Eigenfunctions and eigenvalues of the product of two exponential kernels

Question

Consider the following exponential kernel:

$k(x_1, x_2) = \exp\left(\frac{|x_1 - x_2|}{L}\right)$,

which is symmetric and non-negative definite. By virtue of Mercer's theorem, we have

$k(x_1, x_2) = \sum_{i = 1}^\infty \lambda_i \phi_i(x_1) \phi_i(x_2)$

where $\lambda_i$ and $\phi_i$ are the eigenvalues and eigenfunctions of $k$, respectively. Now, consider the following product:

$K((x_1, y_1), (x_2, y_2)) := k(x_1, x_2) k(y_1, y_2) = \exp\left( -\frac{|x_1 - x_2|}{L} - \frac{|y_1 - y_2|}{L}\right)$.

Since the product of two symmetric, non-negative definite kernels is another kernel with the same properties, Mercer's theorem still applies.

The question is: Having computed $\lambda_i$ and $\phi_i$ of $k$, what can we say about the eigenfunctions and eigenvalues of $K$?

Thank you.

Regards, Ivan

Answer 1

You have two independent sets of variables, so it is a tensor product, not a Hadamard product, no? I am not sure I understand Suvrit's comment...

I think you just get

$K((x_1,y_1),(x_2,y_2)) = k(x_1,x_2)k(y_1,y_2) = \sum_{i=1}^\infty \lambda_i \phi_i(x_1)\phi_i(x_2)\sum_{j=1}^\infty \lambda_j \phi_j(y_1)\phi_j(y_2)$

$=\sum_{i,j=1}^\infty \lambda_i \lambda_j \phi_i(x_1)\phi_j(y_1) \phi_i(x_2)\phi_j(y_2)$

so that the eigenvalues of $K$ are just products of the eigenvalues of $k$,

$\Lambda_{i,j} = \lambda_i \lambda_j$.

And the eigenfunctions are simply products, too, i.e.

$\Phi_{i,j}(x,y) = \phi_i(x)\phi_j(y)$.

If you want sums over one index, you can now use your favorite enumeration of $N\times $,

$\Lambda_1 = \lambda_1 \lambda_1$,

$\Lambda_2 = \lambda_1 \lambda_2$, $\Lambda_3 = \lambda_2 \lambda_1$,

$\Lambda_4 = \lambda_1 \lambda_3$, $\Lambda_5 = \lambda_2 \lambda_2$, $\Lambda_6 = \lambda_3 \lambda_1$,

etc. (and similarly for the $\Phi$'s).