Welcome to MathOverflow

Question

1

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!

flag	asked Dec 9 2011 at 22:54 plus111906146933132709337posts 11●1

What you call SPD is usually called PSD. – Igor Rivin Dec 9 2011 at 23:15

1

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 Magnusson Dec 10 2011 at 15:33

Igor Rivin · Answer 1 · 2011-12-09 23:22:24Z UTC

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).

psd · Answer 2 · 2011-12-09 23:09:11Z UTC

0

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.

answered Dec 9 2011 at 23:09

psd
221●1●5

hoxide · Answer 3 · 2011-12-10 05:58:25Z UTC

0

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.

answered Dec 10 2011 at 5:58

hoxide
117●4

Intrinsic measure on symmetric semi positive definite matrices?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.