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This is also the space of real, symmetric bilinear forms in $\Bbb R^n$.

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    $\begingroup$ The bilinear form should still be positive definite. $\endgroup$
    – S. Carnahan
    Oct 12, 2009 at 2:54
  • $\begingroup$ Better in what sense? $\endgroup$ May 25, 2012 at 19:23

6 Answers 6

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Two possible answers:

  • Standard jargon is SPD (for "symmetric positive-definite").

  • This isn't exactly a "name," but the n x n symmetric positive-definite matrices are exactly those matrices A such that the bilinear function (x, y) -> yTAx defines an inner product on Rn. Conversely, every bilinear function is of that form for some A, so with some abuse of terminology, you could equate the set of those matrices with the set of inner products on Rn.

There are many other ways to characterize SPD matrices, but that's the only one I can think of at the moment that can be summarized as a single noun phrase.

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Note that this space is not a vector space, but is a convex cone in the vector space of nxn matrices (it is closed under addition and multiplication by positive scalars). Hence people sometimes refer to the "positive semidefinite cone".

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  • $\begingroup$ Yes, I was going to post that as well. $\endgroup$ Oct 23, 2009 at 19:18
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This is the symmetric space of GL_n(R)

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How about $M_n(\mathbb{R})^+$? I have seen $S^+$ or $S_+$ used to denote the set of positive linear transformations in a set $S$ of linear transformations on an inner product space, but this was in the context of operator algebras.

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    $\begingroup$ And in words, "the positive part of $M_n({\bf R})$". $\endgroup$
    – Nik Weaver
    Jul 7, 2013 at 3:16
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    $\begingroup$ There is no standard symbol for it, so, whatever the OP decides to use, they'd better define it in their paper. $\endgroup$ Jul 7, 2013 at 7:20
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For starters, since they're real I'd say symmetric instead of self-adjoint.

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It is often usefull to know that this set can be identified with the set of non-singulat covariance matrices of random vectors with values in $\mathbb(R)^n$.

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