Diagonalizable nilpotent matrix

The only diagonalizable nilpotent matrix is the zero matrix. Indeed, if $A^p=0$ and $Av=\lambda v$ with $v\ne 0$, then ${\lambda }^p=0$, so $\lambda =0$. A basis of eigenvectors would make every diagonal entry zero and therefore $A=0$. Nonzero nilpotent matrices require nontrivial Jordan blocks, even if their characteristic polynomial has only one distinct root.