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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 $x \in R(AA^T)$.

hzh
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    Clearly $\operatorname{range}(AA^T) \subseteq \operatorname{range}(A)$, so you need to rule out proper containment. For this, assuming that we're working with finite dimensional vector spaces, it suffices to show that $A$ and $AA^T$ have the same rank. –  Oct 08 '18 at 02:57
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    Assuming $A\in\mathbb R^{m\times n}$, suppose, $x\in R(A) \implies Ay = x, y\in\mathbb R^n$. Write $y=\underbrace{y_1}{\in\mathcal N(A)} + \underbrace{y_2}{\in\mathcal N(A)^\perp = R(A^T)} \implies A^Tw = y_2, w\in\mathbb R^m$. Finally $x = Ay = \underbrace{Ay_1}_{0} + Ay_2 = AA^Tw \implies x\in R(AA^T)$. – Owen Murphy Oct 07 '21 at 18:28

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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.

Van Latimer
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