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How can I prove that the exponential $\exp(K)$ of a kernel function $K$ is again a kernel? I think it can be proved using Taylor expansion but I am not sure how.

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    $\begingroup$ You almost have it! You surely need the Taylor expansion (en.wikipedia.org/wiki/Matrix_exponential). You need to prove that is positive definite. If A > 0, are its powers positive? $\endgroup$
    – jpmuc
    Commented Dec 29, 2017 at 12:04
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    $\begingroup$ everything you need is in Dougal's answer here stats.stackexchange.com/questions/35634/… $\endgroup$
    – jld
    Commented Dec 29, 2017 at 14:33
  • $\begingroup$ Sometimes, $K$ is not a kernel and you still have to prove that the exponential function is a valid kernel, say $\exp(-\theta|x-x'|)$. $\endgroup$
    – zzzhhh
    Commented Jan 1, 2023 at 9:36

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Using Taylor expansion around $0$:

$$ \exp(K) = \exp(0) + \exp(0)K + \frac{\exp(0)}{2!}K^2 + \frac{\exp(0)}{3!}K^3 + ...\\ \exp(K) = 1 + K + \frac{1}{2}K^2 + \frac{1}{6}K^3+... $$

we can see that the exponential of a kernel is just an infinite series of multiplications and additions of that kernel.

Using the fact that addition and multiplication of kernels yield valid kernels:

$$ K' = \alpha K_1 + \beta K_2\\ K' = K_1K_2 $$

we can conclude that the exponential of a kernel is a kernel.

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