Given matrix $A \in \mathbb{R}^{n \times n}$, how do I solve $x^{T}Ax = 0$ for $x \in \mathbb{R}^n$?
Obviously, a zero vector is always a solution and if $A$ is positive or negative definite there is no other solution. However, I'm interested in the cases, where $A$ is neither. Just from plotting a few examples, I believe the solution in the two dimensional case should usually describe one or two lines, but an analytical solution eludes me.
The question Solving quadratic equations of the form $x'(A-B)x = 0$ seems to be closely related, but it only asks, if there is a solution, not how it looks like and is asking for the complex case. And, truth be told, I don't quite understand the answer anyway.
Hint: Proof that $x^T A x = x^T A_+ x$ for all $x$, where $A_+ = \frac{1}{2}(A+A^T)$ is the symmetric part of $A$. Then you can apply the spectral theorem.