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



Eigenfunctions and eigenvalues of the product of two exponential kernels
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).


