# Does equal spectrum imply unitary equivalence? [closed]

Knowing that unitarily equivalent self-adjoint operators have the same spectrum (proven here) I was wondering if the “inverse” implication holds. Namely

Let $$A$$, $$B$$ – self adjoint operators acting on Hilbert spaces $$H_A$$ and $$H_B$$. Their spectras are equal $$\sigma(A) = \sigma(B)$$. Is it true that operators $$A$$ and $$B$$ are unitarily equivalent $$A=U^{\ast}BU$$, where $$U:H_A \longrightarrow H_B$$ – unitary operator?

• Consider diagonal matrices with different allocations of the same numbers along the diagonal. Jun 20, 2021 at 19:28
• This is a Problem Statement Question, and therefor does not fit the quality standards of this website. Please add context, as how it stands I have to vote this question to get closed. Jun 20, 2021 at 19:38
• @Cornman I added some context. Jun 20, 2021 at 19:44
• I retracted my vote to close. Please add next time the more context right away, as this keeps this website fresh and the answers and questions researchable and of more quality! Jun 20, 2021 at 19:47

No: the issue is that the spectrum can have "multiplicity". For instance, in the finite dimensional case, $$\sigma(A)=\sigma(B)$$ means that $$A$$ and $$B$$ have the same set of eigenvalues, but they could have those eigenvalues with different multiplicities, and thus fail to be unitarily equivalent. (For a really simple example, you could have $$A=0$$ and $$B=0$$ but $$H_A$$ and $$H_B$$ have different dimensions!)