locally nilpotent operator.

Dec 2009
1,506
434
Russia
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
Aug 2009
976
387
Israel
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).
 
Dec 2009
1,506
434
Russia
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
Aug 2009
976
387
Israel
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 \(\displaystyle T^k\)).