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Let $g(x) = e^x + e^{-x}$. For $x_1 < x_2 < \dots < x_n$ and $b_1 < b_2 < \dots < b_n$, I'd like to show that the determinant of the following matrix is positive, regardless of $n$:

$\det \left (\begin{bmatrix} \frac{1}{g(x_1-b_1)} & \frac{1}{g(x_1-b_2)} & \cdots & \frac{1}{g(x_1-b_n)}\\ \frac{1}{g(x_2-b_1)} & \frac{1}{g(x_2-b_2)} & \cdots & \frac{1}{g(x_2-b_n)}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{g(x_n-b_1)} & \frac{1}{g(x_n-b_2)} & \cdots & \frac{1}{g(x_n-b_n)} \end{bmatrix} \right ) > 0$.

Case $n = 2$ was proven by observing that $g(x)g(y) = g(x+y)+g(x-y)$, and $g(x_2 - b_1)g(x_1-b_2) = g(x_1+x_2 - b_1-b_2)+g(x_2-x_1+b_2-b_1) > g(x_1+x_2 - b_1-b_2)+g(x_2-x_1-b_2+b_1) = g(x_1-b_1)g(x_2-b_2)$

However, things get difficult for $n \geq 3$. Any ideas or tips?

Thanks!

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2 Answers 2

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At first, we prove that the determinant is non-zero, in other words, the matrix is non-singular. Assume the contrary, then by the linear dependency of the columns there exist real numbers $\lambda_1,\dots,\lambda_n$, not all equal to 0, such that $F(x_i):=\sum_j \frac{\lambda_j}{g(x_i-b_j)}=0$ for all $i=1,2,\dots,n$. But the equation $F(x)=0$ is a polynomial equation with respect to $e^{2x}$ and the degree of a polynomial is less than $n$. So, it can not have $n$ distinct roots.

Now we note that the matrix is close to an identity when $x_i=b_i$ and $b_i$'s are very much distant from each other, and the phase space of parameters $\{(x_1,\dots,x_n,b_1,\dots,b_n):x_1<\dots<x_n,b_1<b_2<\dots <b_n\}$ is connected. Thus the sign of the determinant is always plus.

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    $\begingroup$ Please clarify for the profane as me the argument about connectedness of phase space and positiveness of determinant. $\endgroup$
    – Evgeny Kuznetsov
    Commented Jul 8, 2018 at 18:15
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    $\begingroup$ The sign of a determinant is a constant continuous function on a connected space. Thus its codomain is connected and consists of unique point. In other words, the sign is always the same. $\endgroup$
    – Fedor Petrov
    Commented Jul 8, 2018 at 19:15
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    $\begingroup$ Or you can see the determinant is a continuous function on a connected domain. Had there been values at two point with opposite signs, the intermediate value theorem would stipulate that there is a point on any curve connecting the two points in the domain with the determinant at that point being zero, which contradicts the non-singularity of the matrix. $\endgroup$
    – Hans
    Commented Jul 9, 2018 at 3:22
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To complement Fedor's answer, here is more explicit proof.

Let the original matrix be $G$. Let $D_x :=\text{Diag}(e^{x_1},\ldots,e^{x_n})$. Then, we can write \begin{equation*} G = D_x C D_b,\quad\text{where}\ C = \left[ \frac{1}{e^{2x_i}+e^{2b_j}}\right]_{i,j=1}^n. \end{equation*} To prove that $\det(G)>0$ it thus suffices to prove that $\det(C)>0$. Notice now that $C$ is nothing but a Cauchy matrix, and by explicitly writing its determinant out (under the hypotheses on $x$ and $b$) we can easily conclude that $\det(C)>0$.

Remark. The above argument actually proves that $\text{sech}(x-y)$ is a Totally positive kernel (because the $k(x,y) := 1/(x+y)$ is known to be a TP kernel).

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  • $\begingroup$ The same argument also works for any function $g$ for which we can write $g(x-y) = \frac{k(x)+k(y)}{h(x)h(y)}$, where $k$ is monotonic and positive, and $h$ is positive. I wonder if this class of functions only contains essentially exponential functions... $\endgroup$
    – Suvrit
    Commented Jul 9, 2018 at 15:04
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    $\begingroup$ Yes, only exponential. Substitute $x=y$, get $k(x) =Ch^2 (x) $, then $h(x) /h(y) +h(y) /h(x) $ depends on $x-y$, by monotonicity $h(x) /h(y) $ depends on $x-y$, $h(x) =h(y) f(x-y) $, $h(y) f(z) =h(y+z) =h(z) f(y) $, $f/h=const$, $H(x+y) =H(x) H(y) $ for $H=h\cdot const$, $H$ is exponential. $\endgroup$
    – Fedor Petrov
    Commented Jul 9, 2018 at 15:44
  • $\begingroup$ Ah true! thanks Fedor for the quick resolution. $\endgroup$
    – Suvrit
    Commented Jul 9, 2018 at 15:54
  • $\begingroup$ Great answer! What was the motivation for looking for the matrix factorization? $\endgroup$
    – Sandeep Silwal
    Commented Jul 10, 2018 at 0:15
  • $\begingroup$ @SandeepSilwal thanks! Notice that $g(x-y)= \frac{e^{2x}+e^{2y}}{e^{x}e^y}$, thus it is clear that the $e^{x+y}$ part is superfluous and can be omitted. More generally, this is just a standard useful observation: Hadamard products with rank-1 matrices can be "factored out" into matrix-products with diagonal matrices. $\endgroup$
    – Suvrit
    Commented Jul 10, 2018 at 0:29

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