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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

2 Answers2

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Thm: For ${ A \in \mathbb{R} ^{M \times N} ,}$ ranges ${ \mathcal{R}(A ^T A) = \mathcal{R}(A ^T) }.$

Pf: Any element of ${ \mathcal{R}(A ^T A) }$ looks like ${ A ^T A v }$ and is hence in ${ \mathcal{R}(A ^T) .}$ So ${ \mathcal{R}(A ^T A ) \subseteq \mathcal{R}(A ^T) .}$

Conversely any element of ${ \mathcal{R}(A ^T) }$ looks like ${ A ^T v ,}$ and we are to show ${ A ^T v \in \mathcal{R}(A ^T A) }.$
Let ${ (A ^T A) x }$ be the projection of ${ A ^T v }$ onto ${ \mathcal{R}(A ^T A). }$ Now ${ A ^T v - (A ^T A)x }$ is orthogonal to ${ \mathcal{R}(A ^T A) .}$ Explicitly, $${ (A ^T (v - Ax)) ^T A^T A = 0 .}$$ The equation is ${ (v - Ax) ^T A A ^T A = 0 ,}$ and hence gives ${ (v - Ax) ^T A A ^T A {\color{red}{A ^T (v - Ax)}} = 0 }$ that is ${ (A A ^T (v-Ax)) ^T (A A ^T (v-Ax)) = 0 }$ that is $${ A A ^T (v - Ax) = 0 .}$$ This again gives ${ {\color{red}{(v - Ax) ^T}} A A ^T (v - Ax) = 0, }$ that is ${ (A ^T (v - Ax) ) ^T (A ^T (v - Ax)) = 0 }$ that is $${ A ^T (v - Ax) = 0 .}$$ So ${ A ^T v = A ^T A x \in \mathcal{R}(A ^T A), }$ as needed.

Venkata Karthik Bandaru
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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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