Lemma 41.9.1.label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, and $\phi \in L(S; M_{n}(\complex))$. Define
\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]
then:
- (1)
Let $\xi = \sum_{j = 1}^{n} e_{j} \otimes e_{j} \in \complex^{n} \otimes \complex^{n}$, then
\[\dpn{a, \tilde \phi}{M_n(S)}= \dpn{\phi_n(a)\xi, \xi}{\complex^n \otimes \complex^n}\]for all $a \in M_{n}(S)$.
- (2)
The mapping $\phi \mapsto \tilde \phi$ is a linear isomorphism between $M_{n}(S)^{*}$ and $L(S; M_{n}(\complex))$.
Proof. (1): For any $a \in A$ and $1 \le k, l \le n$,
\begin{align*}\dpn{\phi_n(E_{kl} \otimes a)\xi, \xi}{\complex^n \otimes \complex^n}&= \sum_{i, j = 1}^{n} \dpn{\phi_n(E_{kl} \otimes a)(e_j \otimes e_j), (e_i \otimes e_i)}{\complex^n \otimes \complex^n}\\&= \sum_{i, j = 1}^{n} \dpn{E_{kl} e_j, e_i}{\complex^n}\dpn{\phi(a)e_j, e_i}{\complex^n}= \phi(a)_{kl}\end{align*}
(2): Let $\phi_{0}: M_{n}(S) \to \complex$ be a linear map, and let
\[\phi: S \to M_{n}(\complex) \quad a \mapsto \sum_{i, j = 1}^{n} \dpn{E_{ij} \otimes a, \phi_0}{M_n(S)}\cdot E_{ij}\]
then for any $1 \le k, l \le n$ and $a \in A$,
\[\dpn{E_{kl} \otimes a, \tilde \phi}{M_n(S)}= \phi(a)_{kl}= \dpn{E_{kl} \otimes a, \phi_0}{M_n(S)}\]
Therefore the mapping $\phi \to \tilde \phi$ is an isomorphism.$\square$
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