I have seen quite a number of questions regarding similar issues, like this and this. However, all the answers were trying to approach the topic via a non-straightforward way, that is to prove the statement by proving $N(A) = N(AA^T)$. This method is fine and do be easy to understand.
But I am actually wondering if there is a straightforward way that we can prove this?
Like if we assume $x \in R(A)$, then if we can somehow show $x \in R(AA^T)$ holds, we proved the statement.
I'd like to do this but can't quite push $x \in R(A)$ towards $x \in R(AA^T)$.
I remember having trouble with this question myself. The key is that the kernel of a matrix is the orthogonal complement to the range of its transpose.
Let $x\in R(A)$, so $x = Ay$ for some $y$. We seek a $z\in R(A^T)$ such that $x=Ay = Az$, that is, such that $z -y \in A$’s kernel. Well, the orthogonal projection of $y$ onto $R(A^T)$ does the trick for $z$! This is precisely because of the first paragraph.