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For ench $n\geq1$, $B(\mathcal{H})$ is $\ast$-isomorphic to $\mathbb{M}_n(B(\mathcal{H}))$.

Thanks to the one who tell me the proof or tell me where I can find the proof.

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  • $\begingroup$ what is $\mathbb{M}_n(B(H))$? The set of all $n \times n$ matrices with entries from $B(H)$? Sorry, I don't know the very attractive font you use for "$H$"! What is it? $\endgroup$ – Robert Lewis Aug 11 '13 at 5:43
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    $\begingroup$ I believe the font for $\mathcal H$ is "\mathcal" $\endgroup$ – Devlin Mallory Aug 11 '13 at 7:40
  • $\begingroup$ @Devlin Mallory: thanks, I'll $\mathcal{CHECK IT OUT!}$ $\endgroup$ – Robert Lewis Aug 11 '13 at 8:16
  • $\begingroup$ @ Robert Lewis: Yes! It is the set of all $n \times n$ matrices with entries from $B(\mathcal{H})$. $\endgroup$ – Zhonghua Wang Aug 11 '13 at 8:24
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I assume you mean why is $B(\mathcal{H}^n)$ $*$-isomorphic to $\mathbb{M}_n(B(\mathcal{H}))$, otherwise taking $\mathcal{H} = \mathbb{C}$ we'd have an isomorphism from $\mathbb{C}$ to $\mathbb{M}_n(\mathbb{C})$. It's a standard question that you'll find in texts about operator spaces and operator algebras. If I recall it features in Paulsen's book.

Here's my working of this problem when I looked at it. My apologies it's a little extensive.

Let $\mathcal{H}^{(n)}$ denotes the direct sum of $n$ copies of $\mathcal{H}$, then we can put an inner product on $\mathcal{H}^{(n)}$ making it a Hilbert space as follows $$ \left\langle \left( \begin{array}{c} \xi_1 \\ \vdots \\ \xi_n \end{array} \right), \left( \begin{array}{c} \eta_1 \\ \vdots \\ \eta_n \end{array} \right) \right\rangle = \langle \xi_1, \eta_1 \rangle + \dots + \langle \xi_n, \eta_n \rangle, $$ where $\left( \begin{array}{c} \xi_1 \\ \vdots \\ \xi_n \end{array} \right)$ and $\left( \begin{array}{c} \eta_1 \\ \vdots \\ \eta_n \end{array} \right)$ are in $\mathcal{H}^{(n)}$ with each entry in $\mathcal{H}$.

We show that any element in $M_n(B(\mathcal{H}))$ defines a bounded linear operator on $\mathcal{H}^{(n)}$. Let $(a_{ij})_{i,j=1}^n \in M_n(B(\mathcal{H}))$, that is the entries $a_{ij}$ are in $B(\mathcal{H})$. Define a map $\psi : M_n(B(\mathcal{H})) \to B(\mathcal{H}^{(n)})$ by $$ \psi(a)(\xi) = \left( \sum_{j=1}^n a_{kj} \xi_j \right)_{k=1}^n \qquad \forall \, \xi = \left( \xi_k \right)_{k=1}^n \in \mathcal{H}^{(n)}. $$ We have that this map is one-to-one as say $\psi(a) = \psi(a')$, then $$ \left( \sum_{j=1}^n a_{kj} \xi_j \right)_{k=1}^n = \psi(a)(\xi) = \psi(a')(\xi) = \left( \sum_{j=1}^n a'_{kj} \xi_j \right)_{k=1}^n $$ for all $\xi \in \mathcal{H}^{(n)}$ and we see that we must have $a = a'$.

Defining the canonical projections $p_i : \mathcal{H}^{(m)} \to \mathcal{H}$ we can define an inverse $a \mapsto (p_iap_j^*)_{i,j} \in M_n(B(\mathcal{H}))$ and thus we have a surjection.

Finally letting $b_{ij} = (a_{ij})^*$ we get $$ \psi((a)^*)(\xi) = \psi(b)(\xi) = \left( \sum_{j=1}^n b_{ij} \xi_j \right)_{i=1}^n = \left( \sum_{j=1}^n (a_{ij})^* \xi_j \right)_{i=1}^n. $$ We need to work out what $\psi(a)^*$ would correspond to. We have \begin{align*} ([\psi(a)^*](\xi), \eta)_{\mathcal{H}^{(n)}} &= (\xi, \psi(a)(\eta))_{\mathcal{H}^{(n)}} = \sum_{i=1}^n (\xi_i, (\psi(a)(\eta))_i)_\mathcal{H} = \sum_{i=1}^n \left( \xi_i, \sum_{j=1}^n a_{ij} \eta_j \right)_\mathcal{H} \\ &= \sum_{i,j=1}^n (\xi_i, a_{ij} \eta_j)_\mathcal{H} = \sum_{i,j=1}^n ((a_{ij})^* \xi_i, \eta_j)_\mathcal{H} = \sum_{j=1}^n \left( \sum_{i=1}^n (a_{ij})^* \xi_i, \eta_j \right)_\mathcal{H} \\ &= \left( \left( \sum_{i=1}^n (a_{ij})^* \xi_i \right)_{j=1}^n, \eta \right)_{\mathcal{H}^{(n)}} \end{align*} and thus we have $$ (\psi(a)^*)(\xi) = \left( \sum_{j=1}^n (a_{ji})^* \xi_j \right)_{i=1}^n $$ and therefore $$ (\psi(a)^*)(\xi) = \left( \sum_{j=1}^n (a_{ji})^* \xi_j \right)_{i=1}^n = \left( \sum_{j=1}^n a^*_{ij} \xi_j \right)_{i=1}^n = \psi(a^*)(\xi). $$ Thus we have a $*$-isomorphism $M_n(B(\mathcal{H})) \to B(\mathcal{H}^{(n)})$.

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Firstly, I assume you mean that the Hilbert space $H$ is countably infinite dimensional (because otherwise your statement is not true - take $H = \mathbb{C}$). Assuming that, you can check the following :

  1. $M_n(B(H)) \cong B(H^n)$ where $H^n$ is the n-fold direct sum.
  2. Any two countably infinite-dimensional Hilbert space is abstractly isomorphic (map one countable basis to another - this will give you a unitary)
  3. Hence, $H \cong H^n$, which implies that $B(H) \cong B(H^n)$.
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