I have to design a matrix A that solves a linear system:
$y = A x$
The requirement is that A is an RBF kernel, i.e., it has the following structure:
$ a_{ij} = \alpha \exp\left (\frac{-d_{ij}^2}{2\sigma^2}\right)$, where d is the distance between each element in the matrix.
My first insight was:
Given that V is the eigenbasis (orthogonal eigenvectors for a fixed $\sigma$ and $\alpha$) of A. Then, $A = V diag(e) V^{-1}$, where vector e contains all the eigenvalues of A.
The problem simplifies to
$y = V diag (V^{-1} x) e $
So, for a given $\sigma$ and $alpha$, I can apply least squares to find the eigenvalues e:
$ e = V diag (V^{-1} x) \backslash y $
By finding the eigenvalues, I can then backengineer as $A = V diag (e) V^{-1}$
My question is: Will this procedure result in a RBF kernel A? A linear combination of the eigenbasis of a RBF kernel does result in a RBF kernel?
Is there a way to perform such least squares that would garantee the structure of matrix A?
Best.
Best.