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Let $A \in M_{m \times n}(F)$, prove $$ det(I_m+AA^t)=det(I_n+A^tA) $$

I don't need the full answer, maybe a hint. I've tried using sylvesters identity but I can't solve either way.

Edit: I tried to replicate the proof from, (M*18): https://users.math.yale.edu/~auel/courses/370f06/docs/solutions1.pdf but I wondered if there were others proofs, that is why I asked for a hint, this is not a homework.

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closed as off-topic by Dando18, Alexander Gruber May 10 at 5:36

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    $\begingroup$ Hint: Singular values. $\endgroup$ – Theo Bendit May 9 at 23:34
  • $\begingroup$ We haven't cover such topic, we are trying to use properties of the determinant as a alternating $n-$linear function. $\endgroup$ – ipreferpi May 9 at 23:38
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    $\begingroup$ Put $A, A^T$ and $I$ into a larger matrix as blocks, use invertible transformations on this large matrix, and then take the determinant. $\endgroup$ – trisct May 9 at 23:44
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    $\begingroup$ $det |I+AA^t|= det |(I+AA^t)^t|=det |I+A^t A|$ $\endgroup$ – Dr Zafar Ahmed DSc May 10 at 6:22
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    $\begingroup$ @DrZafarAhmedDSc $(I+AA^t)^t=I+AA^t$. $\endgroup$ – logarithm May 10 at 10:00
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Here's a sketch. Assume without loss of generality that $m < n$. Writing $A = A' \oplus 0$ (over some basis) with $A'\in M_{m\times m}$, reduce to the case of $m = n$. If $A$ is invertible, then $$\det (1 + AA^t) = \det A(A^{-1} + A^t) = \det(A^t) \det ((A^{-1})^t + A)^t = \det(1 + A^t A).$$ But the given equation is a polynomial in $A_{ij}$, and the space of invertible $A$ is dense.

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  • $\begingroup$ I didn't understand the last line, why worry to say such $A$ is dense? $\endgroup$ – ipreferpi May 9 at 23:54
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    $\begingroup$ We assumed that $A$ is invertible midway through the proof sketch. The displayed equation doesn't make sense otherwise; but the result holds for general $A$ if it holds for invertible $A$ by continuity. $\endgroup$ – anomaly May 9 at 23:55
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For square matrices, take determinant on both sides of $$\begin{pmatrix}I&0\\A^T&I\end{pmatrix}\begin{pmatrix}I+AA^T&A\\0&I\end{pmatrix}\begin{pmatrix}I&0\\-A^T&I\end{pmatrix}=\begin{pmatrix}I&A\\0&I+A^TA\end{pmatrix}$$ For non-square matrices you can pad it with zeros to a square matrix and use the formula above.

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