Originally Posted by

**Juneu436** Hi all, another question from my text is confusing me.

Question: A linear operator L is nilpotent if some positive power $\displaystyle L^k=0$. Prove that L is nilpotent iff there is a basis of V such that the matrix of L is upper triangular, with all diagonal entries zero.

Any ideas? Thanks in advance.

EDIT: If L is nilpotent, then its eigenvalues would all be zero. If we then write L is upper triangular form then $\displaystyle L^k$ will have $\displaystyle \lambda_i^k $ on the diagonal, thus $\displaystyle L^k=0 \implies \lambda_i=0$.

Am I correct here?

June