# locally nilpotent operator.

#### 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!

#### Defunkt

MHF Hall of Honor
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).

• Also sprach Zarathustra

#### Also sprach Zarathustra

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. #### Defunkt

MHF Hall of Honor
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 $$\displaystyle T^k$$).

• Also sprach Zarathustra