Recall that the commutator of two linear operators, \(S\) and \(T\), is \([S,T]=ST-TS\) and that they are said to commute if \([S,T]=0\).

Theorem Two normal operators, \(S\) and \(T\), commute if and only if there is an orthonormal basis consisting of common eigenvectors of \(S\) and \(T\). In this case \(S\) and \(T\) are said to be simultaneously diagonalisable since their respective matrix representations with respect to such a basis are diagonal.

Proof The if is trivial so we focus on the only if. \(S\) and \(T\) have spectral decompositions of the form,
\begin{equation*}
S=\sum_{i=1}^r\mu_iQ_{\mu_i}\quad\text{and}\quad T=\sum_{i=1}^s\lambda_iP_{\lambda_i},
\end{equation*}
where the \(\mu_i\) and \(\lambda_i\) are the distinct eigenvalues of \(S\) and \(T\) respectively and \(Q_{\mu_i}\) and \(P_{\lambda_i}\) the projectors onto the corresponding eigenspaces. Now, if any operator, \(A\), commutes with a normal operator, \(T\), then the eigenspaces, \(V_{\lambda_i}\), of \(T\) are \(A\)-invariant since if, \(\ket{\lambda}\in V_\lambda\), then, \(TA\ket{\lambda}=AT\ket{\lambda}=\lambda A\ket{\lambda}\), so, \(A\ket{\lambda}\in V_\lambda\). This means that both, \(V_{\lambda_i}\), and its orthogonal complement, \(V_{\lambda_i}^\perp\), are \(A\)-invariant, so we have a direct sum decomposition, \(V=V_{\lambda_i}\oplus V_{\lambda_i}^\perp\), with corresponding orthogonal projectors, \(P_{\lambda_i}\) and \(P_{\lambda_i}^\perp\), such that for any \(\ket{v}\in V\) we can write, \(P_{\lambda_i}A\ket{v}=P_{\lambda_i}A(P_{\lambda_i}\ket{v}+P_{\lambda_i}^\perp\ket{v})=AP_{\lambda_i}\ket{v}\). That is, \(A\) commutes with each of the projection operators of the spectral decomposition of \(T\). It follows that if the two normal operators, \(S\) and \(T\), commute then so too must their respective projectors, \(Q_{\mu_i}P_{\lambda_j}=P_{\lambda_j}Q_{\mu_i}\). Now define \(R_{ij}=Q_{\mu_i}P_{\lambda_j}\), clearly \(R_{ij}^2=R_{ij}\), \(R_{ij}R_{kl}=0\) unless \(i=k\) and \(j=l\), and \(R_{ij}^\dagg=R_{ij}\) so the \(R_{ij}\) are orthogonal projectors which moreover satisfy,
\begin{equation}
\sum_{i,j}R_{ij}=\sum_{i=1}^rQ_{\mu_i}\sum_{j=1}^sP_{\lambda_j}=\id_V,
\end{equation}
because, \(\sum_{i=1}^rQ_{\mu_i}=\id_V\) and \(\sum_{j=1}^sP_{\lambda_j}=\id_V\). Thus, since, \(\img R_{ij}=V_{\mu_i}\cap V_{\lambda_j}\), we have
\begin{equation}
V=\bigoplus_{i,j}V_{\mu_i}\cap V_{\lambda_j},
\end{equation}
and choosing an orthonormal basis for each \(V_{\mu_i}\cap V_{\lambda_j}\) we obtain, as desired, an orthonormal basis for \(V\) consisting of common eigenvectors of \(S\) and \(T\). Equivalently, we note that \(S\) and \(T\) have spectral decompositions with respect to the projectors \(R_{ij}\), \(S=\sum_{i=1}^r\mu_iQ_{\mu_i}=\sum_{i,j}\mu_iR_{ij}\) and \(T=\sum_{j=1}^s\lambda_jP_{\lambda_j}=\sum_{ij}\lambda_jR_{ij}\).\(\blacksquare\)