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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
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