One way: the minimal polynomial of has the form . Now, apply a well known result about canonical forms.
Hi all, another question from my text is confusing me.
Question: A linear operator L is nilpotent if some positive power . 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 will have on the diagonal, thus .
Am I correct here?
June
Perhaps a more direct way is--well, the direct way. Namely, since can be upper triangularized, say and so if and only if , but from there I bet you can conclude since the diagonal entries of are where are the eigenvalues, with multiplicity, of
EDIT: And, this is precisely what you said. Missed that. So, correct!