Suppose $T,U$ are bounded operators on a Hilbert space $\mathcal{H}$ with $T$ normal and $U$ unitary. Now we define $S = U^*TU$.
Then I want to find the spectral resolution of the identity $E_S$ for $S$ in terms of spectral resolution of the identity $E_T$ for $T$.
The term "spectral resolution" refers to unique resolution of the identity in the following "Borel calculus" result which I also believe is useful to show the claim above (here $\sigma(T)$ denotes the spectrum of $T$):
If $T$ is a normal, bounded operator on a Hilbert space, then there is a unique resolution of the identity $E$ such that $$ T = \int_{\sigma(T)} id_{\sigma(T)} \ dE. $$
I have managed to verify that $S$ becomes a normal operator as well as the fact that $\sigma(S)= \sigma(T)$. So we can indeed use this result on $S$ and get $E_S$ satisfying $$ S = \int_{\sigma(S)} id_{\sigma(S)} \ dE_S. $$ Hence we have that $$ \int_{\sigma(T)} id_{\sigma(T)} \ dE_S =\int_{\sigma(S)} id_{\sigma(S)} \ dE_S =S = U^*TU = U^*(\int_{\sigma(T)} id_{\sigma(T)} \ dE_T )U $$ but I am not sure if this gives me anything. Can anyone help me?