Show that $\mathrm{rank}(AA^TA)=\mathrm{rank}(A)$.

Question

I have shown that $\mathrm{rank}(A^TA)=\mathrm{rank}A$, which clearly suggests $\mathrm{rank}(A^TAA^TA)=\mathrm{rank}A$, while I wonder if it is true for $AA^TA$ i.e. $\mathrm{rank}(AA^TA)=\mathrm{rank}(A)$ (as suggested by inequalities)? Here $A$ are not necessarily full rank.

Answer 1

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