I have a square matrix with elements $B_{ij} = e^{-c(i-j)^2}$ for some $c > 0$, i.e. a sort of matrix of Gaussian distances. How would I go about proving that this matrix is positive definite?
I did a quick check by numerically computing the Cholesky decomposition for sizes up to $1000\times1000$ for a few values of $c$. They seem to work, so I think that the matrix is positive definite.
Using Gaussian characteristic function we see that $e^{-cx^{2}}=\int e^{itx} d\mu(t)$ for some probability measure $\mu$. [$\mu$ is, in fact, normal]. Thus, $\sum c_j\overline {c_k}B_{jk}=\sum c_j\overline {c_k} e^{-c(j-k)^{2}}=\sum c_j \overline {c_k} e^{it(j-k)} d\mu(t)=\int |\sum c_je^{ijt}|^{2}d\mu (t) \geq 0$.