Let $V$ be a finite-dimensional normed vector space. Let $L:V\rightarrow V$ be a linear operator and let $v\in V$. Assume that there is a sequence $\{n_i\}_{i=1}^\infty\subset\mathbb{Z}$ such that $L^{n_i}v\rightarrow 0$. Show that $L^nv\rightarrow 0$ as $n\rightarrow \infty$.
My try: If $v$ is an eigenvector of $L$, then we have $$ 0=\lim L^{n_i}v=\lim \lambda^{n_i}v $$ which implies that $|\lambda|<1$. Then we have $$ \lim L^nv=\lim \lambda^nv=0. $$ How to prove when $v$ is not eigenvector?
Hints. By embedding $V$ in a finite-dimensional vector space over $\mathbb C$ if necessary, one may assume that the underlying field is complex. Since all norms are equivalent on a finite-dimensional vector space, one may reduce the problem to the special case where $V=\mathbb C^N$ for some $N$ and the matrix of $L$ is a Jordan block for an eigenvalue $\lambda$. In this case, if $v_k$ is the last nonzero entry of $v$, the $k$-th entry of $L^mv$ is $\lambda^mv_k$ for every positive integer $m$.
