Bounds for a parameter-dependent quadratic form

Question

Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix, $x(t)\in\mathbb{R}^n$, and $t\in\mathbb{R}^+$. I'm interested in finding upper and lower bounds for the following quadratic form $$ f(x(t)) = x(t)^{\intercal}A x(t)$$ assuming that $\Vert x(t) \Vert \leq \beta$, $\beta\in\mathbb{R}^+$. Any help will be appreciated!

Answer 1

For any symmetric matrix $A$ there holds $$\lambda_{\max(A)}= \max\{x^TAx\,: \,\|x\|\le 1\}, \quad \lambda_{\min(A)}= \min\{x^TAx\,: \,\|x\|\le 1\}$$ (see part hessian) Therefore by homogeneity $$ \lambda_{\min(A)}\|x\|^2\le x^TAx\le \lambda_{\max(A)}\|x\|^2$$ i.e. $$ \lambda_{\min(A)}\|x(t)\|^2\le x(t)^TAx(t)\le \lambda_{\max(A)}\|x(t)\|^2\qquad (*)$$ Thus $$\sup \{x(t)^TAx(t)\,:\,\|x(t)\|\le \beta\}\le \lambda_{\max(A)}\sup_t\|x(t)\|^2\le \lambda_{\max(A)}\beta^2$$ Moreover the equality occurs when $x(t)\equiv \beta\, v_{\max(A)}$, where $v_{\max(A)}$ is the unit eigenvector corresponding to the eigenvalue $\lambda_{\max(A)}.$

By $(*)$ we get $$\inf_t x(t)^TAx(t)\ge \lambda_{\min(A)}\inf_t\|x(t)\|^2$$ The equality occurs when $x(t)=\|x(t)\|\,v_{\min},$ where $v_{\min}$ denotes the unit eigenvector corresponding to $\lambda_{\min(A)}$. The bound is equal $0$ if $\|x(t)\|$ is not bounded away from $0.$