Kernel representation as dot product. Consider the following kernel $k(x,y) = \prod_{i=1}^n (1 + x_i y_i)$. We want to represent it like $k(x,y) = \langle \phi(x), \phi(y) \rangle$, where $\phi(x)$ is vector depends only on $x$.
It differs from the polynomial kernel since we can't easily separate them. Does there any chance to separate here?
 A: Let us take the case $n=2$.
The matrix associated to your kernel is
$$K=\begin{pmatrix}2&1\\1&2\end{pmatrix}$$
with eigenvalues $3$ and $1$, both non-zero.
Using the criteria mentionned for example here, we can conclude to the non-separability.
Explanation: The general entry of $K$ is $K(e_i,e_j)$ where $(e_1,e_2,\cdots e_n)$ is the canonical basis of $\mathbb{R}^n$; for example
$$K_{1,1}=K(e_1,e_1)=K((1,0),(1,0))=(1+1\times 1)(1+0\times 0)=2$$
What we have done in the case $n=2$ extends at once to the general case because $n \times n$ matrix $K$ with $K_{i,j}=1$ if $i \ne j$ except on the diagonal where $K_{i,i}=2$ has eigenvalues:
$$\lambda_1=n+1 \ \text{(multiplicity 1) and} \ \ \lambda_k=1 \ (\text{multiplicity} \  n-1)$$
Remark: We have used eigenvalues $\lambda_i$ instead of singular values $\sigma_i$ because, for real symmetric matrices, $\sigma_i=|\lambda_i|$ ; therefore the criteria of a single non-zero singular value is the same as a single non-zero eigenvalue.
none of them zero.
