# Is the zero a valid kernel?

It seems to me that if $c\geq0$ then $k(x,y)=c$ is a valid kernel since no rules are broken.

However, one of the rules of constructing a kernel disagrees with this. The rule is $k(x,y)=ck'(x,y)$, where it must be true that $c>0$ (this rule is taken from Bishop, Pattern Recognition And Machine Learning). Why can't $c$ be 0? The zero matrix is symmetric and positive semidefinite, so isn't it true that $k(x,y)=0$ should be a valid kernel?

• To what "rules" do you refer? This matters, because perhaps your rules differ from those of other people. Some, for instance, require that a kernel integrate to unity. None of your examples has this property.
– whuber
Oct 30, 2016 at 21:04
• I'm also mildly confused, since the constant zero is a valid kernel by Mercer's theorem, yet zero is not a proper inner product since it does not meet the requirement that <x, x> is greater than zero. Jul 13, 2020 at 2:40

$k(x, y) = 0$ is a positive semidefinite kernel, as is easy to check from the definition (as you've done). It's not, however, a very useful kernel.
Bishop's list of the ways to construct one kernel from another is not an exhaustive list, as you've discovered. They're not necessary for being a kernel; they're just a set of conditions where you know how to construct one kernel from another. His rule would still be valid if he replaced the condition $c > 0$ with $c \ge 0$, but I suppose he didn't bother because the difference can only get you the zero kernel, which is useless anyway.