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

The matrix norm for an n-by-n 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 2-by-2 case.

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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 closed-form evaluation

$$4^n \prod_{k \leq n} \binom{2k}{k}^{-1}.$$

This basically follows from the evaluation of the Selberg beta integral Sn(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
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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 !!

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 When using your formula I get 1/3 Pi^2 for 2x2 matrices and the values for n=3,4 are different from the ones you give as well. Is there a small typo somewhere (or am I just messing up the calculation)? – Armin Straub Oct 22 2009 at 20:20 I doubled-checked and the general formula seems corrects (anyway you can derive it from the paper I quoted). But you are right, there was a typo for n=4 (now corrected). By the way, you should find c_2=pi^2/4 ; c_3=pi^4/9 ; c_4=pi^8/144. – Guillaume Aubrun Oct 23 2009 at 21:47 Looking at the paper I found the typo: c_n = n! 4^{-n} ... Also, your quite right; the integral does have a nice closed form coming from writing it as a Selberg integral. I put details into a new answer. – Armin Straub Oct 28 2009 at 22:10 Oops you're right ... – Guillaume Aubrun Oct 28 2009 at 22:48

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}]


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).

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 My apologies, it is actually easy to see that the matrix norm ball does not sit inside the Euclidean one. The identity matrix clearly does the job. – Mike Hartglass Oct 21 2009 at 1:58 Nice example. At least it sits inside the max norm unit ball (filling it out by an ambitious 41% ...). – Armin Straub Oct 21 2009 at 11:12

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}.

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 There should be a square root outside that expression, right? – jc Oct 21 2009 at 0:16 yes, I forgot the square root (or at least the lack of a square root is a typo in Conway's book). – Mike Hartglass Oct 21 2009 at 2:03

Yes, O(n) is the n(n-1)/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

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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:

a->(w+x)/\sqrt{2},d->(w-x)/\sqrt{2},c->(y-z)/\sqrt{2},b->(y+z)/\sqrt{2}.


Then a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2. And the determinant (ad-bc) = (1/2)*(x^2+y^2-w^2-z^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 4-space). Then the norm squared is:

(1/2)*(r1^2+r2^2 + \sqrt{ (r1^2+r2^2)^2 - (r1^2-r2^2)^2 })


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 4-D integral can be reduced to 2D using r1 and r2, since dx dy = 2π r1 dr1, dw dz = 2π r2 dr2:

(4π^2) \int_{region} dr1 dr2 r1 r2


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 ≤ 2-2\sqrt{2}r1+r1^2=(\sqrt{2}-r1)^2. Hence r2≤ \sqrt{2}-r1, and we can evaluate the r2 integral:

(4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 \int_{r2=0}^{\sqrt{2}-r1} r2 dr2
= (4π^2) \int_{r1=0}^\sqrt{2} dr1 r1 (\sqrt{2}-r1})^2/2
= (4π^2) (1/6)


This yields 2π^2/3, as Armin found.

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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.

By scaling, the volume of the set {|A|≤K} is VKn2. 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)-n2/2VKn2 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 λn, λn-1,…λ1, which are the eigenvalues of R. This can be done in such a way that λk2 has the χ2k-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.

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

O(n) is the n-dimensional orthogonal group

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 Is vol(O(n)) the n(n-1)/2-dimensional measure of the set of orthogonal n by n matrices? I would like to see how you came up with this. – Darsh Ranjan Oct 21 2009 at 8:15