Definition of the joint spectrum of Hilbert space operators

I see in this paper the following definition: How $${\bf A''}$$ is defined? Is there a relation between $$\sigma({\bf A})$$ and $$\sigma_H({\bf A})$$? Note that $$\sigma_H({\bf A})$$ is defined as:

Definition: $$(\lambda_1,\lambda_2,\cdots,\lambda_n)\notin \sigma_H({\bf A})$$ if there exist operators $$U_1,\cdots,U_n,V_1,\cdots,V_n \in \mathcal{B}(\mathcal{H})$$ such that $$\sum_{1\leq k \leq n}U_k(A_k-\lambda_k I)=I\;\hbox{and}\;\;\sum_{1\leq k \leq n}(A_k-\lambda_k I)V_k =I.$$

The double commutant $$A''$$ is defined as $$(A')'$$, where $$A'=\{T\in B(H):\ TA_j=A_jT,\ j=1,\ldots,n\}$$ and $$A''=\{S\in B(H):\ ST=TS\ \forall T\in A'\}.$$ The double commutant is mostly interesting when the original set contains adjoints, because the commutant of a set that contains its adjoints is a von Neumann algebra. I find it weird the way it's used in the paper, but maybe that's just me.
The Double Commutant Theorem says that, if $$M\subset B(H)$$ is a $$*$$-algebra, then $$M''=\overline{M}^{\rm sot}.$$
• Yes. If the $A_j$ are normal and commuting, then by Fuglede-Putnam you have that $A'$ is a von Neumann algebra and that $A''$ is an abelian von Neumann algebra. That's why the one-sided condition can be used. – Martin Argerami Dec 26 '18 at 13:10