# Is a covariance matrix defined through a Gaussian covariance function always positive-definite? [duplicate]

When using Gaussian processes, the covariance matrix $\mathbf{\Sigma}$ is often defined via a covariance function $K$ as follows $$\mathbf{\Sigma}_{ij} = K(\underline{x}_i, \underline{x}_j)$$ where $\underline{x}_i, \underline{x}_j$ are coordinates of two points in some space of interest, and belong to some finite set of $n$ such points, resulting in a $n \times n$ covariance matrix.

Is it known whether the Gaussian (or 'squared exponential') covariance function $$K(\underline{x}_i, \underline{x}_j) = a^2 \exp{\left(-\frac{(\underline{x}_i - \underline{x}_j)^{\top}(\underline{x}_i - \underline{x}_j)}{2l^2}\right)}$$ will always produce a positive-definite $\mathbf{\Sigma}$ for any chosen set of points and values of $a, l$?

Thanks!

• Positive-definite, No; Nonnegative defiinite (sometimes called positive semidefinite) Yes. – Dilip Sarwate Oct 14 '17 at 22:20
• @DilipSarwate oh interesting - I don't suppose you know a source for a proof of that? Thanks – CBowman Oct 15 '17 at 13:06
• It's got nothing to do with Gaussian random variables or processes per se.. All covariance matrices (no matter what the (finite-variance) random variables are) must be positive semi-definite. See this answer of mine to see why this must be so. – Dilip Sarwate Oct 15 '17 at 16:44