Are there any lower bounds on the $\det(I+A^\top A - B^\top B)$? I'm looking for a bound that possibly depends on the 2-norm, Frobenius norm, or trace of $A$ and $B$.
My attempt: Let $\lambda_i$ be the eigenvalues of $C$ and suppose that $\lambda_i>0$. Then,
$$\log(\det(I+C)) = \sum_{i=1}^n \log(1+\lambda_i) \geq \sum_i \frac{\lambda_i}{1+\lambda_i} \geq \sum_i \frac{\lambda_i}{1+\lambda_1} = \frac{\text{tr}(C)}{1+\|C\|_2}.$$
There are two problems with this: (1) For $A^\top A - B^\top B$, we can't assume that $\lambda_i > 0 $, and (2) this inequality is not tight when $\lambda_i$ are not close to $\lambda_1$.