Block Matrices of Operators

Question

I'm trying to prove the following: Consider the vector space of matrices of size $n\times n$ whose entries in $\mathcal B(H)$. Denote this vector space by $M_{n,n}(\mathcal{B(H)})$. We can define involution on $M_{n,n}(\mathcal{B(H)})$ by $$ T^*=[T_{ij}]^*=[T_{ji}^*],\qquad\text{where}\quad T=[T_{ij}]\in M_{n,n}(\mathcal{B(H)}). $$ Thus we have an involutive algebra $M_{n,n}(\mathcal{B(H)})$. I want to prove that there exist a norm on $M_{n,n}(\mathcal{B(H)})$ which makes it a $C^*$-algebra. First I try to prove that there exists a $*$ isomorphism between $M_{n,n}(\mathcal{B(H)})$ and $\mathcal{B}\left(\bigoplus\limits_{k=1}^n\mathcal H\right)$. So I define the following map: $$J:M_{n,n}(\mathcal{A})\to\mathcal{B}\left(\bigoplus\limits_{k=1}^n H\right):[T_{ij}]\mapsto\left((x_1,.....,x_n)\mapsto\left(\sum\limits_{j=1}^n(T_{1j})x_j,\ldots,\sum\limits_{j=1}^n(T_{nj})x_j\right)\right) $$
I proved that this map is $1-1$. My question is : is it onto ? or does it have a closed range? Also , to prove that $J([T_{ij}]^*)= J([T_{ji}^*])=J^*([T_{ij}]))$. I did the following using the definition of the adjoint operator, but I'm not sure where things go wrong: I take $x=(x_1,......,x_n), y=(y_1,.....,y_n)\in \bigoplus\limits_{k=1}^n H $, then
$$\begin{align}(J(T)^*x)(y)= x(J(T)y)&= x(\sum\limits_{j=1}^n(T_{1j})y_j,....,\sum\limits_{j=1}^n(T_{nj})y_j)\\& = \sum\limits_{i=1}^n \left [x_i(\sum\limits_{j=1}^n(T_{1j})y_j,....,\sum\limits_{j=1}^n(T_{nj})y_j)\right] \\&=\sum_{i=1}^n \sum_{j=1}^n (T_{ij}y_j)(x_i)\\&=\sum_{j=1}^n \sum_{i=1}^n (y_j)(T^*_{ij}(x_i)) \end{align}$$

Answer 1

No problem, it is onto. So $M_n(B(H))$ is canonically *-isomorphic to $B(H^n)$. One way to alleviate a little bit the algebra is to introduce the projections $$ p_j:H^n\longrightarrow H\qquad(x_i)\longmapsto x_j. $$ Note that we have $$ p_i^*:H\longrightarrow H^n\qquad y\longrightarrow (0,\ldots,0,y,0,\ldots,0) $$ where $y$ is in $i$th position. Then $J:M_n(B(H))\longrightarrow B(H^n)$ is given by $$ J((T_{ij}))=\sum_{i,j}p_i^*T_{ij}p_j. $$ It makes it straightforward to check it is a *-homomorphism. And it is not difficult to show that it is invertible with inverse given by $$ B(H^n)\ni T\longmapsto(p_iTp_j^*)\in M_n(B(H)). $$ The key things to notice are that $p_jp_i^*=\delta_{ij}\rm{Id}_H$ and $\sum_i p_i^*p_i=\rm{Id}_{H^n}$.

With this *-isomorphism, one can pull back the $C^*$-norm of $B(H^n)$ to make $M_n(B(H))$ a $C^*$-algebra equipped with the norm $$ \|(T_{ij})\|:=\|J((T_{ij}))\|=\|\sum_{i,j}p_i^*T_{ij}p_j\|. $$