1. The problem statement, all variables and given/known data How would I go about proving that if a linear operator [tex]T\colon V\to V[/tex] has all eigenvalues equal to 0, then [tex]T[/tex] must be nilpotent? 3. The attempt at a solution I know that this follows trivially from the Cayley-Hamilton theorem (the characteristic polynomial is [tex]x^n[/tex] and hence [tex]T^n=0[/tex]), but, unfortunately, I'm not allowed to use that theorem. How else could I go about proving this? I've been trying to show that any vector in [tex]V[/tex] is also in the kernel of some power of [tex]T[/tex] and hence [tex]T[/tex] is nilpotent, but I'm getting stuck.