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.

  • $\begingroup$ 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

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, $S^2 = 0$.

  • $\begingroup$ +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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.