The question is originated from an exercise of Paulsen's book "Completely Bounded maps and operator algebras" (P115 Exercises 8.2).
Let $A$ be a C*-algebra with unit, $H$ be a Hilbert space and let $\Phi: M_2(A)\rightarrow B(H\oplus H)$ be a completely positive map. Then we have a minimal Stinespring representation $(\pi_1, V, K_1)$ for $\Phi$. Prove the Hilbert space $K_1$ can be decomposed as $K_1=K\oplus K$ in such a way that the *- homomorphism $\pi_1:M_2(A)\rightarrow B(K\oplus K)$ has the form
$$\pi_1\left[\left[
\begin{array}{ccc}
a & b \\
c & d\\
\end{array}\right]\right]=\left[
\begin{array}{ccc}
\pi(a) & \pi(b) \\
\pi(c) & \pi(d)\\
\end{array}\right],$$
where $\pi: A\rightarrow B(K)$ is a unital *-homomorphism.
My attempt: In this exercise, I do not know how to construct the Hilbert space $K$ such that $K_1=K\oplus K$. Because the Stinespring representation is minimal (i.e. $\overline{\pi_1(M_2(A))V(H\oplus H)}=K_1$), I guess $K=\overline{\pi_1(M_2(A))V(H\oplus{0})}\cong \overline{\pi_1(M_2(A))V(0\oplus H)}$. However, I am not sure whether it is true.
Best Answer
Your approach sounds a priori like a good idea, but it cannot work because you have no control over $V$.
Instead, you have that $\pi_1(A)$ is non-degenerate by the minimality condition. This implies that $\pi_1(1)=1$ (because it is a projection with trivial kernel). Now let $$ K=\pi_1\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)\,K_1. $$ Then $$ K_1=K\oplus\pi_1\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)K\simeq K\oplus K.$$
Define
$$\pi(a)=\pi_1\left(\begin{bmatrix}a&0\\0&0\end{bmatrix}\right).$$ Then, writing $U=\pi_1\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)$, \begin{align} \pi_1\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right) &=\pi(a)\pi(E_{11})+\pi(b)\pi(E_{12})+\pi(E_{21})\pi(c)+\pi(E_{21})\pi(d)\pi(E_{12})\\[0.3cm] &=\pi(a)+\pi(b)U+U\pi(c)+U\pi(d)U\\[0.3cm] &\simeq\begin{bmatrix}\pi(a)&\pi(b)\\\pi(c)&\pi(d)\end{bmatrix} \end{align}