Diagonalizable nilpotent matrix
= Diagonalizable nilpotent matrix
The only <diagonalizable> <nilpotent matrix> is the zero <matrix>. Indeed, if $A^p=0$ and $Av=\lambda v$ with $v\ne0$, 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.