# Lower boundary of quadratic form

I have a quadratic form $x^TAx$ where $x$ is an $n \times 1$ vector and $A$ is a positive definite matrix in the sense that it has only positive eigenvalues. Am I right to say that $||x^TAx|| \ge \lambda_{min}(A) ||x||^2$ where $\lambda_{min}(A)$ is the smallest eigenvalue of $A$? If not, is there another lower boundary?

• It is correct - except that you should replace $\|x\|$ by $\|x\|^2$. – Friedrich Philipp Mar 19 '16 at 15:35
• @FriedrichPhilipp yeah, right, my mistake – Controller Mar 19 '16 at 15:39
• You never said that $A$ is symmetric. By your definition, is the matrix $$\pmatrix{1&10\\0&1}$$ positive definite? If so, then the answer is no. – Omnomnomnom Mar 19 '16 at 15:44
• @Omnomnomnom $A$ is not symmetric. In this case, is there another lower boundary? – Controller Mar 19 '16 at 15:47
• @Controller if you mean what I think you mean (as in, you're using the definition that I've given), then the correct lower bound is $$x^TAx \geq \frac 12 \lambda_{min}(A + A^T)\|x\|^2$$ – Omnomnomnom Mar 19 '16 at 16:24

The bound you've given works specifically in the case that $A$ is symmetric and positive definite. For the "asymmetric, positive definite" case, we have the more general bound $$x^TAx = x^T \left( \frac{A + A^T}{2} \right)x \geq \frac 12 \lambda_{min}(A + A^T)\|x\|^2$$