Computing (a lower bound) on $\det(A^\intercal A +I)$ 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 $n = 1, 2$. 
 A: 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.
A: More generally, one has
$\textbf{Proposition}$. Let $A\in M_n(\mathbb{C})$ st. $spectrum(A)=(\lambda_i)$. Then
$\det(A^*A+I_n)\geq \Pi_i(|\lambda_i|^2+1)$ with equality only when $A$ is normal.
$\textbf{Proof}$. Up to a change of orthonormal basis, we may assume that $A$ is triangular. In a second step, follow the user1551's proof.
