1
$\begingroup$

Title basically says it all: Is the space of positive (semi- or not) definite correlation matrices Polish?

As an aside, I'm interested in general comments/references about the space(s).

Edit: For the sake of completeness, a $p\times p$ positive definite correlation matrix $C$ is a real symmetric matrix with ones on the diagonal and x'Cx>0 for any nonzero $x\in \mathbb{R}^p$. I updated the question to include positive semidefinite correlation matrices as well (ie relaxing to $x'Cx\geq0$) because I suppose it's interesting too :)

  • 0
    Descriptive set theory saves the day once more! Huzzah!2011-07-11

1 Answers 1

5

Let me do it this way:

  • The symmetric matrices are closed in all $p \times p$-matrices (obvious), hence they are Polish.
  • The positive definite matrices are an open cone in the symmetric matrices, hence they are Polish as well (open subsets of a Polish space are Polish — this is a bit easier to prove than the fact on $G_{\delta}$'s).
  • Finally, since the functions $f_{i}:A \mapsto a_{ii}$ are continuous, their simultaneous pre-image of $1$ is closed in the positive definite matrices, hence the positive definite correlation matrices are Polish.

The positive semi-definite case is even simpler, as the conditions are all closed.

Finally, I can only recommend working through the first few sections of Kechris, as most of these arguments become rather simple once one gets used to them.

  • 0
    Oooh that does look nice - I've arrived here from studying Bayesian nonparametric statistics, which takes a level of abstract math I never quite got to...2011-07-11