Rank of transpose of a matrix. How do I prove that $\operatorname{Rank}(A)=\operatorname{Rank}(A^TA)$?
I know that $\operatorname{Rank}(A)=\operatorname{Rank}(A^T)$ but how do I prove it?
Would somebody explain it to me please?

Clearly, $Ax=0$ implies that $A^TAx=0$, and thus
$$
\text{Null}\,(A^TA)\supset\text{Null}\,A
$$
and hence rank$(A)\ge$ rank$(A^TA)$.
If $A^TAx=0$, then $x^TA^TAx=0$, and thus $\|Ax\|=0$, which implies that $Ax=0$, and this provides that rank$(A)\le$ rank$(A^TA)$.
