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 says Reinstate Monica Oct 2 '15 at 12:18

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 Oct 2 '15 at 10:43
  • $\begingroup$ If you do it according to this definition, you get nonsymmetric kernel - which is the main problem here. $\endgroup$ – Alexey Zaytsev Oct 2 '15 at 10:56
  • $\begingroup$ ya, you are right $\endgroup$ – pkj Oct 2 '15 at 11:06
  • $\begingroup$ basis function should be same for $\mathrm{x}_1 $ and $\mathrm{x}_2$ $\endgroup$ – pkj Oct 2 '15 at 11:13

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