Show that $u^TAu=0\Rightarrow u\in \operatorname{Ker}(A)$ for $A\geq 0$

Question

I'm reading a proof that assumes $u^TAu=0\Rightarrow u\in \operatorname{Ker}(A)$ for $A\geq 0$. I feel like the proof should be short, but I can't figure it out. I've tried writing A as a Gram matrix but couldn't get anywhere.

Answer 1

Use $A\geq 0$ to note that there exists a symmetric $B\geq 0$ such that $A=BB=B^TB$. Then: $$ 0=u^TAu=u^TB^TBu=||Bu||^2\implies Bu=0\implies Au=B^T(Bu)=0. $$