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.

  • $\begingroup$ 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 $S,T$. $\endgroup$ – s.harp Aug 7 '19 at 18:16
  • $\begingroup$ 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? $\endgroup$ – A_P Aug 7 '19 at 18:22
  • $\begingroup$ @A_P as far as I am aware that is directly the definition of "simultaneously diagonalisable" :) $\endgroup$ – 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.


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