A question about a linear bounded operator (in hilbert space) I am reading a book about C*-algebra. When i study von Neumann algebras in this book, i meet with a problem. In the book, 
If $H$ is a Hilbert space, we write $H^{(n)}$ for the orthogonal sum of n copies of $H$. If $a\in M_{n}(B(H))$ (an operator matrice), we define $\phi(a)\in B(H^{(n)})$ by setting 
$$\phi(a)(x_{1}, x_{2},..., x_{n})=(\Sigma_{j=1}^{n}a_{1j}(x_{j}),...,\Sigma_{j=1}^{n}a_{nj}(x_{j})),$$ 
for all $(x_{1}, ...,x_{n})\in H^{(n)}$. It is easy to verify that the map 
$$\phi: M_{n}(B(H)) \rightarrow B(H^{(n)}), a\rightarrow \phi(a),$$
is a *-isomorphism.
My question is how to verify that $\phi$ is a *-isomorphism. I suppose it is not an easy conclusion for me .
 A: I will use the notation
$$
(x_1,\ldots,x_n)=\bigoplus_{k=1}^nx_k.
$$
So
$$
\left\langle\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\right\rangle=\sum_{k=1}^n\langle x_k,y_k\rangle.
$$
With the same proof as for complex matrices, one shows that $\phi(a)^*=\phi(a^*)$, where $(a^*)_{kj}=a_{jk}^*$:
$$
\langle\phi(a)\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\rangle=\langle\bigoplus_{k=1}^n\sum_{j=1}^na_{kj}x_j,\bigoplus_{k=1}^ny_k\rangle=\sum_{k=1}^n\sum_{j=1}^n\langle a_{kj}x_j,y_k\rangle
=\sum_{j=1}^n\sum_{k=1}^n\langle x_j,a_{kj}^*y_k\rangle=\sum_{j=1}^n\langle x_j,\sum_{k=1}^na_{kj}^*y_k\rangle=\langle\bigoplus_{j=1}^nx_j,\bigoplus_{j=1}^n\sum_{k=1}^na_{kj}^*y_k\rangle=\langle\bigoplus_{j=1}^nx_j,\phi(a^*)\bigoplus_{j=1}^ny_j\rangle
=\langle\phi(a^*)^*\bigoplus_{k=1}^nx_k,\bigoplus_{k=1}^ny_k\rangle.
$$
So $\phi(a^*)^*=\phi(a)$ for all $a$, which implies that $\phi(a^*)=\phi(a)^*$. 
We have
$$
\phi(a+b)(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^n(a+b)_{kj}x_j=\bigoplus_{k=1}^n\sum_{j=1}^n\,a_{kj}x_j+b_{kj}x_j\\=\bigoplus_{k=1}^n\sum_{j=1}^n\,a_{kj}x_j+\bigoplus_{k=1}^n\sum_{j=1}^n\,b_{kj}x_j=\phi(a)(\bigoplus_{k=1}^nx_k)+\phi(b)(\bigoplus_{k=1}^nx_k)=[\phi(a)+\phi(b)]\bigoplus_{k=1}^nx_k,
$$
showing that $\phi$ is linear. Also,
$$
\phi(ab)(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^n(ab)_{kj}x_j=\bigoplus_{k=1}^n\sum_{j=1}^n\sum_{h=1}^na_{kh}b_{hj}x_j=\bigoplus_{k=1}^n\sum_{h=1}^na_{kh}\sum_{j=1}^nb_{hj}x_j
=\phi(a)(\bigoplus_{k=1}^n\sum_{j=1}^nb_{hj}x_j)=\phi(a)\phi(b)(\bigoplus_{k=1}^nx_k),
$$
and then $\phi(ab)=\phi(a)\phi(b)$. 
If $\phi(a)=0$, then $\sum_{j=1}^na_{kj}x_j=0$ for any $k$ and any $x_1,\ldots,x_n$. From here we can see that $a_{kj}=0$ for all $k,j$. That is, $a=0$. So $\phi$ is injective. 
Finally, let $T\in B(H^{(n)})$. Let $\pi_j$ be the map $\displaystyle\bigoplus_{k=1}^nx_k\mapsto x_j$. Then define $T_{kj}\in B(H)$ by
$$
T_{kj}x=\pi_k T(0,\ldots,0,x,0,\ldots,0),
$$
where the $x$ is in the $j^{\rm th}$ place in the $n$-tuple. Then
$$
\phi((T_{kj})_{kj})(\bigoplus_{k=1}^nx_k)=\bigoplus_{k=1}^n\sum_{j=1}^nT_{kj}x_j=\sum_{j=1}^n\bigoplus_{k=1}^nT_{kj}x_j=\sum_{j=1}^n\bigoplus_{k=1}^n\pi_kT(0,\ldots,0,x_j,0,\ldots,0)\bigoplus_{k=1}^n\pi_kT(\bigoplus_{k=1}^nx_j)=T(\bigoplus_{k=1}^nx_k).
$$
And so $\phi$ is surjective. 
A: I answered a similar question previously, you may (or may not) find this helpful A theorem about operators in Hilbert sapce.
