- Show that a nilpotent operator is diagonalizable if and only if it is the zero operator.
Solution
If \(T\) is nilpotent and \(Tv=\lambda v\) with \(v\ne0\), then for some \(k\), \[ 0=T^kv=\lambda^k v, \] so \(\lambda=0\). Thus a nilpotent operator has only the eigenvalue \(0\).
If \(T\) is also diagonalizable, it has a basis of eigenvectors, and every basis vector has eigenvalue \(0\). Therefore \(T\) vanishes on a basis and hence \[ T=0. \] Conversely, the zero operator is nilpotent and its matrix in every basis is diagonal. Hence \[ \boxed{T\text{ nilpotent and diagonalizable}\iff T=0}. \]