Prove that linear combination of kernels is not a kernel for real-valued coefficients Let's say that $k_1(\textbf{x}, \textbf{x}')$, $k_2(\textbf{x}, \textbf{x}')$: $\mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{R}$ are kernels.
It can then be proven that the linear combination of two kernels (or any finite number of kernels) is also a kernel, i.e. :
$$
k_3(\textbf{x}, \textbf{x}') := \alpha k_1(\textbf{x}, \textbf{x}')  + \beta k_2(\textbf{x}, \textbf{x}') 
$$
is a kernel. However, this only holds for $\alpha, \beta \in \mathbb{R}^+$.
Why does this break for non-positive values?
 A: Thanks to @TAIsup I figured it out.
Let $k_1(\textbf{x}, \textbf{x}')$ and $k_2(\textbf{x}, \textbf{x}')$ be kernels.
Then define $k_3(\textbf{x}, \textbf{x}')$ as follows:
$$
k_3(\textbf{x}, \textbf{x}') = -1 \cdot k_1(\textbf{x}, \textbf{x}') + 0 \cdot k_2(\textbf{x}, \textbf{x}') = -k_1(\textbf{x}, \textbf{x}')
$$
We know that the kernel matrix of $k_1$ has to be positive semidefinite, i.e.
$$
\forall \textbf{v} \in \mathbb{R}^n: \textbf{v}^TK\textbf{v} \geq 0
$$
For kernel-matrix $K_1$ of kernel $k_1$ this means:
$$
\forall \textbf{v} \in \mathbb{R}^n: \textbf{v}^TK_1\textbf{v} \geq 0
$$
Now, if we instead plug in the kernel-matrix of $k_3$ we get a contradiction:
$$
\begin{aligned}
\forall \textbf{v} \in \mathbb{R}^n: \textbf{v}^TK_3\textbf{v} &= \textbf{v}^T (-K_1) \textbf{v}  = -(\textbf{v}^T K_1 \textbf{v}) \geq 0
\end{aligned}
$$
since $\textbf{v}^TK_1\textbf{v} \geq 0$. The only case where this still works is when $K_1$ is full of zeros, but since the statement has to be disproven, one counterexample is enough, which would be any matrix except the zero-matrix.
