# Thread: locally nilpotent operator.

1. ## locally nilpotent operator.

Let $\displaystyle T:V \to V$ nilpotent linear transformation over some field, let assume that $\displaystyle dimV=n$.
Prove that $\displaystyle T$ nilpotent if and only if the characteristic polynomial of $\displaystyle T$ is $\displaystyle x^n$

Thank you!

2. Originally Posted by Also sprach Zarathustra
Let $\displaystyle T:V \to V$ nilpotent linear transformation over some field, let assume that $\displaystyle dimV=n$.
Prove that $\displaystyle T$ nilpotent if and only if the characteristic polynomial of $\displaystyle T$ is $\displaystyle x^n$

Thank you!
Remember, T is nilpotent if there exists some $\displaystyle k \in \mathbb{N}$ such that $\displaystyle T^k = 0$. Use this along with the Cayley-Hamilton theorem to prove this fact (the <= side should be trivial).

3. Originally Posted by Defunkt
Remember, T is nilpotent if there exists some $\displaystyle k \in \mathbb{N}$ such that $\displaystyle T^k = 0$. Use this along with the Cayley-Hamilton theorem to prove this fact (the <= side should be trivial).
No... I thought about it and couldn't do it.

4. If $\displaystyle x^n$ is the char. polynomial of T then by Cayley-Hamilton, $\displaystyle T^n=0$ therefore T is nilpotent by definition.

For the other direction, simply prove that T can not have any non-zero eigenvalues (use the fact that if $\displaystyle \lambda$ is an eigenvalue of T then $\displaystyle \lambda^k$ is an eigenvalue of $\displaystyle T^k$).