How to show that a map is positive (positive-preserving) iff it's dual is also positive?

Question

I was reading UTX: introduction to matrix analysis and application. And in section 2.6 it mentioned the dual $\alpha^*$, $\alpha^*:\mathbb{M}_n\rightarrow\mathbb{M}_k$, of a linear map $\alpha$ between matrices, $\alpha:\mathbb{M}_k\rightarrow\mathbb{M}_n$.

Which is defined by:

$\mathrm{Tr} \alpha(A)B=\mathrm{Tr} A\alpha^{*}(B) \ \ \ \ A\in\mathbb{M}_k, B\in\mathbb{M}_n$

In the same paragraph the author also fleetingly mentioned that it is "easy to see", that a map $\alpha$ can be positive (sending positive matrices to positive one) iff it's dual $\alpha^*$ is also positive.

But I'm struggling to find the connections between the trace used to define dual and the positivity preserving property of the map.

Anyone has idea why this is true?

Answer 1

Suppose that $\alpha:\mathbb{M}_k\to\mathbb{M}_n$ is positive, i.e., $\alpha(A)$ is positive for all positive $A\in\mathbb{M}_k$.

Let $A$ and $B$ be one-dimensional projections, i.e., projections on vectors $\xi$ respectively $\psi$. Then they are positive elements. Since $\alpha$ is positive it maps $A$ to a positive matrix $\alpha(A)$. For the trace on $\mathbb{M}_k$ we can choose any orthonormal basis $(e_j)^k_{i=1}$ of $\mathbb{C}^k$, so in particular we can choose $e_1 = \xi$. Similarly, for the trace on $\mathbb{M}_n$ we can choose any orthonormal basis $(f_j)^n_{j=1}$ of $\mathbb{C}^n$, so in particular we can choose $f_1 = \psi$. This gives us \begin{equation*} \begin{aligned} \langle\xi,\alpha^*(B)\xi\rangle &= \sum^k_{i=1}\langle e_i,A\alpha^*(B)e_i\rangle \\ &= \text{Tr}A\alpha^*(B) \\ &= \text{Tr}\alpha(A)B \\ &= \sum^n_{j=1}\langle f_j,\alpha(A)B f_j\rangle \\ &= \langle\psi,\alpha(A)\psi\rangle \\ &\geq 0 \end{aligned} \end{equation*} by the positivity of $\alpha(A)$ due to $\alpha$ and $A$ being positive. As this holds for any one-dimensional projections $A$ and $B$, and positive elements are of the form $\sum_i x_i Q_i$ where the $x_i$ are positive numbers and the $Q_i$ are projections on vectors $\varphi_i$ such that the $\varphi_i$ form an orthonormal basis, the result follows.

Answer 2

First of all, note that it suffices to show that $\alpha$ being positive implies that $\alpha^*$ is positive, since $\alpha^{**} =\alpha$.

Second, note that the set of positive (semidefinite) matrices is forms a self-dual closed cone. That is, we can say that a matrix $M$ is positive if and only if $\operatorname{tr}(MP)\geq 0$ for all positive matrices $P$.

Now, suppose that $\alpha : \Bbb M_k \to \Bbb M_n$ is a positive map. Denote $\langle P,Q\rangle = \operatorname{tr}(PQ)$. We note that for all positive matrices $P \in \Bbb P_n$ and $Q \in \Bbb P_k$, we have $$ \operatorname{tr}(\alpha^*(Q)P) = \operatorname{tr}(P\alpha^*(Q)) = \operatorname{tr}(\alpha(P)Q)\geq 0. $$ Thus, if $Q \in \Bbb P_k$, then $\alpha^*(Q) \in \Bbb P_n$, which is what we wanted to show.