Let A be any C* algebra,prove $M_n(A^{**})\cong M_n(A)^{**}$

Question

Let A be any C* algebra,prove $M_n(A^{**})\cong M_n(A)^{**}$,where $A^{**}$ is the double dual of $A$.Who can give me any hints or any reference books about the isomorphism about matrix algebras. I'd like to thank in advance anyone who takes some time to help me out.

Answer 1

It is enough to show that $M_n(A^*)\simeq M_n(A)^*$, since then $$ M_n(A^{**})\simeq M_n(A^*)^*\simeq M_n(A)^{**}. $$ A canonical way to do this is to map $\phi:M_n(A^*)\to M_n(A)^*$ by $$ \phi(x)(b)=\sum_{k,j} x_{kj}(b_{kj}). $$ This map is obviously linear. It is easy to see it's one-to-one by evaluating on matrices with a single nonzero entry. It remains to check it is onto. So if $f:M_n(A)\to\mathbb C$ is a linear functional, define functionals $f_{kj}:A\to\mathbb C$ by $$ f_{kj}(a)=f(a\otimes E_{kj}), $$ where $E_{kj}$ denotes the matrix unit with its nonzero entry in the $k,j$ entry. Let $x_f=[f_{kj}]\in M_n(A^*).$ Then, for $b\in M_n(A)$, $$ \phi(x_f)b=\sum_{k,j} f_{kj}(b_{kj})=\sum_{k,j}f(b_{kj}\otimes E_{kj})=f(\sum_{k,j}b_{kj}\otimes E_{kj})=f(b), $$ so $\phi(x_j)=f$ and $\phi$ is onto.