I understand that the point of using the kernel trick is to project the problem onto a higher dimensional space, where the problem is linearly separable. In this explanation, https://www.quora.com/What-is-the-kernel-trick, it states that the inner product $\langle x,y \rangle$ will be equal to $\langle \phi(x),\phi(y)\rangle$. Understanding this equality seems to be key to understanding how this trick works.

My question is, how do know that our function $\phi$ will preserve the inner product and what are the conditions for this to happen?

I have tried google searching this and despite many references to the kernel trick, I do not believe that this has been answered anywhere.


1 Answer 1


It doesn't mean that $\langle x, y \rangle = \langle \phi(x), \phi(y)\rangle$.

Kernel method in general means that for an algorithm that involve $\langle x, y\rangle$, we can replace it with a function $K(x,y)=\langle \phi(x), \phi(y)\rangle$ where computing $K(x,y)$ is easy. It is known that for $K$ that satisfies Mercer's theorem, there is a corresponding $\phi$ which is the map to the higher dimensional space.

We do not need to know $\phi$ explicitly and $\phi$ can be very complicated.

  • $\begingroup$ what gives you the right to replace with $K(x,y)$? Also are you saying that the equality in the guys post is by chance? $\endgroup$
    – Trajan
    Commented Sep 22, 2019 at 8:11
  • $\begingroup$ for some algorithms, we might find that it just involve inner product with some features space, in that case, rather than computing $\langle x, y \rangle$, that is working with $x$ directly, we might like to perform a transformation $\phi(x)$, and then work with it. Since it just involves inner product, what matters is just $\langle \phi(x), \phi(y) \rangle$, and we can compute it via $K(x,y)$ and we don't really care about $\phi$ explicitly sometimes. As for the "by chance" thing, I might need you to be more explicit. $\endgroup$ Commented Sep 22, 2019 at 8:17
  • $\begingroup$ what is the justification for be able to swap to $K(x,y)$? $\endgroup$
    – Trajan
    Commented Sep 22, 2019 at 9:59
  • $\begingroup$ The justification is rather than working with $x$ directly, we want to work with $\phi(x)$, for example, rather than working with $x$, you might want to work with $(x, x^2)$. Then you want to work with inner product of $(x, x^2)$ and $(y, y^2)$, this is denote as $K(x,y)= \langle (x, x^2), (y, y^2) \rangle$. $\endgroup$ Commented Sep 22, 2019 at 10:05
  • $\begingroup$ Yes but why can you swap $x$ for $\phi(x)$? What are the requirements on $\phi$? $\endgroup$
    – Trajan
    Commented Sep 22, 2019 at 10:06

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