I have the function $K: \mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$, $$K(x,t)=\exp(x_1t_1)$$
Where $x_1$ and $t_1$ are the first components of $x$ and $t$ respectively. I have to determine whether it is a valid kernel (ie, the kernel can be written as an inner product in some feature space) and derive the associated feature map representation.
Here is my attempt (using this question):
To prove that $K(x,t)$ is a valid kernel, we have to show that $K(x,t)$ is symmetric:
We have that $\exp(x_1t_1)=\exp(t_1x_1)$, hence $K(x,t)=K(t,x)$ is symmetric.
Now I would like to show that $K(x,t)$ is positive definite, by inspecting for an "inner-product" representation:
We have that $K(x,t)=\exp(x_1t_1)$ can be represented as $\phi(x)^T\phi(t)$ where $$\phi(x)=(1,x_1,\sqrt{\frac{1}{2}}x_1^2,\sqrt{\frac{1}{2}}x_1^3,...)=\left(\sqrt{\frac{1}{i!}}x_1^i: i\in\mathbb{N}\right)$$
Hence $K(x,t)$ can be represented as an inner product and is hence positive semidefinite. Hence $K(x,t)$ is a valid kernel.
Good job.
Actually if you can show the existence of such $\phi$, the rest are consequences of it.
We let $$\phi(x) = \left(1, x_1, \frac{x_1^2}{\sqrt{2!}}, \frac{x_1^3}{\sqrt{3!}}, \ldots \right)= \left(\frac{x_1^i}{\sqrt{i!}}: i \in \mathbb{N}_0 \right)$$
then
$$\langle \phi(x), \phi(t)\rangle = \sum_{i=0}^\infty \frac{x_1^i t_1^i}{i!}=\sum_{i=0}^\infty \frac{(x_1t_1)^i}{i!}=\exp(x_1t_1)=K(x,t),$$
Hence it is a valid kernel.