In this question:
Why is the last box marked as true when the decision boundary will be linear in the high-dimensional space? There is even a slide about this in the course, which comes in the slide immediately after the function in the question is stated.
Thanks for your time and sorry for the huge pictures (couldn't find how to scale them).
If the data is not transformed using a non-linear kernel function (as is this case, because we are still working with the data x and not the feature expanded version phi), then SVM gives linear decision boundaries in the original space. You can check the SVM slides in week 7 (slide 24, 25) for a short reminder.
In the end SVM actually always finds linear decision boundaries in whatever space you transform your data to. Let's say you have a 2 dimensonal x. You transform this using a non-linear kernel to 8 dimensions. Now, the decision boundary that SVM finds will be linear in this new 8 dimensional space (not the original space!). But if you look at this decision boundary in the original low dimensional space, it will be curvy.
I hope this clarifies things! Best,
Sena
Thanks for your reply, this makes perfect sense. The key point was in the "subject to" constraint as you mentioned.
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