Why is a map defined as $A \rightarrow V^\dagger A V$ for all V completely positive?

Question

In this paper by M. D. Choi, he claims,

For each $n * m$ matrix $V$, it is evident that the map: $M_n \rightarrow M_m$ with $A \rightarrow V^\dagger A V$ is completely positive.

$M_x$ denotes all $x*x$ complex matrices.

I can see why such a map would be positive, as one can diagonalise write the matrices such that $A$ is diagonal, and we know that $V^\dagger V$ is positive, so multiplying would cause it to remain positive.

But how would one prove that such a map is completely positive?

Answer 1

One equivalent criterion for the positivity of a complex square matrix $M$ is that $v^\dagger Mv\ge0$ for every vector $v$. Now, let $x^\dagger=(x_1^\dagger,x_2^\dagger,\ldots,x_p^\dagger)$ and $y^\dagger=(x_1^\dagger V^\dagger,x_2^\dagger V^\dagger,\ldots,x_p^\dagger V^\dagger)$, where $x_i\in\mathbb C^m$ for each $i$. Then $$ x^\dagger \left(V^\dagger A_{ij}V\right)_{1\le i,j\le p}\,x =y^\dagger \left(A_{ij}\right)_{1\le i,j\le p}\,y \ge0 $$ whenever the block matrix $\left(A_{ij}\right)_{i,j\in\{1,2,\ldots,p\}}$ is positive. Hence $A\mapsto V^\dagger AV$ is completely positive.

Answer 2

If $\phi(A) = V^\dagger A V$, then we find that $$ \begin{align} (\phi \otimes \operatorname{id}_{M_k})(A \otimes B) &= \phi(A) \otimes B = (V^\dagger A V) \otimes B \\ & = (V \otimes \operatorname{id}_{\Bbb C^k}) (A \otimes B)(V \otimes \operatorname{id}_{\Bbb C^k})^\dagger \end{align} $$ so that $(\phi \otimes \operatorname{id}_{M_k})(C) = (V \otimes \operatorname{id}_{\Bbb C^k}) C(V \otimes \operatorname{id}_{\Bbb C^k})^\dagger$. It follows that $\phi \otimes \operatorname{id}_{M_k}$ is a positive map for all $k$, which is to say that $\phi$ is completely positive.