# Kernel with complex vectors

I have a question regarding my machine learning lecture where we had to decide whether $$K(x,y)=x_1y_1-x_2y_2$$ is a valid kernel (e.g. for a SVM). My intuition would say that it is a valid kernel since we can display it with: $$\Phi(x)=(x_1, ix_2)\implies K(x,y)=\Phi(x)\Phi(y)$$ with $i$ being the imaginary number. Is that right?

Kernels are considered valid if they are positive-definite, so we must have $$\sum_{i=1}^n \sum_{j=1}^n c_i c_j K(x_i, x_j) \ge 0$$ for all $n \in \mathbb{N}$ and $c_1, \dots, c_n \in \mathbb{R}$ and (in your case) all $x^1, \dots, x^n \in \mathbb{R}^2.$
Letting $n=1$ and $c_1 = 1$ and $x^1 = (0, 1)$ shows that $K$ is not positive-definite, since $$K\big((0,1), (0,1)\big) = 0\times 0 - 1\times 1 = -1.$$
The problem with your approach is that you define $\Phi$ on $\mathbb{C}$ but write $K$ as the dot product for real-valued vectors. The feature map argument for positive definite vectors only works if the product is an inner product on $\Phi$'s codomain. The dot product for real-valued vectors is not an inner product on $\mathbb{C}^2$ (because $i\cdot i = -1 < 0$). You would have to use the generalized dot product $x \cdot y = \sum_i x_i \overline{y_i}$, or another inner product on $\mathbb{C}^2$.