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Let matrices $A, B \in \Bbb C^{n \times n}$ be Hermitian, and $A - B \succeq 0$. Is $A B^{-1} - I \succeq 0$? If so, why?

What about the positive definite case? ($A - B \succ 0$)

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    $\begingroup$ Is $A B^{-1}$ symmetric? $\endgroup$ Mar 14, 2021 at 2:23

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That's not true. I'm assuming $A -B \succeq 0$ means $A-B$ is Hermitian positive semi-definite (and definite for $A-B \succ 0)$. Take $A=I_2$, $B=\begin{pmatrix} 0.5 & 0 \\\\ 0 & -2 \end{pmatrix}$.

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  • $\begingroup$ Or, based on your answer: $ 1 - (-2) >0$, but $-1/2 -1 <0$. $\endgroup$
    – peter a g
    Mar 14, 2021 at 3:49

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