# Why is $\mathbb{R}^3$ not the Reproducing Kernel Hilbert Space defined by kernel $k(x,y)=(x_1y_1+x_2y_2)^2$

Let the input space be $\mathcal{X}=\mathbb{R}^2$, and kernel function $k(x,y)=\langle x,y \rangle^2 = (x_1y_1+x_2y_2)^2$.

I can write

\begin{align} k(x,y) &= x_1^2 y_1^2 + x_2^2 y_2^2 + 2 x_1 y_1 x_2 y_2 \\ &= \begin{bmatrix} x_1^2 & x_2^2 & \sqrt{2}x_1 x_2 \end{bmatrix} \begin{bmatrix} y_1^2 \\ y_2^2 \\ \sqrt{2}y_1 y_2 \end{bmatrix}\\ \end{align}

The feature map $\phi: \mathbb{R}^2 \to \mathbb{R}^3$ is then $\phi(x)=k_x=k(\cdot, x) = (x_1^2, x_2^2, \sqrt{2}x_1 x_2)$. Since $\phi: \mathcal{X} \to \mathcal{H}$ maps input vector to the RKHS defined by $k$, I conclude that $\mathcal{H} = \mathbb{R}^3$.

But page 41 of this lecture presentation says otherwise, i.e. $\mathbb{R}^3$ is not the RKHS defined by kernel $k(x,y)=(x_1y_1+x_2y_2)^2$. If not, then what is the RKHS defined by kernel $k$?

The RKHS defined by that kernel is not the "whole" $\mathbb{R}^3$, but a two dimensional manifold defined in it. Note that, for instance, the vector $(1,1,1)$ is not in the image of $\phi(x)$.
• My impression is that $\mathcal{H}$ is not defined by $\{\phi(x) | x \in \mathcal{X} \}$, but all functions of the form $f(x)=ax_1^2+bx_2^2+cx_1x_2$ (hence $\mathcal{H}=\mathbb{R}^3$) . See these slides, p.36-37 – Yibo Yang Mar 12 '17 at 0:11