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?

  • $\begingroup$ It is correct - except that you should replace $\|x\|$ by $\|x\|^2$. $\endgroup$ – Friedrich Philipp Mar 19 '16 at 15:35
  • $\begingroup$ @FriedrichPhilipp yeah, right, my mistake $\endgroup$ – Controller Mar 19 '16 at 15:39
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Mar 19 '16 at 15:44
  • $\begingroup$ @Omnomnomnom $A$ is not symmetric. In this case, is there another lower boundary? $\endgroup$ – Controller Mar 19 '16 at 15:47
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    $\begingroup$ @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 $$ x^TAx \geq \frac 12 \lambda_{min}(A + A^T)\|x\|^2 $$ $\endgroup$ – 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 $$ x^TAx = x^T \left( \frac{A + A^T}{2} \right)x \geq \frac 12 \lambda_{min}(A + A^T)\|x\|^2 $$

  • $\begingroup$ Hello @Omnomnomnom, do you mind showing a proof for this? $\endgroup$ – Hunle Mar 31 '18 at 1:20
  • $\begingroup$ @Hunle the proof is a bit too long for a comment and I'd rather not edit such an old answer. If you're interested, I suggest you post a new question that points to my answer. If I find the time, I might put down an answer myself. $\endgroup$ – Omnomnomnom Mar 31 '18 at 4:32

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