Corollary 37.4.4.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^{*}$-subalgebra with $I \in A$, and $x \in H$ be a cyclic vector for $A$, then $x$ is also a separating vector for $A$.
Proof. Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by Proposition 37.4.3, it is also separating for $A$.$\square$
Post a Comment