Proving the Nilpotency of Square Triangular Matrices with Zero Diagonal Entries

[redyelloworange]

Homework Statement

Prove that any square triangular matrix with each diagonal entry equal to zero is nilpotent

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?

Thanks for your help!

 

[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 A_ij=0 for i>j-k and B_ij=0 for i>j-l, then (AB)_ij=0 for i>j-k-l (or something like that).

 

[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).