Commutant of a set of operators in $B(H)$

Question

Let $\mathcal{T} \subset B(T)$ be a set of operators.

Then we define the commutatnt $\mathcal{T}' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}\}$.

I'm trying to show $\mathcal{T}' = \mathcal{T}'''$.

Attempt:

By definition, we have that $\mathcal{T}'' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}' \}$, and $\mathcal{T}''' = \{ S \in B(H) : ST =TS, \forall T \in \mathcal{T}''\}$.

"$\subset$":

Let $S \in \mathcal{T}$. Then we have $ST=TS$ for all $T\in \mathcal{T}$.

Then I'm a bit confused on what to do next, because the $\mathcal{T}'''$ I wrote above doesn't really give me a clear idea of what an element in $\mathcal{T}'''$ looks like.

Any help will be appreciated!

Thank you.

Answer 1

We make two trivial observations:

(1) $A \subseteq B \implies B' \subseteq A'.$

(2) $A \subseteq A''$.

Your claim then follows:

By $(2)$, $A \subseteq A''$. Hence, by $(1)$, $A'''= (A'')' \subseteq A'$.

On the other hand, another application of $(2)$ yields $A' \subseteq (A')''= A'''$.

Answer 2

Lemma. Let $X$ be any set and let $\lozenge$ be a symmetric relation on $X$ (e.g. $x\mathrel{\lozenge} y \Leftrightarrow x y= yx$, on a ring, or $x\mathrel{\lozenge} y \Leftrightarrow x\perp y$, on an inner-product space). For each subset $S\subseteq X$ define $$ S^\lozenge = \{x\in X: x\mathrel{\lozenge} s, \text{ for all $s$ in S} \}. $$ Then $$ S^\lozenge = ((S^\lozenge)^\lozenge)^\lozenge, $$ for any $S$.

Proof. First notice that $$ S\subseteq (S^\lozenge)^\lozenge \tag 1 $$ for a pretty elementary reason (which nevertheless sounds a bit like a tongue-twister): every element in $S$ is $\lozenge$ to anything that is $\lozenge$ to every element in $S$.

Plugging in $S^\lozenge$ in place of $S$ in (1), we get $S^\lozenge \subseteq ((S^\lozenge)^\lozenge)^\lozenge$.

Next observe that $$ S_1\subseteq S_2 \Rightarrow S_2^\lozenge\subseteq S_1^\lozenge, $$ and if this is applied to (1), we get $ ((S^\lozenge)^\lozenge)^\lozenge\subseteq S^\lozenge. $ QED

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This is based on this answer.