Lemma 41.9.5.label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, $\phi \in L(S; M_{n}(\complex))$, and
\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]
then $\phi$ is positive if and only if $\tilde \phi|_{M_n(\complex)_+ \otimes S_+}\ge 0$.
Proof. ($\Rightarrow$): Let $a \in M_{n}(\complex)_{+}$ and $b \in S_{+}$. Using the Spectral Theorem, assume without loss of generality that $a$ is a rank one projection. That is, there exists $\alpha = (\alpha_{1}, \cdots, \alpha_{n}) \subset \complex^{n}$ such that $a = \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} E_{ij}$. As such,
\[\dpn{a \otimes b, \tilde \phi}{M_n(S)}= \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} \phi(b)_{ij}= \dpn{\phi(b)\alpha, \alpha}{\complex^n}\ge 0\]
($\Leftarrow$): Let $b \in S_{+}$, $\alpha = (\alpha_{1}, \cdots, \alpha_{n}) \in \complex^{n}$, and $a = \alpha\alpha^{*}$, then
\[\dpn{\phi(b)\alpha, \alpha}{\complex^n}= \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} \phi(b)_{ij}= \dpn{a \otimes b, \tilde \phi}{M_n(S)}\ge 0\]
$\square$
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