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I have read that the dimensionality of the feature map of the Gaussian Kernel is infinite. However I saw another post (Kernel SVM) stating that the feature map for a Kernel SVM maps to a dimensionality of at most the size of the training sample.

These seem contradictory to me, as surely if you used the Gaussian Kernel for SVM we would then be in a vector space of infinite dimension.

My main question is, does the number of components of the $\phi$(x) dictate the dimensionality of the feature space?

For example, does this correspond to a feature space of dimensionality 4 $$ \phi(x_i) = \left (1, 2x_i, (5x_i)^2, x_i^{\frac{1}{2}} \right) $$

Bill0
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Yes to your question on whether the dimensionality of feature space is 4. For Gaussian kernel, it's infinite dimension as you have mentioned as well. You can learn in more depth from Professor Yaser Abu-Mostafa's lecture. You can jump straight to around 26:50 if you are already familiar with the subject.

Starz
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