# Determinant and eigenvalues of a specific matrix

This came up in a conversation with an engineer friend of mine.

Let $c>0$ be a constant. Let $A_{ij}$ be an $n$ by $n$ matrix with entries $$A_{ij} = e^{-c(i-j)^2}.$$ Is there a name for this matrix? What is known, perhaps approximately or asymptotically as $c$ and $n$ change about the determinant and the eigenvalues of it? Are there other functions of $(i-j)$ for which the answer is more explicit?

• If you let $e^{-c}$ be $x$, then the matrix has polynomial entries in $x$, and experiment seems to indicate that the determinant is a product of cyclotomic polynomials (for instance, if $n = 6$, we get $-(x-1)^{15}(x+1)^{15}(x^2+1)^6 (x^2-x+1)^3(x^2+x+1)^3(x^4+1)^2(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$. (In particular, the power of $(x \pm 1)$ seems to be $n$ choose $2$.) Jan 2, 2014 at 2:18
• It's an example of a Toeplitz matrix. The entries are Gaussian. So I googled Gaussian Toeplitz matrix which gave the following paper sciencedirect.com/science/article/pii/0024379592902559 , which should answer your question. (I haven't read the paper myself so can't say for sure.) Jan 2, 2014 at 2:19
• This is also a totally positive matrix, in the sense of Gantmakher-Krein. Which implies that all eigenvalues are real and many qualitative properties of eigenfunctions Jan 2, 2014 at 2:44
• This is also an aproximation to the heat kernel in some sense Jan 2, 2014 at 2:45
• Formula (3.1) says the determinant is $$(1-a^2)^{n-1} (1-a^4)^{n-2} \cdots (1-a^{2(n-2)})^2 (1-a^{2(n-1)}) = \prod_{m=1}^{n-1} (1-a^{2m})^{n-m}$$ where $a = e^{-c}$ is what Abhinav called $x$. The $n=6$ case agrees with Abhinav's factorization, and yes, the valuations at $1-a$ and $1+a$ are both $n \choose 2$. (In the paper the size of the matrix is called $k$, not $n$.) Jan 2, 2014 at 3:10

\begin{equation*} k(x_i,x_j) = e^{-c(x_i-x_j)^2}\qquad x_1,\ldots,x_n \in \mathbb{R}. \end{equation*} Clearly, the matrix $A = [k(x_i,x_j)]$ is semidefinite. If the $x_i$s are unique, then it is strictly positive definite (this requires some more work to prove).
The Gaussian kernel is of course the canonical example of a translation invariant kernel, so searching more for such kernel matrices should bring up more examples, namely, matrices of the form $A_{ij} = k(x_i,x_j)=\varphi(x_i-x_j)$, for suitable $\varphi$.
Incidentally, in the paper mentioned by @Lucia, the smallest eigenvalue for the Gaussian-Toeplitz $A$ is bounded from below by (assuming $A$ is $n\times n$): \begin{equation*} \lambda_n \ge \frac{\det(A)}{n!}. \end{equation*} It seems possible to improve this to $\lambda_n \ge \frac{\det(A)}{2^{n-1}}$ by using the technique mentioned in my answer here.