Commutant

Remarks

Example: let $D \subseteq M_n(\C) = B(\C^n)$ be the algebra of diagonal matrices. If $T$ 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 $D' = D$. Likewise, $M_n(\C)' = \C I$, since commuting with all matrix units forces $T$ to be scalar.

Non-example: the matrix unit $E_{12}$ does not lie in $D'$, because $E_{12} \operatorname{diag}(1, 2) = 2E_{12}$ while $\operatorname{diag}(1, 2) E_{12} = E_{12}$.

For all subsets, $S \subseteq S''$, $S_1 \subseteq S_2$ implies $S_2' \subseteq S_1'$, and $S''' = S'$. The commutant $S'$ is always a unital subalgebra of $B(H)$ that is closed in the weak operator topology. If $S$ is closed under adjoints, then $S'$ is also closed under adjoints, so $S'$ is a von Neumann algebra. The same definition in any ring or algebra gives the centralizer of a subset.