Theorem 41.9.3 (Krein’s Extension Theorem).label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, and $\phi: S \to M_{n}(\complex)$ be completely positive, then there exists a completely positive map $\Phi: A \to M_{n}(\complex)$ such that $\Phi|_{S} = \phi$.
Proof. Let
\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]
then by Theorem 41.9.2, $\tilde \phi$ is a positive linear functional. As $S$ is an operator system, $\normn{\tilde \phi}_{S^*}= \tilde \phi(1_{A})$. By the Hahn-Banach Theorem and Lemma 41.9.1, there exists $\Phi \in L(A; M_{n}(\complex))$ such that $\tilde \Phi|_{M_n(S)}= \tilde \phi$ and $\Phi|_{S} = \phi$. By Theorem 39.9.2, $\tilde \Phi$ is positive, so $\Phi$ is completely positive by Theorem 41.9.2.$\square$
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