|
2 | added 69 characters in body | ||
|
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 $|\lambda|=1$. |
||||
|
|
1 |
|
||
|
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. |
||||
