Does $v^TAv = 0$, imply $Av = 0$ for $A$ be a symmetric positive semi-definite matrix, $v \in\mathbb{R}^n$? Let $A$ be a real symmetric positive semi-definite matrix. if  $v^TAv = 0$, can we deduce that $Av = 0$ for $v \in\mathbb{R}^n$?
My try:
Since $A$ is a real positive semi-definite matrix we can write $A=B^TB$ for some matrix $B$. Then
$$0 = v^TAv = v^TB^TBv =(Bv)^TBv= \langle Bv,Bv\rangle \implies Bv =0\implies B^TBv =Av=0  $$

This is true even in infinite-dimensional inner product spaces. It can be proved by using basic properties of inner product. Sophisticated tools like spectral decomposition or existence of square root are not needed.
Suppose $A$ is a (weakly) positive operator and $\langle Av,v\rangle=0$. Then for any vector $u$,
$$
0\le\langle A(tu+v)\,,tu+v\rangle
=t^2\langle Au,u\rangle+
2t\operatorname{Re}\langle Av,u\rangle
$$
for all real numbers $t$. Therefore $\operatorname{Re}\langle Av,u\rangle$ is necessarily zero. In particular, when $u=Av$, we get $0=\operatorname{Re}\langle Av,Av\rangle=\langle Av,Av\rangle$. Hence $Av=0$.
