# Maps which are both completely positive and positive

Definition:
A linear map $$f:\mathbb C^n\to \mathbb C^n$$ is called positive if $$\langle fa,a\rangle\ge0$$ for all $$a\in \mathbb C^n$$. Equivalently, $$f\in M_{n}(\mathbb C)$$ is positive if it can be written in the form $$g^*g$$ for some $$g\in M_{n}(\mathbb C)$$. The unique positive element $$g$$ satisfying $$g^2=f$$ is denoted $$\sqrt f$$, and is called the square root of $$f$$.

Definition:
A linear map $$\phi:M_{n}(\mathbb C)\to M_{m}(\mathbb C)$$ is called completely positive if it sends positive elements to positive elements, and the same holds for $$\phi\otimes id_{M_{k}(\mathbb C)}:M_{n\cdot k}(\mathbb C)\to M_{m\cdot k}(\mathbb C)$$ for every $$k\in\mathbb N$$.

Let $$e_{ij}\in M_{n}(\mathbb C)$$ be the elementary matrix with a $$1$$ at $$(i,j)$$ and all other entries zero. Using the basis elementary matrices $$\{e_{ij}\}$$ to identify $$M_{n}(\mathbb C)$$ with $$\mathbb C^{n^2}$$, we can talk about a linear map $$f:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$$ being positive (this has nothing to do with sending positive elements to positive elements).

Question:
Let $$\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$$ be a map which is both positive and completely positive. Is its square root $$\sqrt\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$$ completely positive?

Remark:
Maps which are both positive and completely positive are frequent. Indeed, for every completely positive map $$\phi:M_{n}(\mathbb C)\to M_{m}(\mathbb C)$$, its adjoint $$\phi^{*}:M_{m}(\mathbb C)\to M_{n}(\mathbb C)$$ is also completely positive. So $$\phi^{*}\phi:M_{n}(\mathbb C)\to M_{n}(\mathbb C)$$ is both positive and completely positive.

• Just to clarify: in your last remark, when you identify $M_m({\bf C})$ with its dual, what is your pairing between $M_m({\bf C})$ and $T_m({\bf C})=M_m({\bf C})$? Are you taking ${\rm Tr}(AB)$ or ${\rm Tr}(AB^\top)$? – Yemon Choi Nov 13 '18 at 0:43
• The pairing is the one with respect to which the $e_{ij}$ form an an orthonormal basis. Namely, $\mathrm{Tr}(AB^*)$. – André Henriques Nov 13 '18 at 15:50

No. I use facts about Schur multipliers which can be found, for example in Paulsen's monograph on completely bounded maps. Basically you are searching for a positive semidefinite matrix with positive entries such that the pointwise square root of the matrix is not positive semidefinite. Specifically, let $$A=\left[ \begin{array}{ccc} 1 & 2^{-1/2} & 0\\ 2^{-1/2} & 1 & 2^{-1/2}\\ 0 & 2^{-1/2} & 1 \end{array} \right].$$ Let $$S_A:M_3(\mathbb{C})\rightarrow M_3(\mathbb{C})$$ be the Schur multiplier associated with $$A$$ (i.e. $$S_A$$ takes a matrix to its Schur product with $$A$$). Since $$A$$ is positive semidefinite, the Schur multiplier is completely positive. Since $$S_A(e_{i,j})=r_{i,j}e_{i,j}$$ for some $$r_{i,j}\geq 0$$ and the set $$(e_{i,j})$$ forms an orthonormal basis, this is a positive map. The positive square root is clearly $$S_B$$ where $$B=\left[ \begin{array}{ccc} 1 & 2^{-1/4} & 0\\ 2^{-1/4} & 1 & 2^{-1/4}\\ 0 & 2^{-1/4} & 1 \end{array} \right].$$ But $$B$$ is not a positive semidefinite matrix and therefore $$S_B$$ is not completely positive.