Suppose $tr(A(P\otimes Q))\geq0$ for all semi-definite positive matrices P and Q, does it implies that A is semi-definite positive?

Question

Suppose $\text{tr}(A(P\otimes Q))\geq0$ for all semi-definite positive matrices P and Q, does it imply that A is semi-definite positive? If it is not true, please provide some ideas on restricting $A$ to some special forms in order to make the proposition true.

Answer 1

It is not true in general. A counter-example is $\text{SWAP}$ on $\mathbb{C}^2 \otimes \mathbb{C}^2$. Let $e_0, e_1$ be orthonormal in $\mathbb{C}^2$. Then, \begin{align} \text{SWAP} := \sum_{i, j = 0}^1 e_{i} e_{j}^* \otimes e_{j} e_{i}^*. \end{align} It is unitary, self-adjoint, and satisfies the easily verifiable equality \begin{align} \text{tr} (\text{SWAP} (A \otimes B)) = \text{tr} (AB), \end{align}
for any operators $A, B$ on $\mathbb{C}^2$. This implies in particular that for $P, Q \geq 0$, $\text{tr} (\text{SWAP} (P \otimes Q)) = \text{tr} (P Q) \geq 0$. However, $\text{SWAP}$ has a negative eigenvalue.