Suppose that $A$ is a lower triangular matrix with all positive elements in the lower half. Is it possible to compute $\det(A^\intercal A+I)$ or find a (non-zero) lower bound on it? I'm looking for an approach that uses the elements of $A$ such as diagonal elements (which are the eigenvalues of $A$).
I've tried the following
$$\det(A^\intercal A + I) = \prod_{i=1}^n \lambda_i(A^\intercal A+I) = \prod_{i=1}^n [\lambda_i(A^\intercal A)+1].$$
Can we say that $\prod_{i=1}^n [\lambda_i(A^\intercal A)+1] \geq \prod_{i=1}^n (\lambda_i^2(A) + 1)$? It seems to hold for $n = 1, 2$.
Note that $\lambda_i(A^TA)$ isn't always bounded below by $\lambda_i^2(A)$. E.g. when $A=\left(\begin{smallmatrix}1&0\\ 1&1\end{smallmatrix}\right)$, the smallest eigenvalue of $A^TA$ is $\frac{3-\sqrt{5}}{2}=0.382$, which is smaller than the square of the smallest eigenvalue of $A$ (namely, $1$).
However, it is true that $\prod_i\left(\lambda_i(A^TA)+1\right)\ge\prod_i\left(\lambda_i(A)^2+1\right)$. In other words, when you fix the diagonal entries of $A$, the determinant of $A^TA+I$ is minimised when $A$ is a diagonal matrix (but this infimum is unattainable because your $A$ has a positive lower triangular part). Let $$ A=\pmatrix{L&0\\ v^T&a} \Rightarrow A^TA+I=\pmatrix{L^T&v\\ 0&a}\pmatrix{L&0\\ v^T&a}+I=\pmatrix{L^TL+vv^T+I&av\\ av^T&a^2+1}. $$ Using Schur complement, we get \begin{aligned} \det(A^TA+I) &=(a^2+1)\det\left(L^TL+vv^T+I-\frac{a^2}{a^2+1}vv^T\right)\\ &=(a^2+1)\det\left(L^TL+I+\frac{1}{a^2+1}vv^T\right)\\ &\ge(a^2+1)\det\left(L^TL+I\right). \end{aligned} Hence $\det(A^TA+I)$ is minimised when $v=0$, and in that case, $\det(A^TA+I)=(a^2+1)\det(L^TL+I)$. By applying the same argument recursively, the conclusion follows.