How to show that $\det(I-AA^T) = \det(I-A^TA)$ for a rectangular matrix $A$? Suppose I have a matrix $A_{p \times q}$ and let $I_p$ and $I_q$ be the identity matrices of dimension $p \times p$ and $q\times q$, respectively. I would like to show that $\det(I_p-AA^T) = \det(I_q - A^TA)$. I have thought about taking transposes, but that doesn't work. Is there something I'm missing here?
 A: Consider $\mathbf{M} = \begin{bmatrix} \mathbf{I}_p  & \mathbf{A} \\ \mathbf{A}^\top & \mathbf{I}_q \end{bmatrix}$.
Applying the Schur determinant formula on $\mathbf{M}$ for the top-left and bottom-right blocks results in:
$\det(\mathbf{M}) = \det(\mathbf{I}_p - \mathbf{A}\mathbf{A}^\top) = \det(\mathbf{I}_q - \mathbf{A}^\top\mathbf{A})$.
A: Let the SVD decomposition of $A=UDV^T$, where $U$ and $V$ are orthogonal.
$$AA^T=UDD^TU^T$$
$$I-AA^T=U(I_p-DD^T)U^T$$
Hence $$\det(I-AA^T)=\det(I_p-DD^T)$$
Similarly, we have
$$A^TA=VD^TDV^T$$
$$I-A^TA=V(I-D^TD)V^T$$
Hence $$\det(I-A^TA)=\det(I_q-D^TD)$$
We can see that $$\det(I_p-DD^T)=\det(I_q-D^TD)$$
Edit to explain the equality above:
For the case where $p \leq q$, we can write $D= \begin{bmatrix} \hat{D} & 0\end{bmatrix}$, where $\hat{D} \in \mathbb{R}^{p \times p}.$
$$DD^T=\begin{bmatrix} \hat{D} & 0\end{bmatrix}\begin{bmatrix} \hat{D} \\ 0\end{bmatrix}=\hat{D}^2$$
$$I_p-DD^T=I_p-\hat{D}^2$$
$$D^TD=\begin{bmatrix} \hat{D} \\ 0\end{bmatrix}\begin{bmatrix} \hat{D} & 0\end{bmatrix}=\begin{bmatrix} \hat{D}^2 & 0 \\ 0 & 0\end{bmatrix}$$
$$I_q-D^TD=\begin{bmatrix} I_p-\hat{D}^2 & 0 \\ 0 & I_{q-p}\end{bmatrix}$$
$$\det(I_q-D^TD)=\det(I_{q-p})\det(I_p-\hat{D}^2)=\det(I_p-\hat{D}^2)=\det(I-DD^T)$$
