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Do SVMs really map the initial space to a new space?
The algorithm does not actually transform the data points to a new, high dimensional feature space. Kernelized SVM compute decision boundaries in terms of similarity measures in a high-dimensional feature space without actually doing a transformation.
Why the RBF kernel maps data to an infinitely 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.
Is the data in our original feature space linearly separable?
Yes, it becomes linearly separable, especially if you always use the RBF kernel which maps to infinite dimensional space.
What are the features of a non linear SVM?
Linear SVM: (without mapping) Non-linear SVM: w could be infinite dimensional 14 Kernel vs. features 15 A tree kernel Common kernel functions •Linear : •Polynominal: •Radial basis function (RBF): •Sigmoid:
How are data classes separated in a SVM?
The data classes are separated by a circle, but an SVC cannot find circles directly. However, if the data are transformed using a radial basis function, then in the resulting space, the classes are separated by a linear boundary. It is important to keep in mind the difference between basis expansion and Kernel method / SVM.
Do you need a kernel function for SVM?
Kernel learning methods such as SVM do not need them to work; they only need the kernel function K. It is possible to write down a formula for Φ but the feature space it maps to is quite abstract and is only really used for proving theoretical results about SVM.
Can a SVM be used to project data points?
So it’s not used to project data points, but rather an outcome of the projection. It can be considered a measure of similarity, but in an SVM, it’s more than that. The optimization for finding the best separating hyperplane in F involves h(x) only through the inner-product form.