If $x^T\!Ay=0$ for all $x,y $ in $ \mathbb{R}^n$ then $A=0$ Is it true that if $x^T\!Ay=0$ for all $x,y $ in $ \mathbb{R}^n$ then $A=0$.  If so how do I justify this statement?

Suppose $Ay \neq 0$. Let $x=Ay$, then $x^TAy = \|Ay\|^2 >0$.
Hence $Ay=0$ for all $y$, or equivalently, $A=0$.
