In the Averson's proof of the bicommutant theorem is proved that, if $A$ is a self-adjoint algebra of operators with trivial null space and $T \in A''$, for every $\epsilon>0$, $n=1,2..$ and every choice of $n$ vectors $\xi_1, \xi_2,..\xi_n \in H$ there is an operator $S \in A$ such that $\sum_{k=1}^n \|T\xi_k-S\xi_k\|^2 < \epsilon^2$.
The theorem is proved for n=1.
For the general case I copy down a part in the Averson's proof and then I'll ask some questions.
"Now the case of general $n\geq2$ is reduced to the above by the following device. Fix $n$ and let $H_n=H \oplus H\oplus \ .. \ \oplus H$ be the direct sum of $n$ copies of the undelying Hilbert space $H$. Choose $\xi_1, \ .. ,\xi_n \in H$ and define $\eta \in H_n$ by $\eta = \xi_1 \oplus \ .. \ \oplus \xi_n$. Let $A_n \subseteq L(H_n)$ be the *-algebra of all operators of the form $\{S \oplus \ .. \ \oplus S :S \in A\}$. Thus each element of $A_n$ can be expressed as a diagonal $n \times n$ operator matrix
\begin{pmatrix} S & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & S \end{pmatrix}
where $S \in A$. The reader can se by a straightforwart calculation that an $n \times n$ operator matrix $(T_{ij})$, $T_{ij} \in L(H)$ commutes with $A_n$ if and only if each entry $T_{ij}$ belongs to $A'$. This gives a representation for $A_n'$ as operator matrices and now a similar calculation shows that $(K_{ij})$ commutes with $A_n'$ if and only if $(K_{ij})$ has the form
\begin{pmatrix} K & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & K \end{pmatrix}
with $K \in A''$...... etc."
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I have problem with this part in particular: "The reader can se by a straightforwart calculation that an $n \times n$ operator matrix $(T_{ij})$, $T_{ij} \in L(H)$ commutes with $A_n$ if and only if each entry $T_{ij}$ belongs to $A'$."
If I have a general operator matrix $T= \begin{pmatrix} T_{11} & \cdots & T_{1n} \\ \vdots & \ddots & \vdots \\ T_{n1} & \cdots & T_{nn} \end{pmatrix} $
then to let T commute with $A_n$ it sufficient that $T_{ii}$ belongs to $A'$. It's not necessary that also the other entries that are not in the diagonal belong to $A'$ and this seems to invalidate the second part of the argument about $A''$.
I ask this question because I know that probabily I'm wrong! Can you help me to figure out where?