Conditions such that $x^{\top}Ax>0$

Question

I know that a sufficient condition such that the quadratic form $x^{\top}Ax>0$ is positive is that eigenvalues of $A$ are all positive, but is there any weaker condition , e.g a condition depending on the vector $x$?

Answer 1

If the inequality $x^{\top}Ax > 0$ holds for every vector nonzero vector $x$, we say that $A$ is a positive definite matrix. There are a list of criteria that you can read here. An interesting necessary criterion (that isn't listed in the previous link) is that positive definite matrices must have all their diagonal entries positive.

Answer 2

Regarding your second question about conditions depending on the vector $x$. You can relax the condition of the matrix being positive semidefinite by constraining your vector $x$. For example, a matrix for which $x^T A x \geq 0$ where every entry in $x$ is positive is called copositive. I'm also struggling with a problem like that. You can view the ongoing discussion in this forum.