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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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    $\begingroup$ What you call SPD is usually called PSD. $\endgroup$
    – Igor Rivin
    Commented Dec 9, 2011 at 23:15
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    $\begingroup$ 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. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Dec 10, 2011 at 15:33

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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).

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  • $\begingroup$ I wonder if this means you'd be able to say something about the following (linked) question about integrating over the space of positive-definite symmetric matrices: stats.stackexchange.com/questions/391538/… $\endgroup$
    – Michael Hardy
    Commented Feb 10, 2019 at 19:14
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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.

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SPD usually means Symmetric Positive Definite, not positive semidefinite. And if you really mean semidefinite then the diagonal won't necessarily have all 1's even after normalization, because it might be singular.

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