If $x^T\!Ay=0$ for all $x,y $ in $ \mathbb{R}^n$ then $A=0$

Question

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?

Answer 1

Hint: Take $x$ and $y$ to be standard unit vectors. What happens?

Answer 2

Suppose $Ay \neq 0$. Let $x=Ay$, then $x^TAy = \|Ay\|^2 >0$.

Hence $Ay=0$ for all $y$, or equivalently, $A=0$.