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When we use kernels in SVM to linearly sperate non linear data points by mapping it to 'another dimensions', does this suitable 'another dimensions' always be a higher dimension with respect to original dimension of the data points?

And is it true that we can always find a higher dimensions that can linearly seperate data points in a training set?

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  1. 'Another dimension' can be also a lower dimensional space. You are free to choose your kernel. Example is: $k\big((x_1, y_1), (x_2, y_2)\big) = x_1x_2$

  2. Is it true that we can always find a higher dimensions that can linearly seperate data points? Yes. If you have a finite dataset, you can always find a higher dimension that can linearly seperate data points. The obvious one is mapping n points to a n-dimensional space. If you do any non-trivial mapping (no d points are on a d-1 dimensional hyperspace), then you can always linearly separate any two class assignments.

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  • $\begingroup$ Is your 2nd point valid for multiclass data as well? $\endgroup$ – user1825567 Oct 4 '17 at 13:54
  • $\begingroup$ What do you mean? In multiclass, you can separate any class from rest with hyperline in higher dimensional space. If you want to separate all classes, you would get voronoi diagram that is piecewise linear. $\endgroup$ – Mortezaaa Oct 5 '17 at 8:01

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