Let $T:V\to V$ be a bounded linear operator on a finite vector space $V$. If the sequence $\frac{1}{n}T^n$ converges, can we prove that its limit is the zero operator?
I think that the answer is yes, but I am struggling a bit with the proof. One approach could be to prove that $\|T\|\leq 1$ but I don't know how to proceed. One could also play around with the sequence terms by setting $S_n=\frac{1}{n}T^n$, $S_0:=\lim_{n\to+\infty}S_n$ and observing that $S_{n+1}=\frac{n}{n+1}TS_n$ which gives $S_0=TS_0$ but I don't know if it is helpful.
This is false. Take, for example, on $\Bbb{C}^2$, $$T(x, y) = (x + y, y).$$ That is, $T$ is the operator whose standard matrix is $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.$$ Then $T^n(x, y) = (x + ny, y)$. Let $S(x, y) = (y, 0)$. Then, $$\left\|\left(\frac{1}{n}T^n - S\right)(x, y)\right\| = \left\|\left(\frac{x}{n}, \frac{y}{n}\right)\right\| = \frac{1}{n}\|(x, y)\|,$$ and hence $$\left\|\frac{1}{n}T^n - S\right\| \le \frac{1}{n} \to 0.$$ Thus, we have an example where $\frac{1}{n}T^n \to S \neq 0$.
EDIT: As a bonus, if $\frac{1}{n}T^n \to S$, we may not be able to say $S = 0$, but we can say $S^2 = 0$. We have, $$\left(\frac{1}{n}T^n\right)^2 =\frac{1}{n^2}T^{2n} = \frac{2}{n} \cdot \frac{1}{2n}T^{2n}.$$ Note that $\frac{1}{2n}T^{2n}$ is a subsequence of the convergent, hence bounded sequence $\frac{1}{n}T^n$. Thus $\left(\frac{1}{n}T^n\right)^2 \to 0$, as well as $S^2$. Thus, $S^2 = 0$.