Suppose that $A$ and $B$ are both $p \times p$ non-singular matrices, and we define $B^{-1/2}AB^{-1/2} = \Gamma \Lambda \Gamma^T$ where $\Gamma$ is orthogonal and $\Lambda$ is a diagonal matrix.
Then, let $\Xi = (\xi_1, \ldots, \xi_p) = B^{-1/2}\Gamma$ so that $\xi_i = B^{-1/2}\gamma_i$ for $I = 1, \ldots, p$. Then, by metric eigenvalues and simultaneous diagonalization:
$$ A\xi_i = \lambda_iB\xi_i $$
here, we call $\lambda_1, \ldots, \lambda_p$ the eigenvalues of $A$ in the $B$ metric. From this set-up, I'd like to show that $\lambda_1, \ldots, \lambda_p$ are also the roots of $det(B^{-1}-\lambda A^{-1}) = 0$, i.e., the eigenvalues of $B^{-1}$ in the metric $A^{-1}$.
Is there an easy way to see what is going on here? I'm having trouble visualizing exactly what is going on.