# Using SVD on features before SVM classification, when p >> N

So I am going through Hastie's Elements of Statistical computing, and in section 18.3.5 which deals with computational shortcuts when the number of dimensions $p$ is much larger that the number of samples $N$ the authors mention:

When $p > N$ , the computations can be carried out in an $N$-dimensional space, rather than $p$, via the singular value decomposition [...]

Here is the geometric intuition: just like two points in three- dimensional space always lie on a line, $N$ points in $p$-dimensional space lie in an $(N − 1)$-dimensional affine subspace.

That does indeed make a lot of sense. They proceed to mention that:

This result can be applied to many of the learning methods discussed in this chapter, such as regularized logistic regression, linear discriminant analysis, and support vector machines.

My question is: Can this result be applied when using the RBF or some other non-linear kernel for the SVM?

If yes, then does it make any sense to perform SVM classification on the full-feature dataset, or should one always use the SVD transform of the data when $p \gg N$?

There's no gain. The result lies in the fact that instead of a coefficient vector $\beta_{P\times1}$ you have instead to estimate a $\beta_{N\times1}$, which is smaller.
Consider the linear Kernel/Gram matrix decomposition, and notice $V$ is orthonormal:
$$K = XX^T=UDV^TVDU^T = UD^2U^T=(UD)(DU^T)= RR^T$$
As shown, applying kernels to the SVM you are already estimating $\beta_{N\times1}$ coefficients. So the solutions are the same, and there's no gain whatsoever. Other kernels are functions of the same dot product, reproducing this result.