Showing that $k(x,x') = (x^Tx' -1)^2 $ is not a valid kernel

Question

I would like to show that $k(x,x') = (x^Tx' -1)^2 $ is not a valid kernel.

Is it possible to show the non-positive semi definiteness by considering a special subspace of the feature space?

In other words;

Can I say that for a particular feature space I only consider x,y such that $<\phi(x), \phi(y)> > 1$ when $x \neq y $ and $1$ otherwise (i.e. have $1$ on the diagonal)?

Then for a $2\times2$ matrix, or the second principal minor of any larger symmetric matrix I would have a negative determinant and thus some negative eigenvalues somewhere.

Feels like cheating, apologies if this is really silly, I can't poke a hole in the reasoning yet. Thanks for explaining why it can work or why it is wrong.

Answer 1

I think a kernel function should meet this condition: k(x,z)=∅(x).∅(z)
(x,z)=[(∑x_i.z_i-1)(∑x_j.z_j-1)= ∑∑x_i.z_i.x_j.z_j-2∑x_i.z_i +1] Because of having minus in:-2∑x_i.z_i ,we cannot find a feature map ∅ have you solved this problem?