Fact: Let $A$ be a symmetric matrix. Then for a vector $x$, $x^TAx=0$ if and only if $Ax=0$.

Proof:

Sufficiency: $Ax=0$ implies $x^TAx=0$. Easy

Necessary: ???

Wrong!!! $x^TAx=0$ implies $Ax=0$ only holds for positive (semi) definite matrices! See Horn P400, problem 1 for proofs.

EDIT: if A is p.s.d, we can prove $x^TAx=0$ implies $Ax=0$ using SVD.

$A=U\Sigma U^T=U\sqrt{\Sigma}\sqrt{\Sigma}^TU^T=B^TB$

where $B=\sqrt{\Sigma}^TU$. Then

$x^T A x=x^TB^TBx=y^Ty=0$

implies $y=Bx=0$. Therefore,

$Ax=B^TBx=0$ QED

If A is merely a symmetric matrix, we can’t have $A=U\Sigma U^T$, right?

EDIT: for arbitrary matrix $A$, we have

$A^TAx=0$ if and only if $Ax=0$.