Prove that if two Hermitian operators commute, then the eigenspace projectors in their spectral decompositions also mutually commute? Suppose A and B are Hermitian, with spectral decompositions $A = \sum_i \lambda_i P_i$ and $B = \sum_j \mu_j Q_j$, where $\lambda_i$ ($\mu_j$) are the eigenvalues of $A$ ($B)$ and $P_i$ ($Q_j$) are the corresponding projectors. What's a simple proof that if $[A,B] = 0$, then $[P_i, Q_j] = 0$ for all $i,j$? (I saw this claim as a step in the proof that commuting Hermitian operators are simultaneously diagonalisable, so this latter result cannot be used in the proof).
 A: This result holds in greater generality. Let $H$ be a Hilbert space and $A,B \in \mathbb{B}(H)$ normal operators with spectral decompositions
$$A=\int_{\sigma(A)} \lambda \,dP(\lambda), \quad B=\int_{\sigma(B)} \mu \,dQ(\mu)$$
It is known that if $A,B$ commute, then $f(A)$ and $g(B)$ commute for all Borel functions $f$ and $g$ on spectra $\sigma(A)$ and $\sigma(B)$ respectively.
The eigenprojections $P(\lambda)$ and $Q(\mu)$ are precisely $\chi_{\{\lambda\}}(A)$ and $\chi_{\{\mu\}}(B)$ so they commute.
In the finite-dimensional case, the functions $\chi_{\{\lambda\}}$ and $\chi_{\{\mu\}}$ can even be written as polynomials, as @Paul suggests, namely
$$\chi_{\{\lambda\}}(x) = \prod_{\lambda' \in \sigma(A), \lambda'\ne \lambda} \frac{x-\lambda'}{\lambda-\lambda'}, \quad \chi_{\{\mu\}}(x) = \prod_{\mu' \in \sigma(B), \mu'\ne \mu} \frac{x-\mu'}{\mu-\mu'}$$
A: Prove that $P_i$ is a polynomial in $A$ and that $Q_j$ is a polynomial in $B$. Then you'll be able to get your result.
To show that, use the product of the $X- \lambda _k$ for $k$ other than $i$.
A: The operator $(\lambda I-A)^{-1}$ is defined for all $\lambda$ not in the set of eigenvalues of $A$. And,
$$
   P_k = \lim_{\lambda\rightarrow\lambda_k}(\lambda-\lambda_k)(\lambda I-A)^{-1}
$$
Similarly,
$$
   Q_l = \lim_{\mu\rightarrow\mu_l}(\mu -\mu_l)(\mu I -B)^{-1}
$$
Because $A$, $B$ commute, so do $(\lambda I-A)^{-1}$ and $(\mu I-B)^{-1}$. So it follows from the above that
$$
        P_k Q_l = Q_l P_k.
$$
