Proving a given kernel is not valid

by db18   Last Updated October 14, 2018 23:19 PM

$K(x,z) = e^{\gamma ||x-z||^2} \quad \gamma >0$

I took $x_1 = [1\quad 0]^T$ and $x_2 = [0\quad 2]^T$ which gives $K(x_1,x_2) = 5 = K(x_2,x_1)$ and $K(x_1,x_1) = K(x_2,x_2) = 0$

Thus the Kernel matrix looks like $[0\quad 5;5\quad 0]$ and hence eigenvalues are -5,5. Thus K is not positive definite and hence, invalid.

Is this rationale correct?



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