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  "body": "\n## Statement\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]], let $N \\colon H \\to H$ be a bounded normal operator, that is, $N^*N = NN^*$, and let $T \\colon H \\to H$ be a bounded linear operator.\nIf $TN = NT$, then $TN^* = N^*T$, where $N^*$ is the [[adjoint|6HSJBJ8TTQ]] of $N$.\n\n## Remarks\n\nProof idea (Rosenblum): from $TN = NT$ one gets $T = e^{i\\bar\\lambda N} T e^{-i\\bar\\lambda N}$ for every $\\lambda \\in \\C$.\nSince $N$ 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.\nSo $e^{\\pm iA_\\lambda}$ are unitary and $f$ is a bounded entire function, hence constant by Liouville's theorem.\nDifferentiating at $\\lambda = 0$ gives $N^*T - TN^* = 0$.\n\nExample: 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 $N^* = \\mathrm{diag}(1, -i)$.\n\nNormality is needed: for the nilpotent $N = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$, the operator $T = N$ commutes with $N$, but $TN^* = NN^* = \\mathrm{diag}(1, 0)$ differs from $N^*T = N^*N = \\mathrm{diag}(0, 1)$.\n\nPutnam's generalization: if $M$ and $N$ are normal and $TN = MT$, then $TN^* = M^*T$.\n\n## Explanations\n\n### L1 human\n\nFor matrices that are not normal, commuting with $N$ need not imply commuting with its adjoint.\nFuglede's theorem says that for normal operators the two conditions go together.\nSo the operators that commute with a normal operator form an algebra closed under taking adjoints.\n\n### L2 human\n\nIn finite dimensions the result follows from the spectral theorem.\nAn operator $T$ commutes with $N$ if and only if it preserves each eigenspace of $N$.\nThen it commutes with $N^*$, which acts on the same eigenspaces by the conjugate eigenvalues.\nIn 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 $N$.\nThis makes the commutant of a normal operator a von Neumann algebra.\n",
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