Show that $\mathrm{rank}(AA^TA)=\mathrm{rank}(A)$.
Solution 1:
Yes, the statement holds. In particular, from the fact that $\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$, we have $$ \operatorname{rank}(A^TA) = \operatorname{rank}(A^T[AA^TA]) \leq \operatorname{rank}(A[A^TA]) \leq \operatorname{rank}(A^TA). $$