0
$\begingroup$

I came across the following problem and I am having a hard time thinking about it.

Let $A$ be a $k\times k$ real matrix. Notice that I do not require that $A$ is symmetric, positive definite or anything else. I would like to consider any real matrix $A$ of such dimensions.

Now, I am interested in necessary and sufficient conditions for $\exists x \neq 0$ such that $ x' A x = 0$, where $x \in \mathbb{R}^{k}$.

Is this a known result? Any ideas?

$\endgroup$
0
$\begingroup$

Note that$$x^TAx = \sum_{i,j} A_{ij}x_i x_j = \sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$ The second equality is due to each product $x_ix_j$ occuring twice in the summation when $i\neq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.

$\endgroup$
  • $\begingroup$ If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right? $\endgroup$ – Raul Guarini Dec 8 '18 at 16:13
  • $\begingroup$ @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx \leq 0$. $\endgroup$ – LinAlg Dec 8 '18 at 16:14
  • $\begingroup$ True! Thanks a lot! $\endgroup$ – Raul Guarini Dec 8 '18 at 16:18

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.