# Limit of operators

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.

• 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 $|x| = \|T\|$ – rubikscube09 Mar 26 at 1:03
• @rubikscube09 What if $\|T\| > 1$, but $\|\frac{1}{n}T^n\| \to 0$ anyway? – user759562 Mar 26 at 1:05
• In particular it follows that the norm of the limit is less than or equal to $0$ (strong limits pass inside norms by defintion). – rubikscube09 Mar 26 at 1:05
• @user759562 Ah yes, that's the non-obvious case. I feel silly now. Should probably think about this some more. – rubikscube09 Mar 26 at 1:06

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, $$S^2 = 0$$.
• +1. I would suggest making the example even more clear by writing $T=\begin{bmatrix} 1&1\\0&1\end{bmatrix}$. – Martin Argerami Mar 26 at 3:01
• @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 $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. – user759562 Mar 26 at 3:03
• 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 $M_n(k)$. – Martin Argerami Mar 26 at 3:34