How to prove $\det(I_m+AA^t)=\det(I_n+A^tA)$?

Question

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

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I don't need the full answer, maybe a hint. I've tried using Sylvester's identity but I can't solve either way. I tried to replicate the proof from here, but I wondered if there were other proofs, that is why I asked for a hint, this is not homework.

Answer 1

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.

Answer 2

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.