TITLE: The probability for a symmetric matrix to be positive definite
QUESTION [39 upvotes]: Let me give a reasonable model for the question in the title. In ${\rm Sym}_n({\mathbb R})$, the positive definite matrices form a convex cone $S_n^+$. The probability I have in mind is the ratio $p_n=\theta_n/\omega_n$, where $\theta_n$ is the solid angle of $\Lambda_n$, and $\omega_n$ is the solid angle of the whole space ${\rm Sym}_n$ (the area of the unit sphere of dimension $N-1$ where $N=\frac{n(n+1)}2$). These definitions are relative to the Euclidian norm $\|M\|=\sqrt{{\rm Tr}(M^2)}$ ; this is the most natural among Euclidian norms, because it is invariant under unitary conjugation.

Because $S_2^+$ is a circular cone, I could compute $p_2=\frac{2-\sqrt2}4\sim0.146$ . Is there a known close formula for $p_n$? If not, is there a known asymptotics?

More generally, we may define open convex cones 
$$\Lambda_n^0\subset\Lambda_n^1\subset\cdots\subset\Lambda_n^{n-1}$$
in the following way: $M\mapsto\det M$ is a homogeneous polynomial, hyperbolic in the direction of the identity matrix $I_n$. Thus its successive derivatives in this direction are hyperbolic too. The $k$th derivative defines a "future cone" $\Lambda_n^k$, these cones being nested. For instance, $\Lambda_n^0=S_n^+$. It turns out that this derivative is, up to a constant,  $\sigma_{n-k}(\vec\lambda)$, where $\sigma_j$ is the $j$th elementary symmetric polynomial and $\vec\lambda$ the spectrum of $M$. Therefore $\Lambda_n^k$ is defined by the inequalities
$$\sigma_1(\vec\lambda)\ge0,\ldots,\sigma_{n-k}(\vec\lambda)\ge0.$$
For instance, $\Lambda_n^{n-1}$ is the half-space defined by ${\rm Tr}M\ge0$.
Let us define again $p_{n,k}$ the probability for $M\in{\rm Sym}_n$ to belong to $\Lambda_n^k$. Thus $p_{n,0}=p_n$ and $p_{n,n-1}=\frac12$. 

What is the distribution of $(p_{n,0},\ldots,p_{n,n-1})$, asymptotically as $n\rightarrow+\infty$?

Edit. As Mikaël mentionned, it is equivalent, and easier for calculations, to consider the standard Gaussian measure (GOE) over ${\bf Sym}_n$.

REPLY [9 votes]: I came across this question while preparing a talk on this paper. There we gave the explicit formula for $p_n$ (in the GOE) that the other answers anticipated could be found by random matrix methods.
The formula is given as follows:
First, let $n\geq 1$ be any integer, and define $n':=2 \lceil n/2 \rceil$.
When real symmetric matrices are chosen according to the
$n$-dimensional Gaussian Orthogonal Ensemble, the probability of
positive definiteness is given by:
\begin{equation}p_n=\frac{{\mathrm{
      Pf}}(A)}{2^{n(n+3)/4}\prod_{m=1}^n\Gamma(\frac m
    2)},\label{theorem3}
\end{equation}
where $A$ is the $n'\times n'$ skew-symmetric matrix whose
$(i,j)$-entry $a_{ij}$ is given for $i<j$ by:
\begin{align}\label{pfaffianformula}
a_{ij} = \begin{cases}
\,2^{i+j-2}\Gamma(\frac{i+j}{2})\left(\beta_{\frac{1}{2}}(\frac{i}{2},\frac{j}{2})- \beta_{\frac{1}{2}}(\frac{j}{2},\frac{i}{2})\right) & \mbox{if }~ i<j\leq n, \\[.075in]
\,2^{i-1}\Gamma(\frac{i}{2})  & \mbox{if }~ i<j=n+1.\\[.075in]
\end{cases} 
\end{align}
$($Note that the second case arises only when $n$ is odd.$)$ 
Here $\mathrm{Pf}$ is the Pfaffian, $\Gamma$ denotes
the usual gamma function $\Gamma(s)=\int_{0}^\infty x^{s-1}e^{-x} \mathrm{d}
x$, and $\beta_{t}$ is the usual incomplete beta function $\beta_{t}(i,j)
= \int_0^{t} x^{i-1}(1-x)^{j-1} \mathrm{d} x.$ This formula follows from de Bruijn's identity for evaluating determinantal integrals.
As for the asymptotics, the leading term of $\log p_n$ is indeed $-\frac{\log(3)}{4} n^2$ as noted in Mikael's answer. If you are interested in the leading order asymptotic for $p_n$ it can be found from the results of Borot, Eynard, Majumdar and Nadal. Specialising to the GOE they give:
\begin{align}
p_n \sim \exp\left(-\frac{\log3}{4} n^2 - \frac{1}{2}\log\left(1+\frac{2}{\sqrt{3}}\right)n - \frac{1}{24}\log n + \alpha_1 \right)(1+o(1)).\label{asympexp}
\end{align}
The constant $\alpha_1$ is given by:
\begin{align}
\alpha_1= \frac{\zeta'(-1)}{2} - \frac{\log(2)}{12} -\frac{\log(3)}{16}+ \frac{1}{4}\log\left(1+\frac{2}{\sqrt{3}}\right)\simeq-0.0172,
\end{align}
where $\zeta(z)$ is the Riemann zeta function. 
Edit: The exact formula lends itself easily to numerics. I have added a plot of $\log(p_n)$ for $1\leq n \leq 400.$ Fitting the data for $201\leq n \leq 400$ to $a n^2 + b n + c\log n +d$ we find $$−0.27465n^2 - 0.38382n -0.041504 \log n -0.018061.$$ A numerical approximation for the asymptotic formula is $$−0.27465n^2 - 0.38382n -0.041667 \log n -0.017223,$$ so we see good agreement.