# Prove that ${f(x,y) = x^Txx^Tyy^Ty}$ is a valid kernel

The rules I can use without a proof are the following ones:

The thing that I know that a kernel is valid if it is symmetric, e.g. $${f(x,y) = f(y,x)}$$ and if $${c^TKc >= 0}$$, for each vector $${c \neq 0}$$, but I don't know how to prove it, and don't know how to get the matrix K (if this is the correct way to solve this exercise).

Let $$\phi(\mathbf x)=\mathbf x\mathbf x^T\mathbf x$$ be a nonlinear transformation for $$\mathbf x$$. Then, you can use the fourth rule in your screenshot, because $$k(\mathbf x, \mathbf y)=\phi(\mathbf x)^T\phi(\mathbf y)$$ is a valid kernel since $$k_3(\mathbf x,\mathbf y)=\mathbf x^T\mathbf y$$ is a valid kernel.
• But how do you know you have to start from a value ${\phi(x) = xx^Tx}$? By intuition? Oct 25, 2022 at 22:15
• @Ele975 Because you generally want to write $\mathop{f}\left(x,y\right)$ as an inner product of the form $\langle \mathop{\phi}\left(x\right), \mathop{\phi}\left(y\right)\rangle$. Realizing that you have to take the transpose of $x^\top x x^\top$ to make it match the form of $y y^\top y$ is just a small step then. Oct 25, 2022 at 23:17
I think you can use R3 and get $$xx^Tyy^T$$and then mention that is $$x^2y^2$$