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Suppose that for a given fixed $v$, and two positive definite symmetric matrices $A$ and $B$ (with real-valued entries) it holds that: $$ v^T A v > v^T B v. $$ Does it then hold that $$ v^T A^{-1} v < v^T B^{-1} v $$ for the same $v$?

Context: I am working with some inequalities involving quadratic forms of the form $v^TA^{-1}v$ where $A$ is the expected value of a matrix-valued random variable. I can bound the expectation $v^TAv$ both above and below but not its inverse so easily, so if I could get rid of the inversion it would be helpful, but I'm not convinced that I can. Of course there may also be another way to approach problem.

What I have tried: I considered some basic approaches using spectral decompositions of $A$ and $B$ but since the eigenvectors can be different I didn't get very far. I am not, however, very experienced in this area. I also tried to find a simple counterexample among 2 x 2 matrices. I couldn't but that does not mean that there isn't one!

usr313
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jul 17 '23 at 09:37
  • Perhaps related https://math.stackexchange.com/questions/4677765/prove-or-disprove-the-following-generalization-sum-i-j-1nf-lefta-ia/4678471#4678471 – Barackouda Jul 17 '23 at 10:47
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    suppose you had two symmetric $3\times 3$ matrices each with the same eigenvalues of $\big{\lambda, 1, \lambda^{-1}\big}$ and with $\big \Vert \mathbf v \big\Vert =1$ you could have the case where the first quadratic form $=\frac{1}{2}\big(\lambda + \lambda^{-1}\big)$ and the second one equal $=1 \cdot 1$, then $ \mathbf v^T A^{-1}\mathbf v=\mathbf v^T A\mathbf v\gt 1 = \mathbf v^T B\mathbf v=\mathbf v^T B^{-1}\mathbf v$. Supposing you know spectral theorem, you should be able to fill in the details to make this an explicit counter-example. – user8675309 Jul 17 '23 at 18:11

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This is not correct. Try $n=2$ and $A=I_2$ and $B$ diagonal.

Gérard Letac
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