Draft of today's classes.

src/op/system/complete-upgrade.tex

@@ -242,3 +242,25 @@ The following is a slight generalisation of one form of the above example, but s
     which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive. 
 \end{proof}
 
+\begin{proposition}[Choi-Kraus]
+\label{proposition:choi-kraus}
+    Let $\phi: M_n(\complex) \to M_m(\complex)$ be a completely positive map, then there exists $\bracsn{V_j}_1^{mn} \subset L(\complex^m; \complex^n)$ such that
+    \[
+    \phi(a) = \sum_{j = 1}^{nk}V_j^* a V_j
+    \]
+
+    for all $a \in M_n(\complex)$. 
+\end{proposition}
+\begin{proof}
+    Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\phi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \phi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$. 
+
+    As $M_n(\complex)$ is simple, the only (non-zero) irreducible representation of $M_n(\complex)$ is the identity. Since $(K, \pi)$ is a direct sum of irreducible representations, after appending additional factors, assume without loss of generality that $K = \bigoplus_{k = 1}^{nm}\complex^n$ and that $\pi(a) = I_{M_{mn}(\complex)} \otimes a$ for all $a \in M_n(\complex)$.
+
+    For each $1 \le k \le mn$, let $P_k \in B(K)$ be the orthogonal projection onto the subspace corresponding to the $k$-th summand, and $Q_k \in L(K; \complex^n)$ be the orthogonal projection onto the $k$-th summand itself, then
+    \begin{align*}
+    \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_j\pi(a)P_j)V = \sum_{k = 1}^{mn}V^*Q_j^*aQ_jV \\
+    &= \sum_{k = 1}^{mn} (Q_jV)^* a (Q_jV)
+    \end{align*}
+    
+\end{proof}
+