Say that a $\omega\times \omega$ Hermitian matrix $A$ is positive semidefinite of rank $n$ if there exists a $\omega\times n$ complex matrix $B$ such that $A=B B^\dagger$ where $^\dagger$ denotes the conjugate transpose.


Let $f$ be a real-analytic function that converges in a neighbourhood of the origin in ${\mathbb{C}}$. Develop $f=\sum\_{i,j=0}^\infty c\_{ij} z^i\bar{z}^j$ as a power series in $z$ and $\bar{z}$. Suppose that $f$ is real-valued so that $(c\_{ij})$ is a $\omega\times \omega$ Hermitian matrix. 


Suppose one shows that for any $(a\_k)\in l^2({\mathbb{C}})$, the sum $\sum\_{i,j=0}^\infty c\_{ij} a\_i\bar{a\_j}$ is nonnegative. Does this imply that $(c\_{ij})$ is positive semidefinite of some rank $n\le \omega$?


This characterization of positive semidefiniteness is valid for finite rank Hermitian matrix. But I'm unsure about the convergence conditions in the infinite rank case.



Positive definite Hermitian matrices of countable rank
I'm puzzled as to why you would expect this ... no, let $(c\_{ij})$ be the identity matrix ($c\_{ii} = 1$, $c\_{ij} = 0$ for $i \neq j$). That is positive semidefinite but it has infinite rank.


There's a well-developed theory of positive semidefiniteness for operators on $l^2$. If the operator is bounded, then $\langle Av,v\rangle \geq 0$ for all $v \in l^2$ does imply that
$A = B^\*B$ for some bounded operator $B$.


Edit: I just noticed that you asked for "rank $n \leq \omega$". If you weren't conjecturing finite rank, the question makes more sense. But if $n$ could equal $\omega$, then without boundedness assumptions expressions like $B^\*B$ don't make sense.


So I think the answer I wrote above that ignores the red herring of "rank" may be what you want. If you're really interested in unbounded operators we can talk about them too.


