If $\langle A,B \rangle=0$ for two symmetric matrices, what about $\langle A,B^{-1}\rangle $?

Question

You are given two square real symmetric matrices $A$ and $B$. They are mutually orthogonal $A \perp B$, or in other words their frobenius inner product is zero, or in other words $\langle A ,\, B\rangle_F=0$.

My intuition says $A\perp B^{-1}$, i.e. $\langle A, B^{-1}\rangle_F=0$, but I don't really know how to approach proving such an expression. Can you prove or disprove such a claim?

Answer 1

Take $A = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$ and $B = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}$. Then $\langle A, B\rangle = 0$ but $\langle A, B^{-1}\rangle = \frac{3}{2}$.