Given a channel $Phi(X)=sum_k c_k(X)sigma_k$, are there always $F_kge0$ such that $Phi(X)=sum_k operatorname{tr}(F_k X)sigma_k$?

Question

Fix a finite number of states $sigma_k$, and consider a channel of the form
$$Phi(X)=sum_k c_{k}(X)sigma_k.$$
For $Phi$ to be linear and trace-preserving we must have:
$$c_k(X+X') = c_k(X) + c_k(X'), qquad sum_k c_k(X)=1.$$
In other words, the coefficients must be linear functionals $c_kinmathrm{Lin}(mathcal X)^*$ for all $k$.
Does this imply that there must be some positive operators $F_kge0$ such that $c_k(X)=operatorname{Tr}(F_k X)$ for all $k$ (which in turn would imply $sum_k F_k=I$ and thus that ${F_k}_k$ is a POVM)? What's a good way to show this?

Norbert Schuch · Accepted Answer

The answer is no.
To this end, pick a linearly independent set ${sigma_k}$ which spans the full matrix space (over $mathbb C)$, that is, a basis.  (This is always possible, as the positive operators span the hermitian ones over $mathbb R$.)
Then pick a dual basis $sigma'_ell$ such that
$$
mathrm{tr}[sigma'_ell sigma_k]=delta_{kell} .
$$
Then,
$$
Phi(X) = sum_k mathrm{tr}[sigma'_k X],sigma_k
$$
is the identity channel, which cannot be written as a POVM $F_kge0$ followed by a preparation of $sigma_k$ (as that channel would be entanglement breaking).
(Note that this shows that the dual basis $sigma'_ell$ has non-positive elements. This is not surprising, since otherwise the scalar product $mathrm{tr}[sigma'_ellsigma_k]ge0$ for all $k,ell$.)

Given a channel $Phi(X)=sum_k c_k(X)sigma_k$, are there always $F_kge0$ such that $Phi(X)=sum_k operatorname{tr}(F_k X)sigma_k$?

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