# Intrinsic measure on symmetric semi positive definite matrices?

Consider real matrices. Is there an intrinsic measure on (normalized)symmetric semi positive definite(SPD) matrices? By intrinsic, I mean some measure derived from Haar measure of a group, since a SPD matrix can be generate from $M \in GL(n)$ as $MM^T$ or from anti-symmetic matrices by exponential map. Another possible connection to $O(n)$ is through Schur decomposition or so. Since SPD is not compact, we can normalize to "correlation coefficient" matrices. Let $D^2$ be the diagonal part of SPD $A$. We normalize $A$ as $D^{-1}AD^{-1}$, and it will have all 1's on diagonal.

Thanks!

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What you call SPD is usually called PSD. – Igor Rivin Dec 9 '11 at 23:15
You can define a Riemannian metric $g$ on the dense subspace of positive-definite symmetric matrices by $g(A,B;M) = tr(M^{-1}AM^{-1} {}^tB)$ (ish). This metric is invariant under the transitive action by $GL(n)$, so it gives the subset the structure of a symmetric space. The boundary has measure zero w/r/t the volume form defined by this metric, so it should give a nice measure on your space. This is probably in Helgason somewhere. – Gunnar Þór Magnússon Dec 10 '11 at 15:33

Your space is the symmetric space for $GL(n, \mathbb{R}),$ with all that entails, or, if you prefer, the cone over the symmetric space for $SL(n, \mathbb{R}).$ See, for example, http://math.berkeley.edu/~reb/courses/261/18.pdf (but there is a million other references, including the venerable Helgason).
The non-definite matrix will from a measure zero set. Then $GL(n,\mathbb{R})/O(n,\mathbb{R})$ form a dense subspace in it.