Is this difference of two kernels also a kernel? If we have two linear kernels $k1$, $k2$ over $R^2$ defined as:
$k1(x,y) = x^\mathsf{T}
\begin{bmatrix}
    3 &1 \\
    1 &5\\
\end{bmatrix}y$ and $k2(x,y)=x^\mathsf{T}y$ for all $x,y\in R^2$.
Will $k1-k2$ be a kernel?
I checked symmetric property but how to check if it is PSD or not? Is there some alternative way to show?

Writing it out, we have $$
k_1 - k_2 = x^\top \begin{bmatrix}3 & 1 \\ 1 & 5\end{bmatrix} y - x^\top y \\
=x^\top (A - I)y
$$
and we can verify that $A - I$ is positive definite.
If we form a square matrix $K$ with each element $k_{ij} = k_1(x_i, x_j) - k_2(x_i, x_j)$, we can write
$$
K = X^\top (A - I) X
$$
So we neet to determine if $K$ is psd, which means showing that $u^\top K u \ge 0 \forall u \in \mathbb{R}^2\setminus 0$.
A simple way to do this is to write $(A - I) = R^\top R$, which we know is possible because $A - I$ is symmetric. Then we can write
$$
u^\top X^\top R^\top RXu = \|RXu \|_2^2 \ge 0
$$
because the sums of squares must be non-negative.
