# How to prove the range of $AA^T$ is the same as range of $A$?

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

• 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.
– user169852
Oct 8, 2018 at 2:57
• 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 $x = Ay = \underbrace{Ay_1}_{0} + Ay_2 = AA^Tw \implies x\in R(AA^T)$. Oct 7, 2021 at 18:28

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.