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Let $A$ be a $n\times n$ matrix. Prove that if $\sup_{n\in\mathbb N}\| A^n\| < \infty$, then $\lim_{n\to\infty}\frac{1}{n}\sum^{n-1}_{i=0}A^i$ converges.

This is my attempt. By using the Jordan normal form of $A$, we have $$ T^{-1}AT = \begin{bmatrix} A_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & A_m \\ \end{bmatrix} $$ where $A_1, \dots, A_m$ are Jordan blocks.

Then, we have that $$ T^{-1}A^n T = \begin{bmatrix} B_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & B_m \\ \end{bmatrix} $$, where $B_j = \frac{1}{n}\sum^{n-1}_{i=0}A_j^i$. By using the hypothesis that the operator norm of $A^n$ is bounded I can conclude that $\| B_j\|$ is also bounded. I'm stuck at this part, what can I do use these facts to show that the limit exists?

Keio203
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    first notice that all eigenvalues must be in the unit disk, and those on the unit circle must have blocks of size 1 – Exodd Apr 07 '22 at 21:27
  • @Exodd: Why all the eigenvalues must be in the unit disk? – Keio203 Apr 07 '22 at 21:28
  • because the norm is greater than the spectral radius, and the spectral radius of $A^n$ is the spectral radius of $A$ to the power of $n$ – Exodd Apr 07 '22 at 21:30
  • @Exodd: I got that $\rho(A) \leq 1$, but why if $|\lambda|=1$, then $\lambda$ has multiplicity 1 ? – Keio203 Apr 07 '22 at 21:45
  • It has not multiplicity 1 (in fact you can take $A=I$) but all its Jordan blocks have size 1 – Exodd Apr 07 '22 at 21:46
  • @Exodd. Ok, but I still don't see why the blocks should be size 1 if $|\lambda| = 1$ – Keio203 Apr 07 '22 at 21:57
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    To see it, use the binomial formula to compute $J^k = (\lambda I + N)^k$ when $J$ is a Jordan block. (https://math.stackexchange.com/questions/326688/why-does-the-n-th-power-of-a-jordan-matrix-involve-the-binomial-coefficient). Hence you may assume WLOG that $J = \lambda I + N$ is a Jordan block of dimension $k \geq 2$ and that $|\lambda| < 1$. – Mason Apr 08 '22 at 00:41

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Indeed the observations of Exodd and Mason are the key.

Consider a Jordan block of the form $J=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$. Then $J^k=\begin{pmatrix}\lambda^k & k\lambda^{k-1}\\ 0 & \lambda^k\end{pmatrix}$ by induction for $k\geq 1$. Therefore $J^k\begin{pmatrix}0\\ 1\end{pmatrix}=\begin{pmatrix}k\lambda^{k-1}\\ \lambda^k\end{pmatrix}$. So if $\|\lambda\|=1$, then $\lim_{k\to\infty}\|J^k\|=\infty$. Clearly this can be done for Jordan blocks with $1$s off the main diagonal in higher dimension too (just multiply with $(0,1,0,0,\dots,0)^T$ instead).

This implies that the Jordan normal form of $A$ is $\begin{pmatrix}Id & 0 &0\\ 0& B&0\\ 0 &0&C\end{pmatrix}$ for some diagonal matrix $B$ with all diagonal entries having norm $1$ but being distinct from $1$, a matrix $C$ with $\rho(C)<1$ and the size of the identity block being equal to the multiplicity of the $1$ eigenvalue.

Since $(Id-C)\sum_{k=0}^{\infty}C^k=Id$, it follows that $\frac{1}{n}\sum_{k=0}^{n-1} C^k \to 0$.

Thanks to Exodd for the suggestion about the $B$ component.

Now let $\lambda\in\mathbb{C}\setminus\{1\}$ such that $|\lambda|=1$, i.e. let us take $\lambda=e^{2\pi i\alpha}$ for some $\alpha\in[0,1)$. If $\alpha\in\mathbb{Q}\cap[0,1)$, the sequence $\{\lambda^n\}_{n\in\mathbb{N}}$ is periodic and runs through the values $e^{2\pi i\frac{k}{N}}$ for $1\leq k\leq N$ for some $N$. Since $\sum_{k=0}^{N-1} e^{2\pi i\frac{k}{N}}=0$ (this can be seen geometrically or by multiplying the expression with $1-e^{2\pi i\frac{1}{N}}$), it follows that $\frac{1}{n}\sum_{k=0}^{n-1} \lambda^k\to 0$ as $n\to\infty$.

If $\alpha$ is irrational, one can use Weil's equidistribution theorem or something similar to argue that for an arbitrary $\varepsilon>0$ the frequency of naturals $n$ such that $\lambda^n=e^{2\pi i\beta}$ with $\beta\in[\beta_0-\varepsilon,\beta_0+\varepsilon]$ is the same as the frequency of $n$s such that $\beta\in[\frac{1}{2}+\beta_0-\varepsilon,\frac{1}{2}+\beta_0+\varepsilon]$ for any $\beta_0\in[0,1)$. Once again this implies $\frac{1}{n}\sum_{k=0}^{n-1} \lambda^k\to 0$ as $n\to\infty$. More formally, I think the exact fact that the limit is $0$ follows from Kinchin's identity by taking $f$ to be the exponential, see e.g. https://en.wikipedia.org/wiki/Equidistribution_theorem .

Therefore, $\frac{1}{n}\sum_{k=0}^{n-1} A^k\to\begin{pmatrix}Id & 0 \\0 & 0\end{pmatrix}$ as $n\to \infty$ (in the appropriate Jordan normal form) with the size of the $Id$ block as described before.

AnCar
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    actually no. Only the eigenvalues exactly equal to 1 stay. If you have any other unit norm eigenvalue, it goes to zero – Exodd Apr 08 '22 at 10:10
  • Oh, thanks. You're right. missed this. Will update the answer. – AnCar Apr 08 '22 at 10:39
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    there's a way simpler argument than Weil distribution... if $z\ne 1$ you get $\sum_{i=0}^{n-1} z =(z^n-1)/(z-1)$ that is bounded in absolute value by $2/|1-z|$, so the average goes to zero – Exodd Apr 08 '22 at 12:15
  • fair point, that is indeed much simpler. – AnCar Apr 08 '22 at 13:25