I've been stuck on a linear algebra problem from Axler's Linear Algebra Done Right. The problem statement is as follows:
Suppose $F=\mathbb{C}$, $V$ is finite-dimensional, $T \in \mathcal{L}(V)$, all the eigenvalues of $T$ have absolute value less than $1$, and $\epsilon > 0$. Prove that there exists a positive integer $m$ such that $\left\|T^m(v)\right\| \le \epsilon\left\|v\right\|$ for every $v \in V$. (Source: Axler 6.B. Problem 16)
I have done some work on the problem by proving the statement for diagonalizable $T$. This holds because then there is an eigenbasis $e_1,...,e_n$ so there exists $a_j$ such that $v=a_1e_1+...+a_ne_n$ so $\left\|T^m(v)\right\|=|a_1\lambda_1^{m}e_1+...+a_n\lambda_n^{m}e_n|$ and since there exists all $\lambda_j$ are less than $1$, by WOP, $\exists{m_j}$ such that each $\lambda_j^{m_j} < \epsilon$ so taking $\max(m_1,...,m_j)=m$ gives a value of $m$ such that $\left\|T^m(v)\right\|$ is less than $\epsilon\left\|v\right\|$.
I'm not quite sure how I should generalize to all linear operators $T \in \mathcal{L}(V)$ from here though. Any help is appreciated.
Let $A$ be a matrix of the linear operator $T$ in some basis.
Recall that diagonalizable matrices are dense in the set of all complex matrices, i.e. for every $X\in\mathbb{C}^{n\times n}$ there exists a sequence of diagonalizable matrices matrices $Y_n$ such that for every $n\in\mathbb{N}$ $||X-Y_n||<\frac{1}{n}$.
Let $Y_n$ be a sequence of diagonalizable matrices approximating $A^m$. We have $$||A^m(v)||=||(A^m-Y_n)(v)+Y_n(v)||\leq||A^m-Y_n||||v||+\varepsilon||v||<\left(\frac{1}{n}||v||+\varepsilon||v||\right)\to\varepsilon||v||\mbox{ as }n\to\infty$$ that proves the result for the case of arbitrary $T\in\mathcal{L}(V)$.