1. linear operator nilpotent

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

Question: A linear operator L is nilpotent if some positive power $L^k=0$. 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.

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

June

2. Re: linear operator nilpotent

One way: the minimal polynomial of $L$ has the form $\mu (x)=x^s$ . Now, apply a well known result about canonical forms.

3. Re: linear operator nilpotent

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 $L^k=0$. 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.

EDIT: If L is nilpotent, then its eigenvalues would all be zero. If we then write L is upper triangular form then $L^k$ will have $\lambda_i^k$ on the diagonal, thus $L^k=0 \implies \lambda_i=0$.
Perhaps a more direct way is--well, the direct way. Namely, since $L$ can be upper triangularized, say $L=UTU^{-1}$ and so $L^k=UT^kU^{-1}=0$ if and only if $T^k=0$, but from there I bet you can conclude since the diagonal entries of $T^k$ are $\lambda_1^k,\cdots,\lambda_n^k$ where $\lambda_1,\cdots,\lambda_n$ are the eigenvalues, with multiplicity, of $L$