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!

up vote 30 down vote accepted

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.

  • 1
    Please clarify for the profane as me the argument about connectedness of phase space and positiveness of determinant. – Evgeny Kuznetsov Jul 8 at 18:15
  • 1
    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. – Fedor Petrov Jul 8 at 19:15
  • 4
    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. – Hans Jul 9 at 3:22

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).

  • 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... – Suvrit Jul 9 at 15:04
  • 1
    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. – Fedor Petrov Jul 9 at 15:44
  • Ah true! thanks Fedor for the quick resolution. – Suvrit Jul 9 at 15:54
  • Great answer! What was the motivation for looking for the matrix factorization? – Sandeep Silwal Jul 10 at 0:15
  • @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. – Suvrit Jul 10 at 0:29

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.