Does $\operatorname{Tr}(AB) \ge 0$ imply $A \ge 0$?

Question

Consider a fixed matrix $A \in \Bbb C^{n \times n}$ and the $\Bbb C^{n \times n} \to \Bbb C$ function $B \mapsto \operatorname{Tr}(AB)$. If $A$ is positive definite, then I know that $\operatorname{Tr}(B^{*}AB) = \operatorname{Tr}(ABB^{*}) \ge 0$ for every matrix $B$. Does the converse hold? I.e., if $\operatorname{Tr}(B^{*}AB) \ge 0$ holds for every $B \in \Bbb C^{n \times n}$, does it imply $A$ is positive definite?

Answer 1

Yes, this is true. Indeed, for each vector $x$, you can take $B=B^*$ the projection on $x$ that is the matrix with coefficients $x_i\,\overline{x_j}$. Then $$ x^\top A x = |x|^{-2} \,\mathrm{Tr}(BAB) \geq 0, $$ so $A\geq 0$.