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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $M_n(B(\mathcal{H})) \to B(\mathcal{H}^{(n)})$.


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 $B(H) \cong B(H^n)$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.