Euclidean volume of the unit ball of matrices under the matrix norm

The volume of the unit ball for the spectral norm in nxn real matrices is given by the formula

$$ c_n \int\limits_{[-1,1]^n} \prod_{i < j} |x_i^2-x_j^2| dx_1\dots dx_n $$

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

• 2/3 π² for 2x2 matrices
• 8/45 π⁴ for 3x3 matrices
• 4/1575 π⁸ for 4x4 matrices ...

There might be a closed formula for the integral above. Edit : such a formula appears in Armin's post below !!

Building on the nice answer of Guillaume: The integral

$$ \int_{[-1,1]^n} \prod_{i < j} \left| x_i^2 - x_j^2 \right| \, dx_1 \dots dx_n $$

has the closed-form 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:

• 2/3 π² for n=2
• 8/45 π⁴ for n=3
• 4/1575 π⁸ for n=4

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.

Integrate[If[a^2 + b^2 + c^2 + d^2
 + Sqrt[((b+c)^2 + (a-d)^2) ((b-c)^2 + (a+d)^2)] <= 2, 1, 0],
{a, -1, 1}, {b, -1, 1}, {c, -1, 1}, {d, -1, 1}]

Its answer is: 2π²/3.

For comparison: the volume of the Euclidean ball in R⁴ is π²/2 (which contradicts Mike's final statement that the matrix norm ball sits inside the Euclidean one).