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Can anybody please give me an example of a quasinilpotent operator $T$, i.e. an operator such that $\sigma(T)=\{0\}$ on $l_2$ such that it has finite dimensional but non-trivial kernel and is not compact?

This is probably easy and well known but I just can't figure it out it and I am getting frustrated.

Thanks!

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1 Answer 1

up vote 1 down vote accepted

Take a quasinilpotent operator with trivial kernel and a finite Jordan Block and glue them together.

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Could you please expand on your answer, perhaps by giving a more concrete example? Why is the kernel finite dimensional? Thanks –  Theo Aug 15 '12 at 16:48
    
Sorry, I was in a hurry and made a fundamental typo. I edit the answer. –  abatkai Aug 15 '12 at 17:14
    
Apologies abatkai, could you please be more explicit what you mean by "glue them together"? –  Theo Aug 16 '12 at 2:53
    
@Theo Gluing = taking direct sum. Say, you have a quasinilpotent non-compact operator $T:\ell_2\to \ell_2$ with trivial kernel. Let $H=\ell_2\oplus \mathbb R$. Define $\tilde T:H\to H$ by $\tilde T(v,x)=(Tv,0)$. This operator has 1-dimensional kernel. –  user31373 Aug 17 '12 at 1:59
    
@LeonidKovalev Ahh....so simple and clear. Thank you. –  Theo Aug 17 '12 at 2:39

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