My question is: Over the space of $n\times n$ complex matrices, does there exist a non-positive matrix B such that trace(AB) is not positive if A is not positive? So just to clarify the terms, a positive matrix is a hermitian $(A=A^*)$ and all of its eigenvalues are non-negative so a better term may be positive-semidefinite. And when I say trace(AB) is not positive, I mean it is not a positive real number.
Of course if $A$ is a positive matrix, then for any positive matrix $B$, we must have that trace(AB) must also be positive as the eigenvalues of $AB$ are positive. Now if $A$ is not positive, that can mean that it is hermitian but not all eigenvalues are nonnegative, or that it is not even hermitian. In the first case, it is easy to find a postive matrix $B$ such that trace(AB) is negative since one of the diagonal elements of $A$ must be negative (after a change of basis).
The case I am having trouble with is if $A$ is not even Hermitian. In that case I don't know what form it takes so it is almost impossible to explicitly construct a positive matrix $B$ where $trace(AB) < 0$. It could be that all the eigenvalues of $A$ are positive, but $A$ is still not Hermitian, thus not positive. I am not sure if there are any decomposition theorem that could help us here.
We will restrict to real valued matrices $A,$ but we admit complex valued matrices $B.$
If $A+A^t $ is positive definite, then for any positive definite matrix $B$ there holds $${\rm Re} \,{\rm tr}\, (AB)\ge 0$$ Indeed, first observe that the matrix $\bar{B}$ is also positive definite (just by inspecting the definition). Hence $B_1={\rm Re}\,B$ is positive definite. We have $${\rm Re}\, {\rm tr}\, (AB)=\underline{ {\rm tr}\, (AB_1)}={\rm tr}((AB_1)^t)={\rm tr}(B_1A^t)=\underline{{\rm tr}(A^tB_1)}$$ Therefore adding up the underlined terms gives $$ 2{\rm Re}\, {\rm tr}\, (AB)={\rm tr}((A+A^t)B)\ge 0$$
For example the $2\times 2$ matrix $$A=\begin{pmatrix} 1& 1\\ 0 &1\end{pmatrix}$$ satisfies $A+A^t\ge 0.$
Conversely assume $A+A^t$ is not positive definite. Thus it admits a negative eigenvalue $\lambda$ corresponding to a unit real eigenvector $v_\lambda.$ Let $B$ denote the orthogonal projection onto $v_\lambda.$ Then $B$ is positive definite and has real entries. Moreover $$2{\rm tr}(AB)={\rm tr}((A+A^t)B)=\lambda<0$$