Let A be a quasinilpotent operator on a Hilbert space and let $A^{\*}A$ have finite spectrum.  

Does then follow, that A is nilpotent ?



quasinilpotence and finite spectrum
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. 


