# If $K$ is Hermitian and positive definite, then $K(z,w) = \sum \psi_n(z) \overline{\psi}_n(z)$ for $\psi_n$ analytic in some region of the plane.

While reading Zeros of Gaussian Analytic Functions and Determinantal Point Processes by Hough et al., I came across the following statement:

Since $K$ is Hermitian and positive definite, we can write $K(z,w) = \sum \psi_n(z) \overline{\psi}_n(w)$, where $\psi_n$ are analytic functions on some domain in the plane.

How does this follow? At first it seemed like an instance of spectral theorem, but there is no linearity involved. For context, $K$ is defined as $$K(z,w) = \mathbb{E} \left(\mathbf{f}(z) \overline{\mathbf{f}}(w) \right)$$

where $\mathbf{f}$ is a Gaussian analytic function; for the unfamiliar, you can think of a Gaussian analytic function as an analytic function chosen randomly. From this definition, it's clear that $K$ isn't an inner product, so I'm not sure where this kind of a statement comes from. The use of the term "positive definite" seems to suggest that $K$ should be thought of as some sort of inner product.

I figured it out, but we need some more assumptions. Importantly, $K$ is analytic w.r.t. $z$ and anti-analytic w.r.t. $w.$ This means that we can write $$K(z,w) = \sum\limits_{k,l \geq 0} a_{k,l} z^k \bar{w}^l.$$
Since $K$ is Hermitian and positive definite, we have that $a_{k,l} = \bar{a}_{l,k}$, and that $a_{k,k} > 0$. Define $Z = (1, z, z^2, \ldots),~A = (a_{k,l})_{k,l = 0}^\infty$ and $W = (1, \bar{w}, \bar{w}^2, \ldots)$. Then we have that $$K(z,w) = ZAW^\dagger.$$
Since $A$ is Hermitian positive definite, there exists $P$ unitary and $D$ diagonal with positive diagonal entries so that $A = PDP^\dagger.$ We thus have $$K(z,w) = ZPD^{1/2}D^{1/2}P^{\dagger}W^\dagger = (ZPD^{1/2})(WPD^{1/2})^\dagger.$$
Define $\psi_n(z)$ to be the $n^{\text{th}}$ row of $ZPD^{1/2}$, and the answer follows.