# Simultaneous diagonalization of hermitian operators using the spectral theorem

Let a Hilbert space $$\mathscr{H}$$ be given and two operators $$A,B$$ which are hermitian and commute.

In that setting, for each of them we can apply the spectral theorem to get projection valued measures $$\mu_A$$ and $$\mu_B$$ on their spectra $$\sigma(A),\sigma(B)$$ such that $$A=\int_{\sigma(A)}\lambda d\mu_A(\lambda),\quad B=\int_{\sigma(B)}\lambda d\mu_B(\lambda).$$

In other words, $$A,B$$ are "diagonal" in the "basis of eigenvectors".

Now, in the finite dimensional case, when they commute they can be "simultaneously diagonalized". I want to understand this statement here.

I think Fubini's theorem is key in this.

I mean, my first thought was to consider the product measure $$\mu_A\times \mu_B$$ on $$\sigma(A)\times \sigma(B)$$. What we would like then is to have $$A=\int_{\sigma(A)\times \sigma(B)} \lambda_1 d(\mu_A\times \mu_B)(\lambda_1,\lambda_2),\quad B=\int_{\sigma(A)\times \sigma(B)} \lambda_2 d(\mu_A\times \mu_B)(\lambda_1,\lambda_2)$$

Now, suppose Fubini's theorem holds, in that case

$$A=\int_{\sigma(A)}\lambda_1 d\mu_A(\lambda_1)\int_{\sigma(B)}d\mu_B(\lambda_2)=\int_{\sigma(A)}\lambda d\mu_A(\lambda)$$

and similarly for $$B$$ so that we indeed recover the usual expression.

On the other hand, I think that for projection-valued measures, Fubini's theorem is then tied with the operators commuting. I don't know how to make this precise.

So I ask: how does the spectral theorem generalizes for a (finite) collection of commuting operators so that we generalize the idea that they can be "simultaneously diagonalized"?

Am I correct that the above is the right procedure? If so, why Fubini's theorem can only be applied when the operators commute? If not, what is the correct way to do it?

• Every bounded operator $B$ that commutes with a seladjoint operator $A$ also commutes with the spectral measure $E_A$ for $A$. If $A$ and $B$ are selfadjoint and commute, then their spectral measures commute. – DisintegratingByParts Nov 22 '18 at 0:45

The critical point here is the construction of the product measure (mind you, these are not scalar-valued measures for which the theory is usually developed). However, it can be done. This is exactly the content of the spectral theorem for commuting normal operators (see Schmüdgen. Unbounded Self-Adjoint Operators on Hilbert Space, Theorem 5.21):

Theorem. If $$A_1,\dots,A_n$$ are pairwise commuting normal operators on a Hilbert space, then there exists a unique projection-valued measure $$E$$ on $$\mathcal{B}(\mathbb{C}^n)$$ such that $$A_j=\int_{\mathbb{C}^n}\lambda_j\,dE(\lambda_1,\dots,\lambda_n)$$ for $$j\in\{1,\dots,n\}$$. Moreover, the support of $$E$$ is contained in the product of the supports of the spectral measures of $$A_1,\dots,A_n$$.

If you are dealing with unbounded operators, you have to be a little careful. In this case "commuting" is to be understood as strongly commuting, i.e., the spectral projections of the operators commute.

Finally, I think that the better generalization of diagonalization is the spectral theorem in multiplication operator form. For commuting normal operators it reads as follows.

Theorem. If $$A_1,\dots,A_n$$ are pairwise commuting normal operators on a Hilbert space $$H$$, then there exists a measure space $$(X,\mathcal{B},\mu)$$, measurable functions $$\phi_1,\dots,\phi_n\colon X\to \mathbb{C}$$ and a unitary operator $$U\colon H\to L^2(X,\mu)$$ such that $$A_j=U^\ast M_{\phi_j}U$$ for $$j\in\{1,\dots,n\}$$. Here $$M_\phi$$ denotes the multiplication operator defined by $$D(M_\phi)=\{f\in L^2(X,\mu)\mid \phi f\in L^2(X,\mu)\},\,M_\phi f=\phi f.$$