I have the matrix G and I want to show that G is only positive semi definite and not positive definite. We have a vector $\mathbf{z}=(z_1,...z_n)$, along with $n$ data points $\mathbf{x}_1,...\mathbf{x}_n$. $\Phi$ is a map into dimension $d<n$ : $$z^T Gz=\langle \sum_j^n z_j\Phi(x_j)\, \sum_i^n z_i\Phi(x_j)\rangle$$

I'm thinking of finding a none zero vector $\mathbf{z}$ such that the sum inside is 0, thus giving the equality of positive semi definite. How do I find this vector?

  • $\begingroup$ Well, can $n$ vectors in $K^d$ be linearly dependent? (Where $K$ is the scalar field) $\endgroup$ – user251257 Oct 24 '17 at 17:40
  • $\begingroup$ If you set $\Psi = [\Phi(x_1),\ldots,\Phi(x_n)]\in\mathbb R^{d\times n}$, then you have $G = \Psi^*\Psi$. Now argue that the kernel of $\Psi$ cannot be zero. $\endgroup$ – amsmath Oct 24 '17 at 17:43
  • $\begingroup$ @user251257 They can't be linearly independent. How do I use this ? $\endgroup$ – ChuckP Oct 24 '17 at 19:07
  • $\begingroup$ nvm I think I have it. $\endgroup$ – ChuckP Oct 24 '17 at 19:18

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