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 $f$.

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).

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?

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.

  • $\begingroup$ 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 ${\rm Tr}(AB^\top)$? $\endgroup$
    – Yemon Choi
    Nov 13, 2018 at 0:43
  • $\begingroup$ The pairing is the one with respect to which the $e_{ij}$ form an an orthonormal basis. Namely, $\mathrm{Tr}(AB^*)$. $\endgroup$ Nov 13, 2018 at 15:50

1 Answer 1


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.

  • 1
    $\begingroup$ This is correct. $\endgroup$
    – Nik Weaver
    Nov 13, 2018 at 3:20

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