# Nilpotent operators, matrix

1. May 7, 2007

### redyelloworange

1. The problem statement, all variables and given/known data
Prove that any square triangular matrix with each diagonal entry equal to zero is nilpotent

3. The attempt at a solution
Drawing out the matrix and multiplying seems a little tedious. Perhaps there is a better way?
Is there another way to do this without assuming that the eigenvalues of a nilpotent operator are all 0?

2. May 7, 2007

### StatusX

This might not be the easiest way, but trying a few examples shows that as you raise the matrix to higher powers, the zeros creep up towards the top-right corner one space at a time. This suggests trying to prove a stronger result, that if Aij=0 for i>j-k and Bij=0 for i>j-l, then (AB)ij=0 for i>j-k-l (or something like that).

Last edited: May 7, 2007
3. May 8, 2007

### matt grime

Since it is triangular with zero diagonal you know the eigenvalues are all 0, thus you know the min poly is x^n for some n. Why are you 'assuming' that the eigenvalues of a nilpotent operator are all 0? It is clearly true (over a field), and isn't important for this question, really (you state you don't want to assume nilpotent implies all e-values 0, but we actually need all e-values 0 implies nilpotent).

Last edited: May 8, 2007