1. ## locally nilpotent operator.

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

Thank you!

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

Thank you!
Remember, T is nilpotent if there exists some $k \in \mathbb{N}$ such that $T^k = 0$. 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 $k \in \mathbb{N}$ such that $T^k = 0$. 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 $x^n$ is the char. polynomial of T then by Cayley-Hamilton, $T^n=0$ 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 $\lambda$ is an eigenvalue of T then $\lambda^k$ is an eigenvalue of $T^k$).