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?

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 $\hat{D} \in \mathbb{R}^{p \times p}.$

$$DD^T=\begin{bmatrix} \hat{D} & 0\end{bmatrix}\begin{bmatrix} \hat{D} \\ 0\end{bmatrix}=\hat{D}^2$$

$$I_p-DD^T=I_p-\hat{D}^2$$

$$D^TD=\begin{bmatrix} \hat{D} \\ 0\end{bmatrix}\begin{bmatrix} \hat{D} & 0\end{bmatrix}=\begin{bmatrix} \hat{D}^2 & 0 \\ 0 & 0\end{bmatrix}$$

$$I_q-D^TD=\begin{bmatrix} I_p-\hat{D}^2 & 0 \\ 0 & I_{q-p}\end{bmatrix}$$

$$\det(I_q-D^TD)=\det(I_{q-p})\det(I_p-\hat{D}^2)=\det(I_p-\hat{D}^2)=\det(I-DD^T)$$

• Thanks, can I ask why $\det(I_p-DD^T)=\det(I_q-D^TD)$? I am guessing there is something related to the singular values, but why? – user136503 Dec 19 '16 at 4:52
• Is it because since $D$ is NOT a square matrix, yet is still diagonal, the only way this happens is if $DD^T$ or $D^TD$ has an extra row of $0$'s, and so $I_p-DD^T$ and $I_q-D^TD$ differ by one having an extra row with $1$ at the end, but that doesn't matter since determinant of a diagonal matrix is the product and so we multiply by $1$? Is there a formal way to show this? – user136503 Dec 19 '16 at 5:02
• I added some explaination for the case where $p\leq q$, the case of $p>q$ is similar. $I_p-DD^T$ and $I_q-D^TD$ differs by a block of identity matrix. along the diagonal. – Siong Thye Goh Dec 19 '16 at 7:13

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:

$\det(\mathbf{M}) = \det(\mathbf{I}_p - \mathbf{A}\mathbf{A}^\top) = \det(\mathbf{I}_q - \mathbf{A}^\top\mathbf{A})$.