Proposition 41.11.6.label Let $S$ be a matrix ordered space, $\phi \in \text{hom}(S; M_{n}(\complex))$ and $s \in \text{hom}(M_{n}(S); \complex)$, and
\[s: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]
then the following are equivalent:
- (1)
$s$ is a positive linear functional.
- (2)
$\phi$ is $n$-positive.
- (3)
$\phi$ is completely positive.
Proof, [Proposition 13.2, Po02]. (1) $\Rightarrow$ (3): Let $\xi = (\xi_{1}, \cdots, \xi_{n}) \in \complex^{n}$, $\eta \in (\eta_{1}, \cdots, \eta_{n}) \in \complex^{n}$, and $x \in S$, then
\[\dpn{\phi(x)\xi, \eta}{\complex^n}= \sum_{i, j = 1}^{n} \phi(x)_{ij}\xi_{j} \ol{\eta_i}= s({\eta^* x\xi})\]
Let $m \in \natp$, $\xi = (\xi_{1}, \cdots, \xi_{m}) \in \complex^{m} \otimes \complex^{n}$, and $A = \sum_{i, j = 1}^{m} E_{ij}\otimes a_{ij}\in M_{n}(S)$ with $a \ge 0$, then
\begin{align*}\dpn{\phi_n(A)\xi, \xi}{\complex^m \otimes \complex^n}&= \sum_{i, j = 1}^{m} \dpn{\phi(a_{ij})\xi_j, \xi_i}{\complex^n}= \sum_{i, j = 1}^{m} s(\xi_{i}^{*}a_{ij}\xi_{j}) \\&= s\braks{\begin{bmatrix}\xi_{1} \\ \vdots \\ \xi_{n}\end{bmatrix}^*A\begin{bmatrix}\xi_{1} \\ \vdots \\ \xi_{n}\end{bmatrix}}\ge 0\end{align*}
$\square$
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