Let $PD\_{n}$ be the cone of positive definite $n \times n$ real matrices and let $B$ be the unit sphere in $n \times n$ dimensions. What is the volume of $PD\_{n} \cap B$?


EDIT: Let's assume that $B$ is the unit sphere w.r.t the operator norm: $||A||=\sup\_{||x||=1}{||Ax||}$.



What it is the volume of the unit ball section of the cone of positive definite matrices?
Using the operator norm, as you have defined it, the fraction of the unit ball in real symmetric $n$-by-$n$ matrices that consists of positive definite matrices is $2^{-n(n+1)/2}$. Thus, this fraction shrinks very quickly as $n$ increases.


**Added comment 1**: Actually, I just realized that you don't need to do the calculation below; the ratio is obvious anyway: Consider the map $\Phi:B\_n\to \overline{PD\_n\cap B\_n}$ defined by $\Phi(A) = \tfrac12(A+I\_n)$. This affine invertible map identifies $B\_n$ with $\overline{PD\_n\cap B\_n}$, which has the same volume as $PD\_n\cap B\_n$. Obviously, it multiplies volumes by a factor of $2^{-n(n+1)/2}$, since $\tfrac12n(n{+}1)$ is the dimension of the ambient space of symmetric $n$-by-$n$ matrices. QED.


**Added comment 2**: I note that the OP never did say that he wanted to concentrate on *symmetric* positive definite matrices, which seems bizarre to me, but maybe he really does mean this. I looked online and found that some sources say that a real $n$-by-$n$ matrix $A$ is positive definite if $x\cdot Ax \ge 0$ for all $x\in\mathbb{R}^n$, with equality if and only if $x = 0$. (Of course, this is a condition only on the symmetric part of $A$.) If one does adopt this convention and considers the problem as being posed on the set of all $n$-by-$n$ real matrices with the operator norm, then the problem becomes much harder, and the above scaling trick doesn't work. I don't know the answer in this case for general $n$, but, for $n=2$, there is a fortuitous simplification: If 
$$
A = \begin{pmatrix}a^1\_1&a^1\_2\\ a^2\_1&a^2\_2\end{pmatrix}
 = \begin{pmatrix}r+a&b+s\\ b-s& r-a\end{pmatrix},
$$
then, for the operator norm as defined by the OP, one has a nice formula
$$
\| A\| = \sqrt{r^2+s^2} + \sqrt{a^2+b^2}
$$
and $A$ is positive definite if and only if $r>\sqrt{a^2+b^2}$. (For higher values of $n$, these are not nice at all.)


Now, it's just a calculus problem. The volume of the unit ball turns out to be
$V = 2\pi^2/3$, and the volume of the positive definite matrices turns out to be
$$
V\_+ = 4\pi(53-15\pi)/45
$$ 
with the resulting ratio being 
$$
V\_+/V \approx 0.2493898621\ .
$$
Thus, nearly one-quarter of the unit ball in this case consists of positive definite matrices.




---


Here is how the calculation goes: First, we need to parametrize this unit ball in some way. The easiest way to do this is to write
$$
A = g R g^{-1}
$$
where $g$ is an element of the orthogonal group $\mathrm{O}(n)$ and $R=\mathrm{diag}(r\_1,\ldots,r\_n)$ where the the $r\_i$ belong either to the domain
$$
B = \{(r\_1,\ldots,r\_n)\ |\ 1\ge r\_1\ge r\_2\ge\cdots\ge r\_n\ge-1\}
$$
(in which case $A$ belongs to the unit ball in the operator norm) or 
$$
B\_+ = \{(r\_1,\ldots,r\_n)\ |\ 1\ge r\_1\ge r\_2\ge\cdots\ge r\_n\ge0\}
$$
(in which case $A$ belongs to the part of the unit ball that consists of positive semidefinite symmetric matrices). (Actually, this parametrization is generically $2^n$-to-$1$ from $B\times\mathrm{O}(n)$ to the unit ball in the operator norm and $2^n$-to-$1$ from $B\_+\times\mathrm{O}(n)$ to the postive semi-definite matrices within the unit ball, but that won't matter to the calculation of the ratio of volumes.)


Now, a straightforward calculation using this parametrization shows that the two volumes we want to compute are $\Omega\_n$ (the volume of $\mathrm{O}(n)$) divided by $2^n$ times the integrals over $B$ and $B\_+$ of the following $n$-form:
$$
\phi = \left(\prod\_{i<j}(r\_i-r\_j)\right)\,
\mathrm{d}r\_1\wedge\mathrm{d}r\_2\wedge\cdots\wedge \mathrm{d}r\_n
$$
Fortunately, it is not necessary to compute this integral over either domain because, if you look at the mapping $f:B\_+\to B$ defined by 
$$
f(r\_1,r\_2,\ldots,r\_n) = (2r\_1{-}1,2r\_2{-}1,\ldots,2r\_n{-}1),
$$
you see immediately that 
$$
f^\ast\phi = 2^{n(n+1)/2}\,\phi
$$
Thus, the integral of $\phi$ over $B\_+$ divided by the integral of $\phi$ over $B$ is 
$ 2^{-n(n+1)/2}$, as claimed.


