What are the eigenfunctions and the eigenvalues of the exponential kernel?

The exponential kernel is defined as $$k(x,x')=\sigma^2\exp\left(\frac{||x-x'||}{l}\right)$$ where both $\sigma>0$ and $l>0$.

Mercers theorem tell us that for every kernel function $k(x,x')$ there exists a decomposition in eigenfunctions $\phi_i(x)$ and corresponding eigenvalues $\lambda_i$ such that

$$k(x,x')=\sum_{i=1}^\infty \lambda_i \phi_i(x)\phi_i(x)$$

The Fourier transform $$\mathcal{F}(k)(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} k(r) e^{i\omega r}dr$$ of the function $$k(r)=\sigma^2 \exp\left(\frac{||r||}{l}\right)$$ with $r=x-x'$ is $$\mathcal{F}(k)(\omega)=\frac{\sqrt{\frac{2}{\pi }} \sigma^2 l}{l^2 \omega ^2+1}.$$ How to proceed from here?

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    $\begingroup$ Some information may be found in Gaussian Processes for Machine Learning, section 4.3, "Eigenfunction Analysis of Kernels." $\endgroup$ – Sycorax Jan 13 '16 at 19:15
  • $\begingroup$ Is this a machine learning question or a pure math question? $\endgroup$ – gung - Reinstate Monica Jan 13 '16 at 20:28
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    $\begingroup$ @gung It could be both. I was however hoping that the solution to this problem is better known in the machine learning problem. That's why I posted it here. $\endgroup$ – Julian Karch Jan 14 '16 at 12:14

Assuming the Hamiltonian of your system is the |x> operator, what you are really trying to find is the reciprocal space of x. The easiest way to do this is to take the fourier transform of k(x,x') which by definition is a linear combination of of the k(x,x') states. You can read more here: https://en.wikipedia.org/wiki/Fourier_transform

I also recommend you switch variables to r = x-x' to simplify the math.

  • $\begingroup$ Hey thanks for your comment. I added your suggestion to my original question. How to proceed from there? $\endgroup$ – Julian Karch Jan 14 '16 at 12:29
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    $\begingroup$ Actually I should take a step back. What operator are you investigating? To find eigenvalues and eigenfunctions you need to define this first. I realized I was assuming it was the Fourier Transform operator. $\endgroup$ – Greg Petersen Jan 15 '16 at 19:11

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