It is well known that for a hermitian matrix $A$ we have $\lambda_{min}(A)=min{x\ne 0} <x,Ax>/<x,x>$, which we can see be diagonalizing $A$. Now here is my question about the following I stumbled upon while doing homework:
Let $A,B$ be hermitian positive definite matrices. Then $\lambda_{min}(AB^{-1})=min_{x\ne 0}<AB^{-1}x,x>_{B^{-1}}/<x,x>_{B^{-1}}$ where $<x,y>_{B^{-1}}:=<B^{-1}x,y>$.
I seem to be not able to prove this, so I am starting to doubt its validity. Can anyone shed some light on this?
If you add "min", then it works: $$ \begin{split} \min_{x\neq 0}\frac{\langle AB^{-1}x,x\rangle_{B^{-1}}}{\langle x,x\rangle_{B^{-1}}} &= \min_{x\neq 0}\frac{\langle B^{-1}AB^{-1}x,x\rangle}{\langle B^{-1}x,x\rangle} = \min_{x\neq 0}\frac{\langle B^{-1/2}AB^{-1/2}(B^{-1/2}x),(B^{-1/2}x)\rangle}{\langle (B^{-1/2}x),(B^{-1/2}x)\rangle} \\&= \min_{y\neq 0}\frac{\langle B^{-1/2}AB^{-1/2}y,y\rangle}{\langle y,y\rangle}=\lambda_{\min}(B^{-1/2}AB^{-1/2})=\lambda_{\min}(AB^{-1}). \end{split} $$