I've this question that i hope it has an elegant solution.
Consider the symmetric part of a matrix $\mathcal{H}\{A\} = \dfrac{A+A^{T}}{2}$ where $A\in\mathbb{R}^{n\times n}$. Furthermore, consider that $\mathcal{H}\{A\}<0$, i.e. is negative-definite.
Provided that $A$ is invertible: Is it possible to say that $\mathcal{H}\{A^{-1}\}$ will be also negative-definite?
Thanks in advance!
If the symmetric part of $A$ is negative definite, then you have $x^t (A + A^t) x < 0 $ for all non-zero column vectors $x \in \mathbb{R}^n$, which is equivalent to $x^t A x < 0 $ for all non-zero column vectors $x \in \mathbb{R}^n$. Now do a change of variables $x = A^{-1}y$, and you get that, for all non-zero column vectors $y \in \mathbb{R}^n$, $y^t (A^{-1})^t y < 0 $, which is equivalent to saying that the symmetric part of $A^{-1}$ is negative definite.