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How do I see that if $T$ is a self-adjoint operator with spectral measure $E$ that this implies that $UTU^*$ has the spectral measure $UEU^*?$ Is there an easy way to see this? Somehow I believe that this must be a very simple fact, but I currently do not see how to show it without referring to anything else but the spectral theorem and the functional calculus. In particular, I do not want to use any resolvent formulas (like Stone's one).

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If $T$ is selfadjoint with spectral measure $E$, then $$ T = \int \lambda dE(\lambda). $$ If $E'$ is a spectral measure such that $T=\int\lambda dE'(\lambda)$, then $E'=E$ by uniqueness of spectral measures.

If $U$ is unitary, then $U^*TU$ is selfadjoint, and $$ U^*TU = \int \lambda d(U^*EU). $$ Because $U^*EU$ is a spectral measure and $U^*TU$ is selfadjoint, then $U^*EU$ is the spectral measure associated with $U^*TU$ because of uniqueness of spectral measures.

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