Suppose $A \in M_n(\mathbb R)$ is fixed. If we have ${\bf Tr}(AX) = 0$ for all real, symmetric, positive definite $X$, does it imply $A=0$?
I think the answer is yes. I tried different nonzero $A$'s, but I can always find an $X$ to which the trace is nonzero.
The map $(C,D)\mapsto \text{Tr}(D^TC)$ is an inner product. For any matrix $X$, write $$X = (X+\lambda I)-\lambda I$$for large enough $\lambda >0$. Then both $(X+\lambda I)$ and $\lambda I$ are positive definite and $$(X,A^T) = \text{Tr}(AX) = \text{Tr}(A(X+\lambda I)) - \text{Tr}(A(\lambda I)) = 0 - 0 = 0.$$ Since $(X,A^T) = 0$ for all $X$, we can choose $X = A^T$ to see that $A^T =0$ and so $A = 0$.