Good evening!


Let X be a banachspace and T a bounded linear operator on X.
The cesaro avearges of T are defined as:


$A\_n:=\frac{1}{n} \sum\limits\_{j=0}^{n-1}T^j $


We call T cesaro bounded if: $\sup\_{n \geq 0}\Vert A\_n \Vert<\infty$.


We call T power bounded if: $\sup\_{n \geq 0}\Vert T^n \Vert<\infty$.


E. Hille showed in "Remarks on ergodic theorems, Trans. Amer. Math. Soc. 57, 1945, 246-269" that one can find a cesaro bounded Operator in $\mathcal{L}(L\_1[0,1])$ which is not power bounded.


**Here is my question**: can this be achieved in a finite dimesional setting? 


With best regards, 


Matthias



Cesaro bounded Operator which is not power bounded 
Consider $T = \pmatrix{-1 & 1\cr 0 & -1\cr}$. Then $T^n = \pmatrix{(-1)^n & (-1)^{n+1} n\cr
0 & (-1)^n\cr}$ so $T$ is not power-bounded. But $A\_n = \pmatrix{\frac{1-(-1)^n}{2n} & \frac{(-1)^n}{2} + \frac{1-(-1)^n}{4n}\cr 0 & \frac{1-(-1)^n}{2n}\cr}$ so it is cesaro-bounded.


You could replace $-1$ by any $\lambda \ne 1$ with $|\lambda|=1$.


