Suppose I have a symmetric matrix $A$, and $A$ is neither positive definite nor negative definite. So, can we say that the quadratic form $x^{T}Ax$ has a non trivial root? i.e. other than $x=0$?
I looked up Non-trivial kernel if quadratic form is indefinite this post but the reasoning is quite unclear to me. Can anyone help?
According to a theorem of Sylvester there exists a base change, i.e. an invertible matrix $S$, such that $D:=S^tAS$ is diagonal and the diagonal elements are in the set $\{-1,0,1\}$. If $A$ is indefinite two cases can appear:
At least one $0$ appears among the diagonal elements. Hence $e_k^tDe_k=0$ for some standard basis vector, which gives $x:=Se_k$ as a non-trivial 'root'.
$1$ and $-1$ appear as diagonal elements. Hence $e^tDe=0$ for a certain vector $e$ having exactly one coordintae equal to 1 and exactly one coordintae equal to -1. Again $Se$ then is a non-trivial 'root'.