Suppose $x\in\mathbb{R}^n$, $A\in\mathbb{R}^{n\times n}$.
Does anyone know the answer to the following problems.
(1) $\min\limits_{x\neq0} f(x)=\frac{x^\mathrm{T}A^\mathrm{T}Ax}{x^\mathrm{T}Ax}$, where $\frac{A^\mathrm{T}+A}{2}\succ0$. Note that $A$ is not necessarily symmetric. Can anyone explicitly give the optimal objective value $f(x^*)$?
$f(x^*)=\lambda_{\min}\left\{\frac{A^\mathrm{T}+A}{2}\right\}$, smallest eigenvalue?
(2) $\min\limits_{x\in\mathcal{X}} f(x)=\frac{x^\mathrm{T}A^\mathrm{T}Ax}{x^\mathrm{T}Ax}$, where $\frac{A^\mathrm{T}+A}{2}\succcurlyeq0$ and $\mathcal{X}=\{x\big|x^\mathrm{T}Ax>0\}$. Again, $A$ is not necessarily symmetric. Can anyone explicitly give the optimal objective value $f(x^*)$?
$f(x^*)=\tilde{\lambda}_{\min}\left\{\frac{A^\mathrm{T}+A}{2}\right\}$, smallest non-zero eigenvalue?
1) Note that $A$ is invertible. Let $y = A x$, so $x^T A x = y^T A^{-1} y = y^T \left(\dfrac{A^{-1} + (A^{-1})^T}{2}\right) y$, while $x^T A^T A x = y^T y$. Thus minimizing $f(x)$ is equivalent to maximizing $ \dfrac{y^T B y}{y^T y}$ where $B = \dfrac{A^{-1} + (A^{-1})^T}{2}$, and that maximum occurs when $y$ is an eigenvector for the greatest eigenvalue of $B$, the corresponding value of $x$ being the reciprocal of that eigenvalue.