Would a 100-by-3 dimensional feature matrix be mapped into a 100 dimensional or into a infinite dimensional feature space, if the mapping would not be bypassed by the Gaussian RBF Kernel?
Following this reasoning (https://stackoverflow.com/questions/25082222/the-rbf-kernel-of-support-vector-machine) I would tend to say the feature matrix would be mapped to a inifinite dimensional feature space. Here a summary of the content:
Given a m-by-n feature matrix
X. Each n-dimensional instance
X is used to define a n-dimensional normal distribution function
N1(n), with its center equal to the n-dimensional point x.
N1(p) gives a real number as output for every inputted real number
p. So an instance with a finite number of
n dimensions is mapped into a function (I dont know how to assess the number of dimensions of this function).
K(a,b) is a function capable of computing the dot product
<phi(a).T, phi(b)>. The dot product of two functions is defined as the integral of the multiplication of them. So
<phi(a),phi(b) = int phi(a)*phi(b) da. This integral results in the gaussian rbf kernel
K(a,b) = exp(-gamma*||a-b||^2).
Following this reasoning (https://stackoverflow.com/questions/23581508/why-gaussian-radial-basis-function-maps-the-examples-into-an-infinite-dimensiona?rq=1) I would say the feature matrix would be mapped to a 100 dimensional feature space. Here a summary of the content:
If you have
m distinct instances then the gaussian radial basis kernel makes the SVM operate in an m-dimensional space. We say that the radial basis kernel maps to a space of infinite dimension because you can make m as large as you want and the space it operates in keeps growing without bound.