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Been having a tough time working on this one. I can easily show the inverse, anyone have ideas on how to show this?

$Im(H^T) \subset Im(H^TH)$ where $H \in \mathbb{R}^{m \times n}$

theodw15
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    Does this answer your question? Let $A$ be any matrix. How do I prove that $im(A) = im(AA^T )$.For $A=H^T$ this yields $im(H^T)=im (H^TH)$. – Dietrich Burde Sep 06 '22 at 18:16
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    $\text{Im} (A^{T} A ) \subset \text{Im } (A^{T} ) $, but it directly follows that $\text{kern} (A^{T}) = \text{kern} (A^{T} A)$ (since $Ax = 0$ implies $A^{T} A x = 0$ implies $x^{T} A^{T} A x = 0$ implies $|A x|^{2} = 0$ implies $A x =0$). Thus, $\text{dim Im} (A^{T} A) = \text{dim} (\text{Im} (A^{T}))$, by the rank–nullity theorem. Hence $\text{Im} (A^{T} A ) = \text{Im } (A^{T} ) $ because it is a subspace of the same dimension as the spaces containing him. – R. W. Prado Sep 06 '22 at 18:26
  • @DietrichBurde yep, that's it. thanks! – theodw15 Sep 08 '22 at 02:33
  • @R.W.Prado, that's pretty much what the other question's answer is - I'm glad we're all on the same page. – theodw15 Sep 08 '22 at 02:35

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