The matrix norm for an nbyn matrix A is defined as A=max(Ax) where x ranges over all vectors with x=1, and the norm on the vectors in R^n is the usual Euclidean one. This is also called the induced (matrix) norm, the operator norm, or the spectral norm. The unit ball of matrices under this norm can be considered as a subset of R^(n^2). What is the Euclidean volume of this set? I'd be interested in the answer even in just the 2by2 case.
Building on the nice answer of Guillaume: The integral $$ \int_{[1,1]^n} \prod\_{i < j} x_i^2  x_j^2  dx_1\dots dx_n $$ has the closedform evaluation $$ 4^n \prod_{k \leq n} \binom{2k}{k}^{1}.$$ This basically follows from the evaluation of the Selberg beta integral S_{n}(1/2,1,1/2). Combined with modding out by a typo, we now arrive at the following product formula for the volume of the unit ball of nxn matrices in the matrix norm: $$ n! \prod_{k\leq n} \frac{ \pi^k }{ ((k/2)! \binom{2k}{k})} .$$ In particular, we have:



The volume of the unit ball for the spectral norm in nxn real matrices is given by the formula $$ c_n \int\limits_{[1,1]^n} \prod_{i < j} x_i^2x_j^2 dx_1\dots dx_n $$ where $c_n = n! 4^{n} \prod_{k=1}^n v_k^2$ and $v_k=\pi^{k/2}/\Gamma(1+k/2)$ is the volume of the unit ball in R^n. A much more general formula for calculating all kind of similar quantities appears e.g. here (Lemma 1). The proof is by applying the SVD decomposition as a change of variables. The first values are
There might be a closed formula for the integral above. Edit : such a formula appears in Armin's post below !! 


Concerning the 2x2 case: As Mike points out, you can write down an explicit formula for the norm of the matrix {{a,b},{c,d}}. It takes a good while but Mathematica can then compute the volume you're asking for.
Its answer is: 2π^{2}/3. For comparison: the volume of the Euclidean ball in R^{4} is π^{2}/2 (which contradicts Mike's final statement that the matrix norm ball sits inside the Euclidean one). 


Not that this is too helpful, but in the case of a 2 x 2 matrix A (with diagonal entries a and d and off diagonal entries b and c all real) the norm for the matrix is given by the formula $\frac{1}{2}(a^{2} + b^{2} + c^{2} + d^{2} + \sqrt{(a^{2} + b^{2} + c^{2} + d^{2})^{2}  4D})$ where $D = det(A^{*}A)$. It is a pretty ugly region but at least it can be computed in terms of a, b, c, and d and this unit ball will sit inside the Euclidean ball in R^{4}. 


Yes, O(n) is the n(n1)/2 dimensional space of orthogonal n by n matrices. Vol(O(n)) is its volume. The integrand in the answer is simply the Jacobian of the singular value decomposition, {s_ i} is just the ordered set of the singular value and the integration is performed on the subset bounded by 1. I may just have missed a factor of 1/2^n because of the sign ambiguity in the svd singular values 


I worked out the answer for the 2 by 2 case as well. First, when dealing with 2 by 2 matrices in general, a convenient variable change is:
Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2. And the determinant (adbc) = (1/2)*(x^2+y^2w^2z^2). (Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as R^4 is a cone over the Clifford torus, since x^2+y^2 = w^2+z^2 on a sphere x^2+y^2+w^2+z^2=r^2 implies x^2+y^2 = r^2/2 and w^2+z^2 = r^2/2, which are scaled equations for a flat torus) Let r1^2 = x^2+y^2, r2^2 = w^2+z^2. (These are radial coordinates of two cylindrical coordinate systems filling out 4space). Then the norm squared is:
When this is less than one, this corresponds to the region plotted below: Note that each point in the r1,r2 picture corresponds to a different "torus", x^2+y^2=r1^2, w^2+z^2=r2^2. We can now integrate over the shaded in region, \int_{region} dw dx dy dz. This 4D integral can be reduced to 2D using r1 and r2, since dx dy = 2π r1 dr1, dw dz = 2π r2 dr2:
Now, note that we can rewrite r2 in terms of r1. In particular, after some manipulation of our norm, the shaded in region is defined by r2^2 ≤ 22\sqrt{2}r1+r1^2=(\sqrt{2}r1)^2. Hence r2≤ \sqrt{2}r1, and we can evaluate the r2 integral:
This yields 2π^2/3, as Armin found. 


I had a go at this question, but the method I tried here doesn't quite work out. It does reduce it to upper triangular matrices, although that doesn't seem to be a lot of help for general n. Let your volume be V. By scaling, the volume of the set {A≤K} is VK^{n2}. Now let M be a matrix whose entries are independent normal random variables with mean 0 variance 1. From the density function of the normal distribution, this gives P(M≤K)~(2π)^{n2/2}VK^{n2} in the limit of small K. I'll now calculate this expression in an alternative way. Use the M=QR decomposition, where Q is orthogonal and R is upper triangular, with diagonal elements λ_{n}, λ_{n1},…λ_{1}, which are the eigenvalues of R. This can be done in such a way that λ_{k}^{2} has the χ^{2}_{k}distribution (a quick google search gives this but there's probably better references). The upper triangular parts of R have the standard normal density. We need to calculate R. I was originally thinking that this is the max eigenvalue, but it's not quite that simple. 


By means of singular value decomposition, I think that the general answer for a real n by n matrix should be: Required volume = $$ {\rm vol}(O(n))^2 \int\limits_{0\leq s_n \leq s_{n1}\leq \dots s_1\leq 1}\prod_{i < j < n} (s_ i^2s_ j^2).$$ O(n) is the ndimensional orthogonal group 

