If we let $A \in \mathbb{R}^{n \times n}$, I want to prove that $N(A) = N(A^T A)$.
MY ATTEMPT
I already prove that $N(A) \subseteq N(A^T A)$. I just need clarification for the other direction. We can first take $x \in N(A^T A)$. By definition, $A^T Ax = 0$, where $Ax \in \mathbb{R}^n$. Multiplying both sides by $x^T$, we get $x^T A^T A x = 0$ and this means $(Ax)^T (Ax) = 0$ and so $||Ax||^2 = 0 \Rightarrow Ax = 0$. Hence, by definition $x\in N(A)$.
Is this proof sound?