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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?

Rodrigo de Azevedo
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Bravo
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2 Answers2

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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 $U^T$.

Thomas
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  • Thanks. Does the bounds using singular values fit somewhere inside the ones I have with eigenvalues? I do not see an easy way as we do not have normal matrices... – Bravo Feb 05 '13 at 15:47
  • Sorry I dont quite understand your question. Are you searching for a relation like $\max \sigma < \max \lambda$ or $\max \sigma > \max \lambda$? – Thomas Feb 05 '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 05 '13 at 15:58
  • Oh I don't know aboth that, but you can go this way $$ \max{\min |\sigma|, \min |\lambda|}\leq |x^TAx|\leq \min{\max |\sigma|, \max |\lambda|} $$ – Thomas Feb 05 '13 at 16:02
  • @Bravo perhaps this could help ? https://math.stackexchange.com/questions/127500/what-is-the-difference-between-singular-value-and-eigenvalue – Marine Galantin Dec 30 '22 at 01:23
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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.

noumenon28
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