# Completely positive maps - dimension of the ancilla space

If a map between positive operators $$\Phi: X \rightarrow Y$$ is also completely positive, it is true that $$\Phi\otimes I_A$$ is also a positive map for any choice of ancilla operator space $$A$$.

That is, for any positive semidefinite operator $$\rho \in X$$, my map gives me a positive semidefinite operator $$\Phi(\rho) = \rho' \in Y$$. But my state could be entangled i.e. I could have had $$\rho = Tr_A(\sigma)$$ for some positive semidefinite $$\sigma \in X\otimes A$$. In this case, the map still works $$(\Phi\otimes I_A) \sigma_{XA} = \sigma'_{XA}$$ and gives a positive semidefinite output.

My question is regarding the dimension of $$A$$. I have seen proofs only consider the dimension of $$A$$ to be equal to the dimension of $$X$$. Why is it not required to consider a larger ancilla?

EDIT: To see the proof of what is written in the accepted answer, see https://en.wikipedia.org/wiki/Choi%27s_theorem_on_completely_positive_maps

Given a channel $$\Phi$$, the Choi state $$\sigma_\Phi=(\Phi\otimes I)(|\Omega\rangle\langle\Omega|)$$ with $$|\Omega\rangle=\tfrac{1}{\sqrt{d}}\sum_{i=1}^d|i,i\rangle$$ the maximally entangled state, contains all the information about the channel $$\Phi$$ (i.e., given $$\sigma_\Phi$$, we can reconstruct $$\Phi$$). In particular, if $$\sigma_\Phi$$ is positive semidefinite, this implies that $$\Phi$$ is completely positive. Thus, it is sufficient to choose the dimension of the ancilla equal to the dimension of the system, and it is sufficient to test positivity on a single state on that system+ancilla (e.g. the maximally entangled state $$|\Omega$$, but in fact any pure state with full Schmidt rank will do).
• Thank you. I think you answered my question (I didn't think about the Choi state at all) but I think for completeness, I will put this link here en.wikipedia.org/wiki/…. It has an explicit proof about why n-positive $\Phi$ implies a completely positive $\Phi$. Dec 20, 2018 at 14:53