The Gaussian radial kernel is positive semidefinite

For each sample point put $a_i=c_i\exp (-\mathrm{\parallel }{x}_i{\mathrm{\parallel }}_2^2\mathrm{/}(2{\sigma }^2))$. The identity $\mathrm{\parallel }{x}_i-x_j{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{x}_i{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_2^2-2x_i^T{x}_j$ gives ${\sum }_{i,j}{c}_i{c}_j{k}_{\sigma }(x_i,x_j)={\sum }_{i,j}{a}_i{a}_j\exp (x_i^T{x}_j\mathrm{/}{\sigma }^2)$.

For each fixed pair $i,j$, the exponential series has the absolutely convergent multi-index expansion $\exp (x_i^T{x}_j\mathrm{/}{\sigma }^2)={\sum }_{\alpha \in {\mathbb{N}}_0^p}(x_i^{\alpha }{x}_j^{\alpha })\mathrm{/}({\sigma }^{2\mathrm{\mid }\alpha \mathrm{\mid }}\alpha !)$, since the corresponding series of absolute values is ${\prod }_{r=1}^p\exp (\mathrm{\mid }{x}_{ir}{x}_{jr}\mathrm{\mid }\mathrm{/}{\sigma }^2)<\mathrm{\infty }$. The sample sums are finite, so they can be interchanged with this absolutely convergent series.

Substitution into [step 1.1] yields ${\sum }_{i,j}{c}_i{c}_j{k}_{\sigma }(x_i,x_j)={\sum }_{\alpha \in {\mathbb{N}}_0^p}\frac{1}{{\sigma }^{2\mathrm{\mid }\alpha \mathrm{\mid }}\alpha !}({\sum }_i{a}_i{x}_i^{\alpha }{)}^2\ge 0$. Symmetry of $k_{\sigma }$ is immediate from $\mathrm{\parallel }x-z{\mathrm{\parallel }}_2=\mathrm{\parallel }z-x{\mathrm{\parallel }}_2$, so [F1] proves positive semidefiniteness; zero eigenvalues or repeated points do not affect the sum-of-squares argument.