I'm super new to studying functional analysis and I'm currently reading https://arxiv.org/pdf/1410.7188.pdf (The Functional Analysis of Quantum Information Theory based on lectures by Paulsen, Pisier, and Winter). Anyway, it is claimed that
$$M_n(B(H)) \cong B(H^{(n)})$$ where $$H^{(n)} = H \oplus \cdots \oplus H$$
but I'm not quite sure how to make this identification, although it feels natural.
My attempt:
$M_1(B(H))\cong B(H)$ seems obvious since if we have an element of $B(H)$, call it $T$, then we can put $T$ in a matrix $[T]$ and so consider it an element of $M_1(B(H))$.
If I consider $M_2(B(H))\cong B(H^{(2)})$ it is not clear to me what the map is that connects these spaces (but I figure if I can understand this example then the general one follows).
I think, for starters, I should have a clear idea of what elements of $B(H^{(2)})$ are (and elements of $M_2(B(H))$ seem pretty clear).
Here's my take: If $T\in B(H^{(2)})$ then it is bounded and operates on elements of $H^{(2)}$. Elements of $H^{(2)}$ look like $h=h_1\oplus h_2$ where $h_j \in H$ for $j=1,2$.
A "natural" way for an operator to act on something like $h$ is for us to define how it acts in each component... so presumably $T = T_1\oplus T_1$ and each $T_i$ is bounded since $T$ is. ... and we define $$Th = (T_1\oplus T_2)(h_1\oplus h_2) = (T_1h_1\oplus T_2h_2)$$
So my gut feeling is that this looks like:
$$Th = \begin{pmatrix} T_1&0\\0&T_2 \end{pmatrix}\begin{pmatrix} h_1\\h_2 \end{pmatrix} = \begin{pmatrix} T_1h_1\\T_2h_2 \end{pmatrix} $$
I don't actually know what $T_1\oplus T_2$ (or $T$) should look like, this just feels right.....
But... if this is the case, then I guess the isomorphism that helps me identify the spaces is a map which decomposes $T$ into its components then sends those to the appropriate diagonal matrix... yet this doesn't feel right because $M_2(B(H))$ should consist of more than just diagonal elements.
Thank you for reading.... I wrote out all my thoughts just to illustrate I've at least tried.
It is true that $M_n(B(H)) \cong B(H^n)$. So you sould indeed prove that there are more than just diagonal operators in $B(H^n)$. You can prove that the following map is an isomorphism of $C^*$-algebras: $$ \phi \colon M_n(B(H)) \to B(H^n), $$ where $$ \phi \big ( (T_{ij})_{i,j=1}^n \big)(\xi) = \begin{pmatrix} \sum_{k=1}^n T_{1k} \xi_k \\ \vdots \\ \sum_{k=1}^n T_{nk} \xi_k \end{pmatrix}. $$ Here $\xi \in H^n$ has the form $\xi_1 \oplus \ldots \oplus \xi_n$. Note that $\phi$ lets the matrix $ (T_{ij})_{i,j=1}^n $ act on $\xi$ just as you would expect it, namely by matrix-multiplication.