Is it true that rank$(AA^{*}A) =$ rank$(A) ?$

Question

Is it true that $\operatorname{rank}(AA^{*}A) = \operatorname{rank}(A)$?

I know that $\operatorname{rank}(A^{*}A) = \operatorname{rank}(A)$, but I am not able to prove this one. Neither I could find any counter example to disprove the statement.

Any hint$?$

Answer 1

Hints.

1. Use singular value decomposition of $A$.
2. Alternatively, try to prove that $AA^\ast Ax=0$ if and only if $Ax=0$. For the forward implication, suppose $AA^\ast Ax=0$. Then $\|A^\ast Ax\|^2=x^\ast A^\ast AA^\ast Ax=0$. Hence $A^\ast Ax=0$. Use the same trick again to infer that $Ax=0$.

Answer 2

Also $\text{rank}(AA^*)$ is the same as $\text{rank}(AA^*AA^*)$.