Prove that a normal and nilpotent operator is the zero operator.
Solution
1. Let \(T\) be normal and nilpotent on a finite-dimensional inner product space \(V\).
setup.
2. A normal operator is diagonalizable (over \(\CC\)).
the spectral theorem for normal operators.
3. A nilpotent operator has all eigenvalues equal to \(0\).
if \(Tv = \lambda v\) with \(\lambda \neq 0\), then \(T^n v = \lambda^n v \neq 0\) for all \(n\), contradicting nilpotency.
4. Hence \(T\) is diagonalizable with all eigenvalues \(0\).
2 and 3.
5. Therefore \(T = 0\).
a diagonalizable operator with all eigenvalues \(0\) is the zero operator.
6. Q.E.D.
5.