# 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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