# Bounding $x^TAx$ when A is not a symmetric matrix

If $A$ is a real and symmetric $n\times n$ matrix, then we know $\lambda_{min}||x||^2\le x^TAx \le \lambda_{max}||x||^2\ \forall x\in \mathbb{R}^n$ where $\lambda_{max}$ and $\lambda_{min}$ are the maximum and minimum eigenvalues of $A$ respectively.

However if $A$ is real and non-symmetric, as answered in this question, we identify that $x^TAx=x^T(\frac{A+A^T}{2})x$. So we have: $$Re(\lambda_{min})||x||^2 \le \hat{\lambda}_{min}||x||^2\le x^TAx \le \hat{\lambda}_{max}||x||^2\le Re(\lambda_{max})||x||^2\ \forall x\in \mathbb{R}^n$$ 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?

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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 $U^T$.
Sorry I dont quite understand your question. Are you searching for a relation like $\max \sigma < \max \lambda$ or $\max \sigma > \max \lambda$? – k1next Feb 5 '13 at 15:49
Yeah, I was wondering if there was a connection between $Re(\lambda_{min})||x||^2 \le x^TAx \le Re(\lambda_{max})||x||^2$ and $\sigma_{min}||x||^2 \le x^TAx \le \sigma_{max}||x||^2$· – Bravo Feb 5 '13 at 15:58
Oh I don't know aboth that, but you can go this way $$\max\{\min |\sigma|, \min |\lambda|\}\leq |x^TAx|\leq \min\{\max |\sigma|, \max |\lambda|\}$$ – k1next Feb 5 '13 at 16:02
There seems to be something wrong with your logic when you claim that $\text{Re}(\lambda_\min)\leq \hat{\lambda}_\min$ and $\hat{\lambda}_\max\leq\text{Re}(\lambda_\max)$, so I think the original bound is wrong. A counterexample is $$A=\pmatrix{2&2\\-2&4}, \quad\quad\tfrac{A+A^T}{2}=\pmatrix{2&0\\0&4}.$$ The first has eigenvalues $3 \pm i \sqrt{3}$, while the second has eigenvalues 2 and 4.