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Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum.
Does then follow, that A is nilpotent ?

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2 Answers 2

Yes, if $A$ is a compact operator. (From the spectral theorem you have for some ON basis $(e_n)$ and some ON sequence $(f_n)$ that $Ae_n = \lambda_n f_n$, where $(\lambda_n)$ are in the spectrum of $A^* A$. Your hypothesis implies that only finitely many of the $(\lambda_n)$ are non zero, so $A$ has finite rank.)

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up vote 2 down vote accepted

I found a counterexample :

Let $e_{1},e_{2},...$ be ON basis of the Hilbert space and define A by $Ae_{2n-1} = \sqrt{1-\frac{1}{n^{2}}} \ e_{2n} \ + \ \frac{1}{n} \ e_{2n+1}$ , $\ \ $n=1,2,3,... ,
$Ae_{2n} = 0$ , $\ \ $n=1,2,3,...

Then A is a partial isometry and therefore $A^{*}A$ a projection.
Furthermore A is quasinilpotent but not nilpotent.

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