Solution
= Solution
Apply the <cyclic property of the trace> and $BB^{-1}=I$:
$$
\boxed{\operatorname{tr}(B^{-1}AB)=\operatorname{tr}(ABB^{-1})=\operatorname{tr}A.}
$$
This proves <similarity invariance of trace> without requiring $A$ or $B$ to be symmetric.