Given a symmetric matrix $H$, can we prove $x^THx\le \sigma(H)\|x\|_2^2$?

Question

Given a symmetric matrix $H$, can we prove that $$x^THx\le \sigma(H)\|x\|_2^2$$ where $ \sigma(H)$ is the maximal singular value of $H$?

Answer 1

Yes. Say the eigenvalues are $\lambda_1, \lambda_2,\dots, \lambda_n\in\mathbb{R}.$ The singular values of $H$ are square roots of eigenvalues of $H^TH = H^2$, which are just $|\lambda_1|, \dots, |\lambda_n|.$ Therefore, your inequality is easily true since: $$\dfrac{x^THx}{x^Tx}\leq\max\limits_{1\leq i\leq n}\lambda_i\leq\max\limits_{1\leq i\leq n}|\lambda_i| = \sigma(H),$$

where we used the fact that the Rayleigh quotient takes values in $[\lambda_{min}, \lambda_{max}].$