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
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? 
[EDIT]: Based on the link below and the given $x_1, x_2$, I arrive at $S = 2a_1^2+6a_1a_2+4a_2^2$ Now, how can I choose $a_1,a_2$ to make this negative?
 A: 
Based on the link below and the given $x_1, x_2$, I arrive at $S = 2a_1^2+6a_1a_2+4a_2^2$ Now, how can I choose $a_1,a_2$ to make this negative?

We can factorize $S$ to obtain
$$
S = 2(a_1 + a_2) (a_1 + 2 a_2)
$$
so if we want $S < 0$, we need to choose $a_1, a_2$ such that exactly one of $(a_1 + a_2) <0$ or $(a_1 + 2a_2)<0$. This is because the product of two positive numbers is positive and the product of two negative numbers is positive but the product of a positive number and a negative number is negative.
Since this problem is under-determined, a useful way to go about solving it is to fix $a_1$ at some value and then find $a_2$ that satisfies our criteria. Arbitrarily, I chose $a_1 = -1$. This gives
$$
S = 2(-1 + a_2) (-1 + 2 a_2)
$$
Now I decided to make the second factor negative and the third factor positive. This means we need $a_2 - 1 < 0$ but $2a_2 -1 >0$. Together, these inequalities provide $\frac{1}{2} < a_2 < 1$. We can arbitrarily choose any $a_2$ in that interval.
Thus one solution among many is $a_1 = -1, a_2 = \frac{3}{4}$.
