I need help in a question. I need to show that, given an $ m \times n$ matrix $A$, $\bar x \in Null(A)$ if and only if $\bar x \in Null(A^tA)$.
I found this answer: "Let $X=Null(A)$; then $\forall x \in X, Ax=0$. Assume that $Y=Null(A^tA)$, i.e. $\forall y \in Y, A^t Ay=0$. This implies, $Ay=0$ or $A^tAy=0$ and $Ay \ne 0$. If case (1) is true than we are done: $y \in X$. Now $A^tAy=0 \implies y^tA^tA = 0^t \implies y^tA^t = 0^T$ so (since $A \ne 0$) $Ay=0$. And this is same as case (1). Thus $X=Y$."
But how does this show that it is "if and only if"?
The "only if" part is really trivial:
$x\in Null(A)$ implies $A^tAx=A^t0=0$.
I don´t know where You found this answer to the if part, but it´s weird, try:
If $y\in Null(A^tA)$ then $A^tAy=0$, by definition and thus $|Ay|^2=y^tA^tAy=y^t0=0$ and thus $Ay=0$, since |.| is a norm, i.e. $y\in Null(A)$