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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.

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  • $\begingroup$ The sequence of real numbers $\frac{x^n}{n}$ converges only if $|x|\leq1$ and it must converge to 0. Think about it in terms of the numerator growing exponentially and the denominator growing linearly. Then, properties of operator norm tell us that $\left \|\frac{1}{n}T^n\right\| \leq \frac{1}{n}\|T\|^n$ and thus you can set $|x| = \|T\|$ $\endgroup$ – rubikscube09 Mar 26 at 1:03
  • $\begingroup$ @rubikscube09 What if $\|T\| > 1$, but $\|\frac{1}{n}T^n\| \to 0$ anyway? $\endgroup$ – user759562 Mar 26 at 1:05
  • $\begingroup$ In particular it follows that the norm of the limit is less than or equal to $0$ (strong limits pass inside norms by defintion). $\endgroup$ – rubikscube09 Mar 26 at 1:05
  • $\begingroup$ @user759562 Ah yes, that's the non-obvious case. I feel silly now. Should probably think about this some more. $\endgroup$ – rubikscube09 Mar 26 at 1:06
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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, $S^2 = 0$.

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  • $\begingroup$ +1. I would suggest making the example even more clear by writing $T=\begin{bmatrix} 1&1\\0&1\end{bmatrix}$. $\endgroup$ – Martin Argerami Mar 26 at 3:01
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    $\begingroup$ @MartinArgerami I've always taken issue with equating operators with matrices; they're not the same, and in my mind this has always made things less clear. But, I think you're right that it's good to note that the standard matrix for $T$ is indeed $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. $\endgroup$ – user759562 Mar 26 at 3:03
  • $\begingroup$ Yes, that's a rather obvious point of view for me since my natural environment is infinite-dimensional. But in finite dimension it is super useful to think of $\operatorname{End}(\mathbb k^n)$ as $M_n(k)$. $\endgroup$ – Martin Argerami Mar 26 at 3:34

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