If a kernel matrix is positive semi-definite, how can I prove there exists a $\phi$ s.t $k(x,y) = \phi(x).\phi(y)$
My method: Every real symmetric matrix can be diagonalized, so we can write: $$ K = PDP^T = \begin{bmatrix} \vert & \dots &\vert \\ p_1 & \dots & p_m \\ \vert & \dots &\vert \end{bmatrix} \begin{bmatrix} d_{1} & & \\ & \ddots & \\ & & d_{m} \end{bmatrix} \begin{bmatrix} \text{---} & p_1^T & \text{---} \\ \dots & \dots & \dots \\ \text{---} & p_m^T & \text{---} \end{bmatrix} $$
$$ \begin{bmatrix} \vert & \dots &\vert \\ d_1p_1 & \dots & d_mp_m \\ \vert & \dots &\vert \end{bmatrix} \begin{bmatrix} \text{---} & p_1^T & \text{---} \\ \dots & \dots & \dots \\ \text{---} & p_m^T & \text{---} \end{bmatrix} = d_1 p_1p_1^T + \dots + d_m p_m p_m^T = M $$
$$ M_{ij} = d_1 p_i^{(1)}p_j^{(1)} + \dots + d_m p_i^{(m)} p_j^{(m)} = \phi(x_i).\phi(x_j) $$ $$ \phi(x_i) = (\sqrt{d_1}p^{(1)}_i, \sqrt{d_2}p_i^{(2)}, \dots, \sqrt{d_m}p_i^{(m)}) $$ $$ \phi(x_j) = (\sqrt{d_1}p^{(1)}_j, \sqrt{d_2}p_j^{(2)}, \dots, \sqrt{d_m}p_j^{(m)}) $$ What bothers me is that I haven't used the positive-semi definiteness property of the kernel matrix. Is there a mistake in my proof?
We first start by assuming that the matrix $K(i,j)$ is positive definite. Then the form $$\langle z,w\rangle =\sum_{i,j=1}^mK(i,j)z_i\bar{w}_j>0,\ z,w\in \mathbb{C}^m$$ defines an inner product in $\mathbb{C}^m.$ Let $e_1,_2,\ldots, e_m$ denote the standard basis in $\mathbb{C}^m.$ Let $u_1,u_2,\ldots, u_m$ denote the orthonormal basis with respect to this inner product. Thus $$e_i=\sum_{k=1}^m a_{ik}u_k$$ for some coefficients $a_{ik}.$ Moreover by orthonormality we get $$K(i,j)=\langle e_i,e_j\rangle =\sum_{k=1}^ma_{ik}\bar{a}_{jk}=\varphi_i\bar\varphi_j$$ where $\varphi_l=(a_{l1},a_{l2},\ldots, a_{lm}).$
If the matrix is positive semidefinite there are two options. Either considering the quotient space $\mathbb{C}^m/\ker K$ or considering $K^{(n)}=K+{1\over n}I$ and then taking the accumulation point of $\{\varphi^{(n)}_{i}\}_{i=1}^m,$ the elements obtained for $K^{(n)}.$ The latter is possible as by construction $$\sum_{i,j=1}^m|a^{(n)}_{ij}|^2=\sum_{i,j=1}^m|K^{(n)}(i,j)|^2$$ so the elements $a_{ij}^{(n)}$ form a bounded set in $\mathbb{C}^{m\times m}.$