TITLE: quasinilpotence and finite spectrum
QUESTION [6 upvotes]: Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum.
Does then follow, that A is nilpotent ?

REPLY [2 votes]: 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.