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

1$\begingroup$ The bilinear form should still be positive definite. $\endgroup$ – S. Carnahan♦ Oct 12 '09 at 2:54

$\begingroup$ Better in what sense? $\endgroup$ – Felix Goldberg May 25 '12 at 19:23
Two possible answers:
Standard jargon is SPD (for "symmetric positivedefinite").
This isn't exactly a "name," but the n x n symmetric positivedefinite matrices are exactly those matrices A such that the bilinear function (x, y) > y^{T}Ax defines an inner product on R^{n}. 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 R^{n}.
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.
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".

$\begingroup$ Yes, I was going to post that as well. $\endgroup$ – Ilya Nikokoshev Oct 23 '09 at 19:18
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.

1$\begingroup$ And in words, "the positive part of $M_n({\bf R})$". $\endgroup$ – Nik Weaver Jul 7 '13 at 3:16

1$\begingroup$ There is no standard symbol for it, so, whatever the OP decides to use, they'd better define it in their paper. $\endgroup$ – Federico Poloni Jul 7 '13 at 7:20
For starters, since they're real I'd say symmetric instead of selfadjoint.
It is often usefull to know that this set can be identified with the set of nonsingulat covariance matrices of random vectors with values in $\mathbb(R)^n$.