TITLE: Column Space of AA' is equal to column of A.
QUESTION [5 upvotes]: This is my question. How to show that the column space of matrix A
is just equal to the column space of AA'?.. A' represents the transpose of A.
I know that the column space of AA' is a subset of the column space of A
which is just trivial. But the other way around I still used inclusion 
but it seemed that it is going nowhere. Maybe it can be done by some manipulation
or the barbaric way of doing this. I just want to obtain a simple proof. Anyone?

REPLY [2 votes]: Hint. If you can show that $A'x=0$ whenever $AA'x=0$, then $\operatorname{nullity}(AA')\le\operatorname{nullity}(A')$ and hence $\operatorname{rank}(AA')\ge\operatorname{rank}(A')=\operatorname{rank}(A)$, i.e. the dimension of the column space of $A$ does not exceed the dimension of the column space of $AA'$.
If you really want a super-duper simple proof, use singular value decomposition.