# Checking if a kernel is valid

The kernel is $$K(x,z) = \sum_{i=1}^D (x_i+z_i)$$

My approach was trying to express $$K = \phi(x)^T\phi(z) = (x_1 x_2 ... x_D \quad 1 1 ...1)(1 1 ...1\quad z_1 z_2 ... z_D )^T$$ where $$\phi$$ is 2Dx1 and thus a Kernel.

The solution says:

K is not a kernel. Consider $$x_1 = [1 \quad 0]^T \quad x_2 = [0 \quad 2]^T$$. Their kernel matrix has eigenvalues −1 and 5.

What explains this discrepancy?

Your confusion is that you're using $$\phi$$ to denote two different mappings. The first one maps its input $$v$$ to the vector $$\begin{pmatrix} v & \mathbf{1} \end{pmatrix}$$, and the second one maps it to $$\begin{pmatrix} \mathbf{1} & v \end{pmatrix}$$. In the same way we can prove that $$-1$$ is the square of a real number: $$-1 = x \cdot x = (-1) \cdot (+1)$$.