First, some definitions:

We say that two bounded operators on Hilbert spaces $T:H_1\to H_1$ and $S:H_2\to H_2$ are unitarily equivalent if there is an unitary operator $U:H_1\to H_2$ such that $T=U^*SU$ (where the supscribed $*$ represent the Hilbert adjoint).

I'd already proved the following fact:

If $T$ and $S$ are unitarily equivalent self-adjoint operators, then they have the same spectrum.

Now... what if they are not self-adjoint? I was looking for some sufficient conditions (such as this) and, more important, counterexamples: two unitarily equivalent operators that don't share the same spectrum.

I tried to look for some simple examples (finite ranked and even compact infinite ranked), but had no success.


1 Answer 1


$$ (U^*SU-\lambda I)= U^*(S-\lambda I)U $$ So $S-\lambda I$ is invertible iff $U^*SU-\lambda I$ is invertible, which leaves $\sigma(S)=\sigma(U^*SU)$.

  • 2
    $\begingroup$ Now I noticed that, yesterday, I didn't used self-adjointness in my proof. This is what 16 hours of study can do to your brain... $\endgroup$
    – Filburt
    Commented Jan 28, 2018 at 11:28
  • 1
    $\begingroup$ @Filburt : Only 16 hours! Novice. :) $\endgroup$ Commented Oct 15, 2020 at 5:19

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