$\min(\det(\mathbf{A}))$ for special matrix $\mathbf{A}$ (The construction of matrix $\mathbf{A}$ is not difficult to be understood. You can first jump to A Toy Example to take a glance. Any idea or suggestion would be appealing for me.)

Update #2:
The original problem has been resolved in here.

Update #1:
I get a partial answer and a hint from @Noam D. Elkies in mathoverflow. I may understand his/her partial answer for $(\text{Q}2)$. Are the ideas given by @Noam D. Elkies correct? What should I do to complete the proof? Could you please provide more details about that? Or do you have some other ideas?

The Original Problem:
Given $N,D\in\mathbb{Z}^+~(D\ge N)$ and $\alpha\in\mathbb{R}^+$, the vector $\mathbf{p}$ and the matrix $\mathbf{A}_\mathbf{p}$ are defined as follows:

*

*$\mathbf{p}=[p_1,p_2,\cdots,p_N]$, where $p_i$s are selected from $\{1,2,\cdots,D\}$ and satisfying the condition of $(p_1<p_2<\cdots<p_N)$.


*Given $\mathbf{p}$, there is $\mathbf{A}_\mathbf{p}=[a_{ij}]_{N\times N}$, where $a_{ij}=e^{-\alpha |p_i-p_j|}$.
I am trying to find out the property of $\det(\mathbf{A}_\mathbf{p})$. Based some of my findings, I am confused by the following two subproblems:
$(\text{Q}1)$ Can we conveniently calculate the value of $\det(\mathbf{A}_\mathbf{p})=f(\mathbf{p})$ for a given $\mathbf{p}$? In other words, is there a way to explicitly unfold $\det(\mathbf{A}_\mathbf{p})$?
$(\text{Q}2)$ Is $\mathbf{A}_\mathbf{p}$ positive semi-definite?
$(\text{Q}3)$ Does $\det(\mathbf{A}_\mathbf{p})$ hit its minimal value only when $(p_{i+1}-p_i=1)$? By the way, in this case, $\mathbf{A}_\mathbf{p}$ will become a special symmetric Toeplitz matrix.

A Toy Example:
Given $N=3$, $D=10$ and $\alpha =1$. I construct $\mathbf{p}_1=[3,4,5]$ and $\mathbf{p}_2=[2,5,7]$. Then we have:
$$
\det \left( \mathbf{A}_{\mathbf{p}_1} \right) =\left| \begin{matrix}
 1&  e^{-1}&  e^{-2}\\
 e^{-1}&  1&  e^{-1}\\
 e^{-2}&  e^{-1}&  1\\
\end{matrix} \right|\approx 0.748,
$$
and
$$
\det \left( \mathbf{A}_{\mathbf{p}_2} \right) =\left| \begin{matrix}
 1&  e^{-3}&  e^{-5}\\
 e^{-3}&  1&  e^{-2}\\
 e^{-5}&  e^{-2}&  1\\
\end{matrix} \right|\approx 0.979 > \det \left( \mathbf{A}_{\mathbf{p}_1} \right) .
$$

Some of My Efforts:
I may have the following observations:
$(\text{O}1)$ The diagonal elements of $\mathbf{A}_\mathbf{p}$ are all ones since $|p_i-p_i|=0$.
$(\text{O}2)$ All elements of $\mathbf{A}_\mathbf{p}$ are in $[0,1]$.
$(\text{O}3)$ $\mathbf{A}_\mathbf{p}$ is symmetric since $|p_i-p_j|=|p_j-p_i|$.
$(\text{O}4)$ Actually, the order of $\mathbf{p}_i$s do not affect the value of $\det(\mathbf{A}_\mathbf{p})$.
I guess that $\mathbf{A}_\mathbf{p}$ has the following two properties:
$(\text{P}1)$ The answer of $(\text{Q}2)$ is "Yes", i.e., $\mathbf{A}_\mathbf{p}$ is positive semi-definite.
$(\text{P}2)$ The answer of $(\text{Q}3)$ is "Yes", i.e., $\left[\det(\mathbf{A}_\mathbf{p})=\min\left\{{\det(\mathbf{A}_{\mathbf{p}_k})}\right\}\right] \Leftrightarrow \left[ \forall i, ~p_{i+1}-p_{i}=1 \right]$.
The above conjectures of $(\text{P}1)$ and $(\text{P}2)$ is empirically presented. I write the following Python code to validate them and find that all randomly generated $\mathbf{A}_\mathbf{p}$ satisfy $(\text{P}1)$ and $(\text{P}2)$:
import numpy as np
import random
from scipy import spatial

alpha = 1
N = 10
X = np.arange(N).reshape(N, 1)
X = np.exp(-alpha * spatial.distance.cdist(X, X))
X_det = np.linalg.det(X)
for D in range(N, 1000):
    for i in range(100):
        p = np.array(random.sample(range(1, D + 1), N)).reshape(N, 1)
        A = np.exp(-alpha * spatial.distance.cdist(p, p))
        A_det = np.linalg.det(A)
        if A_det <= 0:
            print(A_det, p.reshape(N,))  # det(A) <= 0
            exit(0)
        if A_det < X_det and abs(A_det - X_det) > 1e-8:
            print(p, p.reshape(N,))  # det(A) < det(X) with numerical tolerance
            exit(0)
print('Done.')

I test many combinations of $\{\alpha, N, D\}$. I see that there is no any case satisfy the conditions of A_det <= 0 and A_det < X_det.

Why I Try to Study $\det(\mathbf{A}_\mathbf{p})$?
I study the entropy of multivariate Gaussian distributions with some special covariance matrices (i.e., the above defined $\mathbf{A}_\mathbf{p}$). The entropy value is related to $\det(\mathbf{A}_\mathbf{p})$ (you can see more details from my previous problems below).
I have made efforts and spent more than 14 days on it. Specifically, I read some textbooks, papers and blogs related to it. Here are some previous problems posted by me: Problem 1, Problem 2, Problem 3 and Problem 4. However, I am still stucked. Now I think that the key step is to resolve the problem I posted above.
I am sorry for occupying much public resource of this platform. But I really want to resolve the problems, especially $(\text{Q}2)$ and $(\text{Q}3)$. Could you please provide help or some tips?
 A: Q3 is actually relatively simple (and the beliw actually implies an answer to  Q1).
You have a matrix with the form
\begin{pmatrix}
1&m&mn\\m&1&n\\mn&n&1\\\end{pmatrix}
or something obtained from this by one row and one column interchange. The determinant is then
$1+m^2n^2-m^2-n^2=(1-m^2)(1-n^2)$
which is smallest when $m$ and $n$ are largest is absolute value. With your criteria for selecting entries $m=n=1/e$ is as large an absolute value as you can get for these parameters, hence the smallest possible determinant.
For the other two questions:
The form given above for the determinant can be used to cobstruct an answer to Q1. To wit, let the elements in your permutation be $k,k+d_1,k+d_1+d_2$ in ascending order. Then in the above you have $m=\exp(-d_1),n=\exp(-d_2)$ and thus the determinant will be
$(1-\exp(-2d_1))(1-\exp(-2d_2))$
which, as noted above, is minimal for $d_1=d_2=1$.
For Q3 note that for any positive whole numbers $k,l$, $|\exp(-k)+\exp(-l)|<1$. So the matrix is diagonally dominant with all positive elements in the diagonals. The guarantees positive definite matrix.
