**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 $k\in\mathbb N$.




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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?
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**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.



Maps which are both completely positive and positive
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. 


