# nilotent proof

• Jun 9th 2008, 05:09 PM
mathisthebestpuzzle
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 $\displaystyle T \in L(v)$ over the Complex. but that is not true because it is not necessarily normal.

HELP!
• Jun 9th 2008, 06:23 PM
NonCommAlg
Quote:

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 $\displaystyle T \in L(v)$ over the Complex. but that is not true because it is not necessarily normal.

HELP!

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

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

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

theorem we have $\displaystyle T^n=0. \ \ \ \square$
• Jun 9th 2008, 07:08 PM
mathisthebestpuzzle
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.