2
$\begingroup$

Just started learning about SVM, kernels and all those things. Got stuck on this question (homework question):

Consider a kernel $K_1$ and its corresponding mapping $φ_1$ that maps from the lower space $R^n$ to a higher space $R^m (m>n)$. We know that the data in the higher space $R^m$ is separable by a linear classifier with the weights vector $w$.

Given a different kernel $K_2$ and its corresponding mapping $φ_2$, we create a kernel $K = K_1 + K_2$.

Can you find a linear classifier in the higher space to which $φ$, the mapping corresponding to the kernel $K$, is mapping?

If yes, find the linear classifier weight vector.
If no, prove why not.

My intuition says no, because $K_2$ may mess my data up. But it makes more sense that it is true, because if the data is already linearly separable in $R^m$, when we will perform $K$ we will arrive at space $p$ that is $p>m$, and if the data is linearly separable in $R^m$, so probably in $R^p$ it will also be linearly separable. I really don't know how to start proving it.

$K$ will probably be $\phi_1(x)\phi_1(y) + \phi_2(x)\phi_2(y)$. Still don't know how to proceed from here. I'm trying somehow to reset all values from $\phi_2$, because I know with $\phi_1$ my data is already linearly separable.

$\endgroup$
  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung May 18 '18 at 0:55
  • $\begingroup$ Hint: We know that $K_1(x, y) = \phi_1(x) \cdot \phi_1(y)$ and $K_2(x, y) = \phi_2(x) \cdot \phi_2(y)$. Given that $K(x, y) = \phi(x) \cdot \phi(y)$ and that $K = K_1 + K_2$, what could $\phi$ look like in terms of $\phi_1$ and $\phi_2$? $\endgroup$ – user20160 May 18 '18 at 5:35
  • $\begingroup$ @user20160 $K$ will probably be $\phi_1(x)\phi_1(y) + \phi_2(x)\phi_2(y)$. Still don't know how to proceed from here. I'm trying somehow to reset all values from $\phi_2$, because I know with $\phi_1$ my data is already linearly separable. $\endgroup$ – sheldonzy May 19 '18 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.