^{-n}|^{1/n} < 1$ for $n \geq N$, hence $|v| \leq \epsilon$. This shows that $v = 0$.

The following works for general (upper-triangular) Jordan blocks of eigenvalue with modulus $\geq 1$. Let $J$ be a block of size $n$ with eigenvalue $\alpha$. We prove that if $\lim_{k\to\infty} J^k v =0$, then $v = 0$. The proof is by induction on $n$. For $n=0$, $v=0$ and we're done. In general, the $n$th entry of $J^k v$ is $\alpha^k v_n$, which tends to zero if and only if $v_n = 0$. Hence $v$ is in the domain of $J$ restricted to the upper $(n-1)\times(n-1)$-block, which is a Jordan block of size one smaller. Now induct to finish the proof.
Essentially, the polynomials $p_i$ you reference are not so complicated since one is always constant.
