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Let $A$ be an $n \times n$ real matrix. prove $A^tA$ and $A^t$ have same rank.

I know it has an answer here but I don't wish to look into a solution. can somebody give me a hint?

I have no idea about the geometric interpretation of transpose since it involves dual spaces. So in fact, I don't have any idea where to start

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    $\begingroup$ One thing to think about is how this problem looks over $\mathbb{C}$, if it doesn't still hold, you know your proof must involve some specific properties about $\mathbb{R}$ that $\mathbb{C}$ doesn't have. $\endgroup$
    – Chris H
    May 3, 2019 at 14:46

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First, $A$ and $A^t$ have the same rank, and it's a bit easier to work with $A$, so that's what I would do.

Second, if two matrices (linear transformations) have the same domain and same kernel, then they must have the same rank.

It's easy to show that $\ker(A)\subseteq \ker(A^tA)$.

As for the other direction, take an $x\in \ker(A^tA)$, and consider the length of $Ax$.

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  • $\begingroup$ This will prove that $A^tA$ and $A^t$ have the same range, not just the same rank. Is it true? $\endgroup$ May 12, 2019 at 11:17
  • $\begingroup$ @VinayDeshpande Yes, $A^tA$ and $A^t$ do have the same range. $\endgroup$
    – Arthur
    May 12, 2019 at 11:26

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