$M_n(B(H))$ is * isomorphic to $B(H^{(n)})$

Question

$H$ is a Hilbert space,$H^{(n)}$ is the orthogonal sum of n copies of H.How to prove $M_n(B(H))$ is * isomorphic to $B(H^{(n)})$.I met with troubles in checking the surjectivity.

Answer 1

There is a natural $*$-morphism $\phi : M_n(B(H)) \rightarrow B(H^n)$ :

If $(T_{ij})_{i,j \in \{1,\cdots,n\}} \in M_n(B(H))$, we define $\phi((T_{ij})_{i,j \in \{1,\cdots,n\}})$ to be the map sending $(v_1,\cdots,v_n) \in H^n$ to $(\sum^n_{j=1} T_{1j}v_j, \sum^n_{j=1} T_{2j}v_j, \cdots , \sum^n_{j=1} T_{nj}v_j) = (\sum^n_{j = 1} T_{ij}(v_j))_{i \in \{1,\cdots,n\}}$.

$\phi$ is easily seen to be injective.

Let $T \in B(H^n)$. Let $\pi_1,...\pi_n : H^n \rightarrow H$ be the projection on the $n$-factor. For $j \in \{1,...,n\}$, let $i_j : H \rightarrow H^n$ be the map sending $v \in H$ to $(0,\cdots, 0, v, 0, \cdots, 0)$ where the $v$ is un the $j$-th coordinate.

For $i,j \in \{1,...,n\}$, let $T_{ij} := \pi_i \circ T \circ s_j$. Then, $T_{ij}$ is a bounded operator on $H$, and $T = \phi((T_{ij})_{i,j \in \{1,...,n\}})$. Indeed, if $v_1,\cdots, v_n \in H$, then $\begin{array}{rcl} T(v_1,\cdots, v_n) &= &T(\sum^n_{j = 1} s_j(v_j))\\ &= &(\pi_1(T(\sum^n_{j = 1} s_j(v_j)),\cdots, \pi_n(T(\sum^n_{j = 1} s_j(v_j)))\\ &= &(\sum^n_{j = 1} T_{ij}(v_j))_{i \in \{1,\cdots,n\}}\\ &= &\phi((T_{ij})_{i,j \in \{1,...,n\}})(v_1,\cdots, v_n).\\ \end{array}$