# Thread: nilotent proof

1. ## nilotent proof

show that T is nilpotent if and only if 0 is the only eigenvalue of T.

I am having difficulty finding a way to solve this problem. I first solved it by assuming that T was diagonalizable because $T \in L(v)$ over the Complex. but that is not true because it is not necessarily normal.

HELP!

2. Originally Posted by mathisthebestpuzzle
show that T is nilpotent if and only if 0 is the only eigenvalue of T.

I am having difficulty finding a way to solve this problem. I first solved it by assuming that T was diagonalizable because $T \in L(v)$ over the Complex. but that is not true because it is not necessarily normal.

HELP!
let T be nilpotent, say $T^m=0$, and $\lambda$ be an eigenvalue of T. so $Tv=\lambda v$, for some $0 \neq v \in V.$

now we have $T^2v= \lambda Tv = \lambda^2 v, \ ... \ , T^m v = \lambda^m v.$ thus $\lambda^m v = 0.$ but $v \neq 0.$ thus $\lambda^m = 0,$ i.e.

$\lambda= 0.$ conversely, suppose all eigenvalues of T are 0, and $\dim V = n.$ then by Cayley-Hamilton

theorem we have $T^n=0. \ \ \ \square$

3. hey thank you so much i appreciate that alot.
i'm stuck on a few more in this section...help is pretty rare in this section i would love more insight.