(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.)




---


**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](https://math.stackexchange.com/questions/4524167/expectations-of-multivariate-gaussian-entropy), [Problem 2](https://math.stackexchange.com/questions/4526630/prove-disprove-the-matrix-convergence-in-probability), [Problem 3](https://math.stackexchange.com/questions/4529767/prove-disprove-mathbbe-ln-det-mathbfa-rightarrow-0-when-bounded) and [Problem 4](https://math.stackexchange.com/questions/4525722/is-the-determinant-of-mathbfa-in0-1n-times-n-bounded). 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?



$\min(\det(\mathbf{A}))$ for special matrix $\mathbf{A}$
**Q1** The determinant is $\prod\_{n=1}^{N-1} (1 - e^{-2\alpha(p\_{n+1}-p\_n)})$.


**Q2** Yes, using the answer to **Q1**.


**Q3** Yes, using the answer to **Q1**.




---




---


The formula for **Q1** is proved by induction on $N$.
The base case $N=1$ is just $\det(1)\_{1\times1} = 1$.
For $N>1$, let $c = e^{-\alpha(p\_N^{\phantom.} - p\_{N-1}^{\phantom.})}$,
and let ${\bf A'\_p}$ be the matrix obtained from ${\bf A\_p}$
by subtracting $c$ times row $N-1$ from row $N$
and then subtracting $c$ times column $N-1$ from column $N$.
Then $\det{\bf A'\_p} = \det{\bf A\_p}$.
But the last row and column of ${\bf A'\_p}$ are all zero
except for the $(N,N)$ entry which is $1-c^2$.
Therefore $\det{\bf A'\_p}$ is $1-c^2$ times the determinant of
the symmetric matrix obtained from ${\bf A\_p}$ by deleting its
$N$-th row and $N$-th column -- which is just ${\bf A\_{p'}}$
where ${\bf p'} = [p\_1, p\_2, \ldots, p\_{N-1}]$.
This completes the induction step and the proof.




---


**Q2** By the product formula, $\det A$ is positive, as is the determinant of
every symmetric minor of $A$ (which is of the same form as $A$ for some
subsequence of $p\_1,\ldots,p\_N$).




---


**Q3** This too follows from the product formula:
each factor $1 - e^{-2\alpha(p\_{n+1}-p\_n)}$ is at least
$1 - e^{-2\alpha}$, with equality if and only if $p\_{n+1} - p\_n = 1$.




---




---


We can also deduce the positive answer to **Q2** by constructing
linearly independent probability distributions $\mu\_i$ on the real numbers
whose covariance matrix is proportional to ${\bf A\_p}$: the distribution $\mu\_i$
chooses integers $p \geq p\_i$ with probability
$(1-e^{-\alpha}) e^{-\alpha(p-p\_i)}$.
If the real numbers $p\_i$ are not required to be integers then we can achieve
the same goal using continuous $\mu\_i$ with distribution functions
$\alpha e^{-\alpha(p-p\_i)}$ supported on $p \geq p\_i$.


