Fuglede's theorem

Remarks

Proof idea (Rosenblum): from $TN = NT$ one gets $T = e^{i\bar\lambda N} T e^{-i\bar\lambda N}$ for every $\lambda \in \C$. Since $N$ and $N^*$ commute, $f(\lambda) = e^{-i\lambda N^*} T e^{i\lambda N^*}$ equals $e^{-iA_\lambda} T e^{iA_\lambda}$ with $A_\lambda = \lambda N^* + \bar\lambda N$ self-adjoint. So $e^{\pm iA_\lambda}$ are unitary and $f$ is a bounded entire function, hence constant by Liouville's theorem. Differentiating at $\lambda = 0$ gives $N^*T - TN^* = 0$.

Example: if $N = \mathrm{diag}(1, i)$ on $\C^2$ and $TN = NT$, then $T$ is diagonal because the eigenvalues of $N$ are distinct, and a diagonal $T$ commutes with $N^* = \mathrm{diag}(1, -i)$.

Normality is needed: for the nilpotent $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, the operator $T = N$ commutes with $N$, but $TN^* = NN^* = \mathrm{diag}(1, 0)$ differs from $N^*T = N^*N = \mathrm{diag}(0, 1)$.

Putnam's generalization: if $M$ and $N$ are normal and $TN = MT$, then $TN^* = M^*T$.

Going deeper

For a student

In finite dimensions the result follows from the spectral theorem. An operator $T$ commutes with $N$ if and only if it preserves each eigenspace of $N$. Then it commutes with $N^*$, which acts on the same eigenspaces by the conjugate eigenvalues. In infinite dimensions, Fuglede's theorem shows that an operator commuting with $N$ commutes with all spectral projections of $N$, and hence with every bounded Borel function of $N$. This makes the commutant of a normal operator a von Neumann algebra.