Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum.
Does then follow, that A is nilpotent ?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.



I found a counterexample : Let $e_{1},e_{2},...$ be ON basis of the Hilbert space and define A by
$Ae_{2n1} = \sqrt{1\frac{1}{n^{2}}} \ e_{2n} \ + \ \frac{1}{n}
\ e_{2n+1}$ , $\ \ $n=1,2,3,... , Then A is a partial isometry and therefore $A^{*}A$ a projection. 


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.) 

