The positive determinant of one special matrix

Question

I try to prove the positive value of determinant for matrix ($n\times n$ for any $n$):

\begin{equation*} ||a_{ij}|| = ||f(x_i - y_j)|| , \text{where}~f = f(\lambda(x - y)) = \exp(-\lambda(x - y)^2),~\lambda > 0,\\ x_1 < ... < x_n,~y_1 < ... < y_n,~x_i \in [0; 1],~y_i \in [0; 1]. \end{equation*}

For $n=2$ this easily to show, but even for $n=3$ the explicitly calculating is cumbersome. I have an idea of proof using method of mathematical induction, but my efforts didn't bring success.

Answer 1

Thanks to all, the problem was solved, for function $exp(-(x-y)^2)$. If we take out off the determinant all the multipliers $exp(−x_i^2) > 0$ and $exp(-y_j^2) > 0$, $i,j=1,\dots,n$, we will get the generalized Vandermonde Matrix, and it is known as positive.

Answer 2

Omnomnomnom is right. Gaussian function is (strictly) positive definite (Wendland's "Scattered data approximation", pp.74,76) and leads to PD matrices. The determinant of a positive definite matrix is always positive.