Why is $M_n(A)$ a von Neumann algebra

Question

I'm trying to verify that for each von Neumann algebra $A$ the algebra of matrices with entries in $A$ is again von Neumann.

I do already know, that those kind of matrix algebras are again C-*-algebras.

I'd prefer a basic argument involving only weak or strong closedness over others.

Thanks!

Answer 1

If $A\subset B(H)$, you consider $M_n(A)\subset B( H^n)$. It is now a standard exercise, after noticing that $B(H^n)$ can be identified with $M_n(B(H))$, that $$ M_n(A)'=\left\{\begin{bmatrix} a&&0\\ &\ddots\\0&&a\end{bmatrix}:\ a\in A' \right\}, $$ and then that $$ M_n(A)''=M_n(A'')=M_n(A). $$

Answer 2

Strong convergence of a net in $M_n(A)$ is equivalent to entrywise strong convergence.

Assume $x_\lambda$ is a net in $M_n(A)$ converging strongly to $x \in B(H^n)$. Let $h,k \in H$ and $1 \leq i,j \leq n$. Let $\xi \in H^n$ be the vector which has $h$ in the $i$-th component and $0$ elesewhere, and let $\eta \in H^n$ which has $k$ in the $j$-th component and $0$ else. Then, in particular $\langle x_\lambda \xi, \eta \rangle \to \langle x \xi, \eta \rangle$. But this just means $\langle (x_\lambda)_{ij} h ,k \rangle \to \langle x_{ij} h, k \rangle$.

It follows that $x_{ij} \in A$.