Let $T$ be a bounded normal operator on Hilbert space $\mathfrak{H}$ and let $E$ be its resolution of the identity. Let $\mathfrak{M}$ be a closed subspace of $\mathfrak{H}$ which is reducing for $T$. Then the restriction $T_\mathfrak{M}$ of $T$ to $\mathfrak{M}$ is a bounded normal operator on $\mathfrak{M}$ and the spectrum $\sigma(T_\mathfrak{M})$ is contained in $\sigma(T)$. What is the relationship between $E$ and the resolution of the identity $E^\mathfrak{M}$ for $T_\mathfrak{M}$?
Here's what I've tried: If $P$ is the projection on $\mathfrak{M}$, I know that $PT=TP$ and that therefore $PE(\omega)=E(\omega)P$ for all Borel subsets $\omega$ of $\sigma(T)$. From that I know that $E(\omega)\mathfrak{M}\subseteq\mathfrak{M}$, so I can define $E':B_\mathfrak{M}\to B(\mathfrak{M})$, where $B_\mathfrak{M}$ are the Borel subsets of $\sigma(T_\mathfrak{M})$, by $E'(\omega)=E(\omega)|_\mathfrak{M}$. I can show that $E'(\omega)$ is a self-adjoint projection on $B(\mathfrak{M})$ and all the other properties necessary to be a resolution of the identity, except I am unable to prove that $E'(\sigma(T_\mathfrak{M}))=I_{B(\mathfrak{M})}$. I was planning to show that $E'$ was a resolution of the identity on $\sigma(T_\mathfrak{M})$ which satisfied $$T_\mathfrak{M}=\int_{\sigma(T_\mathfrak{M})}\!\lambda\,dE'(\lambda),$$ and therefore by uniqueness, $E'=E^\mathfrak{M}$.
The answer is that if $\omega$ is a Borel subset of $\sigma(T)$, then $$E^{\mathfrak{M}}(\omega\cap\sigma(T_\mathfrak{M}))=E(\omega)|_{\mathfrak{M}}.$$ This makes sense because we can show that every self-adjoint projection $E(\omega)$ commutes with the orthogonal projection onto $\mathfrak{M}$, and therefore $E(\omega)\mathfrak{M}\subseteq\mathfrak{M}$. The details can be found in my answer here https://math.stackexchange.com/a/3743317/527829.