2
$\begingroup$

The function is defined as

$k(x,x')=||x||$

Norm in Hilbert Spaces can be defined as $||x||= \sqrt{x^Tx} $. I am not sure about the feature map of this function that how will it be and I am positive that it does not exist. Moreover, kernel function depends both on $x$ and $x'$ but how should we formally prove that condition?

$\endgroup$
8
  • $\begingroup$ Is there a typo in the title? The kernel function depends on both $x,x'$. Also I think it's always nice when the main question includes all details required to answer (instead of "frontloading" some details to the title). $\endgroup$
    – air
    Commented Sep 12, 2020 at 11:19
  • $\begingroup$ No its not a typo! yes agreed it does depends on both $x$ and $x'$ but how to formally show that!. Yeah i am updating the title thanks $\endgroup$ Commented Sep 12, 2020 at 11:21
  • $\begingroup$ a function of two arguments can't be defined by using the same argument twice $\endgroup$
    – carlo
    Commented Sep 12, 2020 at 11:46
  • $\begingroup$ @carlo even if we consider other entity to be 0? $\endgroup$ Commented Sep 12, 2020 at 11:56
  • $\begingroup$ wait... $x'$ can be any vector? I thought it was $x$ translated $\endgroup$
    – carlo
    Commented Sep 12, 2020 at 12:02

1 Answer 1

2
$\begingroup$

$k(\cdot, \cdot)$, depending on the first term, may be any positive number, and is independent on the second term, hence kernel matrix $K$ can be any matrix having equal positive values whitin rows.

Take a matrix $K$ all equal to 0 except for the first row, and make it equal to 1. Also take a vector $c$ equal to 1 in all its value except for $c_1=-1$: $c^T K c < 0$ as long as the dimension of the form is greater than 2.

In conclusion, $k$ is not positive definite.

$\endgroup$
0

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.