**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.