Suppose $A$, $B$, and $C$ are all $m\times m$ real, symmetric matrices. If $tr(AB) > tr(AC)$ holds, what does that imply about $B$ in comparison to $C$?
If every element of $AB$ is larger than every element of $AC$, then I can conclude that $tr(AB) > tr(AC)$, as trace is just the sum of the diagonal elements.
However, even if every element of $B$ is larger than every element of $C$, I don't think it's necessarily true that $tr(AB) > tr(AC)$. So I'm not sure if it makes sense to compare the magnitude of the elements in $B$ to those in $C$.
Since the trace of a square matrix is the sum of its eigenvalues, I was also trying to think of conditions on $B$'s eigenvalues in comparison to $C$'s eigenvalues that would imply $tr(AB) > tr(AC)$. However, is it necessarily true that if $B$'s eigenvalues are larger than $C$'s eigenvalues, that $tr(AB) > tr(AC)$?