## Extended eigenvalues and the Volterra operator.(English)Zbl 1037.47013

If $$A$$ is a bounded operator on a Hilbert space, one says that $$\lambda$$ is an extended eigenvalue of $$A$$ if there is a nonzero operator $$X$$ such that $$XA=\lambda AX$$. The authors show that if $$V$$ is the Volterra operator $$(Vf)(x)= \int_0^x f(y)\,dy$$ in $$L^2(0,1)$$, then the set of extended eigenvalues is precisely the set $$(0,\infty)$$ and the corresponding extended eigenvectors can be chosen to be certain explicit Volterra integral operators. Thus they give an example of a compact quasinilpotent operator for which there is an algebra that strictly contains the commutant of $$A$$ and which has a nontrivial invariant subspace.

### MSC:

 47A65 Structure theory of linear operators 47B49 Transformers, preservers (linear operators on spaces of linear operators) 45P05 Integral operators

### Keywords:

extended eigenvalue; Volterra operator