# Necessary and sufficient conditions for $x'Ax = 0$

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?

## 1 Answer

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.

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