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Let $m,n\in\mathbb N$. Can we show that $\det(A^TA)=\det(AA^T)$?

If $m=n$, the answer is trivially yes, since then $\det(A^TA)=\det(A^T)\det A=\det^2A=\det(AA^T)$.

For the case $m=3$, $n=2$, I've checked the equality by a direction calculation and it actually holds true. So, I think this can be shown in general. But how?

0xbadf00d
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    You cannot. Take for instance $$A = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}.$$ Something slightly weaker holds: for non-square real $A$, (not counting multiplicity) the eigenvalues of $A^T A$ and $AA^T$ agree except possibly for $0$. – Cameron Williams Sep 22 '20 at 15:40
  • However, ranks coincide, see https://math.stackexchange.com/questions/349738/prove-operatornamerankata-operatornameranka-for-any-a-in-m-m-times-n – Peter Franek Sep 22 '20 at 15:53
  • More simply, $A=(1;;0)$. – gpassante Sep 22 '20 at 15:59

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This is false. Try even a $n\times 1$ example. Then $A^\top A$ is $1\times 1$ and the determinant is nonzero, assuming $A\ne 0$. However, $AA^\top$ is $n\times n$ and has rank $1$, so $\det(AA^\top)=0$.

Ted Shifrin
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  • Hah I was about to add this to my comment. $+1$ :) – Cameron Williams Sep 22 '20 at 15:43
  • Well, it seems like you're right. I was trying to find a suitable formula which I could use to find a substitution rule for surface integrals. Based on your background, you might know the answer. Could you please take a look? https://math.stackexchange.com/q/3836020/47771. – 0xbadf00d Sep 22 '20 at 15:44