# If $S, T \in B(X)$ are self-adjoint, compact, and commuting bounded linear operators on a Hilbert space $X$, they can be simultaneously diagonalized

More specifically, there is an orthonormal basis of $$X$$ consisting of common eigenvectors.

So far, I've approached proving this by using the spectral theorem for compact, self-adjoint operators. I know that $$S$$ and $$T$$ can separately be diagonalized. I've also seen a hint that suggests considering the compact operator $$S+ iT$$, but this operator isn't self-adjoint, so the spectral theorem won't apply.

Any suggestions on how to proceed with this proof would be appreciated.

• One thing you could do is look at the $C^*$ sub-algebra of $B(X)$ generated by $S,T$. You will find a self-adjoint $R$ and functions $f, g$ so that $f(R)=S$, $g(R)=T$, ie $R$ is a common root of $S,T$. – s.harp Aug 7 '19 at 18:16
• I'm out of my depth here, but isn't it enough to show that any commuting Hermitian $S$ and $T$ have a common eigenbasis, in which each is diagonalized? – A_P Aug 7 '19 at 18:22
• @A_P as far as I am aware that is directly the definition of "simultaneously diagonalisable" :) – s.harp Aug 7 '19 at 19:06

You can write $$S=\sum_{n=1}^{\infty}\lambda_n P_n$$, where $$\{ \lambda_n \}$$ are the non-zero eigenvalues of $$S$$, and $$P_n$$ is the orthogonal projection onto the eigenspace of $$S$$ associated with $$\lambda_n$$, which is finite-dimensional. The projections $$P_n$$ commute with $$T$$ because $$P_n$$ commutes with everything that commutes with $$S$$. Similarly, $$T=\sum_{m=1}^{\infty}\mu_m Q_m$$. Every $$P_n$$ commutes with every $$Q_m$$. So $$P_nQ_m=Q_mP_n$$ is either $$0$$ or is an orthogonal projection $$R_{n,m}=P_nQ_m$$ such that $$SR_{n,m}=\lambda_nR_{n,m}$$ and $$TR_{n,m}=\mu_m R_{n,m}$$. It can happen that $$R_{n,m}=0$$; after eliminating the trivial products, you're left with orthogonal projections $$P_nQ_m=Q_mP_n$$ with ranges that are finite-dimensional eigenspaces of both $$S$$ and $$T$$. The orthogonal sum of all of these non-zero eigenspaces is the full space. This gives you what you want.
I've been informed that the trick is to note that the eigenspaces of $$T$$ are invariant under $$S$$ (or vice versa). And since, for the eigenvalues $$\lambda_n$$ of $$T$$, we can express the Hilbert space as $$X = \bigoplus_{n=0}^\infty E_{\lambda_n}$$, where $$E_{\lambda_n}$$ is the closed linear span of the eigenvectors associated with $$\lambda_n$$. It follows that we can express $$S$$ as $$S = \bigoplus_{n=0}^\infty S\big|_{E_{\lambda_n}}$$. The diagonalization of $$S$$ using eigenvectors of $$T$$ follows from there.