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  "body": "\n## Definition\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $B(H)$ be the [[algebra of bounded operators|BMDG4JQPEH]] on $H$.\nThe *commutant* of a subset $S \\subseteq B(H)$ is\n$$S' = \\set{T \\in B(H) : TA = AT \\text{ for all } A \\in S}.$$\nThe *bicommutant*, or double commutant, of $S$ is $S'' = (S')'$.\n\n## Remarks\n\nExample: let $D \\subseteq M_n(\\C) = B(\\C^n)$ be the algebra of diagonal matrices.\nIf $T$ commutes with $E = \\operatorname{diag}(e_1, \\dots, e_n)$ for distinct $e_k$, then $(e_i - e_j) T_{ij} = 0$, so $T$ is diagonal; hence $D' = D$.\nLikewise, $M_n(\\C)' = \\C I$, since commuting with all matrix units forces $T$ to be scalar.\n\nNon-example: the matrix unit $E_{12}$ does not lie in $D'$, because $E_{12} \\operatorname{diag}(1, 2) = 2E_{12}$ while $\\operatorname{diag}(1, 2) E_{12} = E_{12}$.\n\nFor all subsets, $S \\subseteq S''$, $S_1 \\subseteq S_2$ implies $S_2' \\subseteq S_1'$, and $S''' = S'$.\nThe commutant $S'$ is always a unital subalgebra of $B(H)$ that is closed in the weak operator topology.\nIf $S$ is closed under adjoints, then $S'$ is also closed under adjoints, so $S'$ is a von Neumann algebra.\nThe same definition in any ring or algebra gives the centralizer of a subset.\n",
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