Symmetries

The Jordan matrix $J(\l)$ is not diagonalizable unless $n=1$

Put $A\colon=J(\l)-\l$, then $\ker A=\lhull{e_1}$ and since $Ae_2=e_1$ it follows that $e_2\in\ker A^2$, i.e. $\ker A\neq\ker A^2$.