I just wanted to know whether a kernel could be defined as follows: $$ k(\mathrm{x}, \mathrm{x}') = x_1 + x_2 \quad \mbox{OR} \quad k(\mathrm{x}, \mathrm{x}') = \left<\begin{bmatrix}x_1\\ x_2 \end{bmatrix},\begin{bmatrix}1 \\ 1\end{bmatrix}\right>$$ where $\mathrm{x} = \left[ \begin{matrix}x_1 \\x_2 \end{matrix} \right]$ and $\mathrm{x}' = \left[ \begin{matrix}x'_1 \\x'_2 \end{matrix} \right]$


  • $\begingroup$ Mercer kernels have two attributes (1) symmetry and (2) the matrices are positive semi-definite. This is obviously not a Mercer kernel, since it lacks both of those properties. However, there is some research into kernels that do not satisfy Mercer's conditions, for example the first hit on a google search: kyb.tuebingen.mpg.de/fileadmin/user_upload/files/publications/… so perhaps this provides some interesting options for your research. $\endgroup$
    – Sycorax
    Commented Oct 2, 2015 at 12:18

1 Answer 1


All function of two arguments from similar spaces are kernels, but introduced kernels lack two important properties that kernels typically have.

For the first definition consider two points $x_1$, $x_2$ and a kernel matrix for these points: $$ \begin{pmatrix} 2 x_1 & x_1 + x_2 \\ x_1 + x_2 & 2 x_2 \\ \end{pmatrix} $$ Determinant is $-(x_1 - x_2)^2 < 0$ if $x_1 \neq x_2$. So, the kernel matrix is not nonnegative-definite for this kernel.

For another definition of kernel you have the function that is not symmetric, so it is a rather strange kernel.

Consequently, introduced kernels cannot be used for example as covariance functions.

  • $\begingroup$ I think you have missinterpreted my question. Note that $ \mathrm{x}\ne x_1,\mathrm{x}' \ne x_2$ and $\mathrm{x}$ is in $\Bbb{R}^2$. Lets try to understand it this way: $$ \mathrm{x}_1 = \left[ \begin{matrix}y_1 \\z_1 \end{matrix} \right], \mathrm{x}_2 = \left[ \begin{matrix}y_2 \\z_2 \end{matrix} \right] $$ where $\mathrm{x}_1$ and $\mathrm{x}_2$ are two points, then by above definition of kernel, kernel matrix should be as follows: $$ \begin{bmatrix} y_1+z_1 & y_1+z_1 \\ y_2+z_2 & y_1+z_1\end{bmatrix}$$ which is a singular matrix and hence it can not be a kernel. Am I Right? $\endgroup$
    – pkj
    Commented Oct 2, 2015 at 10:43
  • $\begingroup$ If you do it according to this definition, you get nonsymmetric kernel - which is the main problem here. $\endgroup$ Commented Oct 2, 2015 at 10:56
  • $\begingroup$ ya, you are right $\endgroup$
    – pkj
    Commented Oct 2, 2015 at 11:06
  • $\begingroup$ basis function should be same for $\mathrm{x}_1 $ and $\mathrm{x}_2$ $\endgroup$
    – pkj
    Commented Oct 2, 2015 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.