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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.

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up vote 2 down vote accepted

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.

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I think I should be using bounded operators instead. Thanks for pointing that out. – user2529 Oct 9 '12 at 5:08
This should be in Reed and Simon, Functional Analysis vol. 1 or in Conway, A Course in Functional Analysis. – Nik Weaver Oct 9 '12 at 5:44

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