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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?

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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.

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  • $\begingroup$ 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

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