Prove that if $\vec{x} \in \operatorname{Null}(A^TA)$ then $\vec{x} \in \operatorname{Null}(A).$

Question

How would you prove that if $\vec{x} \in \operatorname{Null}(A^TA)$ then $\vec{x} \in \operatorname{Null}(A)?$ I have been reading some proofs and am having trouble at a certain step specifically how $(Ax)^T(Ax)=\vec{0}\,$ implies that $Ax=\vec{0}.$

Answer 1

If $A$ is a real $n \times p$ matrix and $x \in \mathbb{R} ^ p$, then $$ Ax=b_{n\times 1}, $$ i.e., $b$ is a vector from $\mathbb{R}^n$. Now, note that $$ (Ax)^T Ax = b^Tb=(b_1,...,b_n) \begin{pmatrix} b_1\\ \vdots \\ b_n \end{pmatrix} =b_1^2+b_2^2+\cdots b_n^2 \ge0, $$ hence if $(Ax)^T(Ax) = b^Tb =0$ then every $b_i = 0$ for all $i=1,..,n$.