Let $A$ be a $m \times n$ matrix with real entries, and let $B$ be a $n \times n$ real symmetric matrix with absolute eigenvalues $\le 1$.
Are there (ideally sharp) bounds for the inequality $$\det(AA^T) \ge \det(ABA^T)$$ in terms of $\det(AA^T)$ and functions of $B$ such as e.g. its rank?
In my particular problem $B$ is idempotent, if that makes for an interesting special case.
One approach is as follows.
Note that $\det(AA^T)$ is the squared product of the singular values of $A$. On the other hand, as I note in my comment, if $B$ is symmetric and nilpotent, $\det(ABA^T)$ can be expressed as $\det((AU)(AU)^T)$ for some matrix $U$ with $r$ orthonormal columns, where $r$ is equal to the rank of $B$. In other words, $\det(A^TBA)$ is the squared product of the singular values of $AU$.
The Cauchy interlacing theorem for singular values implies that if $\sigma_1,\dots,\sigma_m$ denote the (non-zero) singular values of $A$ and $\tau_1,\dots,\tau_m$ denote the singular values of $AU$, then we must have $$ \sigma_k \geq \tau_{k} \geq \sigma_{(n-r) + k} $$ for all $k = 1,\dots,m$. With that, we can consider the fact that $$ \det(AA^T) = \sigma_1^2 \cdots \sigma_m^2, \quad \det(ABA^T) = \tau_1^2\cdots \tau_m^2 $$ To derive a lower bound for $\det(ABA^T)$ in terms of the singular values of $A$ and of $AU$.
Concretely: we have $\tau_k \geq \sigma_{(n-r) + k}$ for all $k = 1,\dots,m+r-n$. With that, we have $$ \det(ABA^T) \geq \sigma_{n-r} \cdot \sigma_{n-r+1}\cdots\sigma_m \cdot \tau_{m+r-n+1}\cdots\tau_m. $$
Proof that $\det(AA^T) \geq \det(ABA^T)$: note that both $ABA^T$ and $A(I - B)A^T$ are both (symmetric and) positive semidefinite. If $\lambda_i$ denotes the $i$th eigenvalue in ascending order, then any two symmetric matrices $P,Q$ will satisfy $$ \lambda_i(P + Q) \geq \lambda_i(P) + \lambda_1(Q). $$ Thus, we have $$ \lambda_i(AA^T) = \lambda_i(ABA^T + A(I - B)A^T) \geq \lambda_i(ABA^T) + \lambda_1(A(I - B)A^T) \\ \geq \lambda_i(ABA^T) + 0. $$ Thus, each eigenvalue of $AA^T$ is greater than or equal to the corresponding eigenvalue of $ABA^T$, and all of these eigenvalues are non-negative. It follows that $\det(AA^T)$ (the product of the eigenvalues of $AA^T$) is greater than $\det(ABA^T)$ (the product of the eigenvalues of $ABA^T$).