If $n \times n$ matrix $A$ is a real and symmetric, then we know that for all $x \in \mathbb{R}^n$
$$\lambda_{\min} (A) \|x\|^2 \le x^T A x \le \lambda_{\max} (A) \|x\|^2$$
where $\lambda_{\max}$ and $\lambda_{\min}$ are the maximum and minimum eigenvalues of $A$, respectively.
However if matrix $A$ is real and non-symmetric, as answered in this question, we identify that
$$x^T A x = x^T \left( \frac{A+A^T}{2} \right) x$$
So, for all $x \in \mathbb{R}^n$, we have
$$\Re(\lambda_{\min} (A)) \|x\|^2 \le \hat{\lambda}_{\min} \|x\|^2 \le x^T A x \le \hat{\lambda}_{\max} \|x\|^2 \le \Re(\lambda_{\max} (A))\|x\|^2$$
where $\hat{\lambda}_{\min}$ and $\hat{\lambda}_{\max}$ are, respectively, the minimum and maximum eigenvalues of $\frac{A+A^T}{2}$.
Is there anything more we could say about this bound? Is it possible to tighten or simplify it somehow?
Best Answer
If we write the singular value decomposition of $A$ as $A= U \Sigma V^T$ with orthogonal matrices $U,\ V^T$ and $\Sigma = diag(\sigma_1,...,\sigma_n)$ where $\sigma_i$ denotes a singular value we can bound $x^TAx$ by the following $$ ||x^T Ax|| = ||x^TU \Sigma V^Tx||=|| u^T \Sigma v||\leq ||u^T|| \max_i{|\sigma_i|}||v||=||x|| \max_i{|\sigma_i|}||x||=\max_i{|\sigma_i|} ||x||^2 $$ with $u=U^Tx$ and $v =V^Tx$ and the fact that $||U^Tx||=||x||$ for any orthogonal matrix $U^T$.