Suppose that two random vectors $x$ and $y$ are uniformly distributed on unit sphere $S_{n-1}$. Is it possible to show that the Kronecker product of $x$ and $y$ is uniformly distributed on a subset of the unit sphere $S_{n-1}\otimes S_{n-1}$?


Yes. This becomes clear upon working through the definitions.

The Kronecker Product

The unit sphere $S^{n-1}$ is the set of unit vectors in the Euclidean space $E^n = \left(\mathbb{R}^n, ||\cdot ||\right)$ where $$||(x_1,x_2,\ldots,x_n)^\prime|| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}$$ is the Euclidean norm.

The "Kronecker product" is the usual tensor product. There are several ways to think about it and compute with it. One is to define it as an $n\times n$ matrix

$$x \otimes y = x\, y^\prime = \pmatrix{x_1y_1 & x_1y_2 & \cdots & x_1y_n \\ x_2y_1 & x_2y_2 & \cdots & x_2y_n \\ \vdots & \vdots & \cdots & \vdots \\ x_ny_1 & x_ny_2 & \cdots & x_ny_n} \in \operatorname{Mat}\left(\mathbb{R}^n, \mathbb{R}^n\right).$$

Another equivalent way unravels the components of this matrix into a vector with $n^2$ components, allowing us to view $x\otimes y$ as an element of $\mathbb{R}^{n^2}$. Notice that the Euclidean metric for $\mathbb{R}^{n^2}$ can be written

$$||Z||^2 = \operatorname{Tr}( Z Z^\prime ).\tag{1}$$

Both are ways of writing the sum of squares of all $n^2$ components.

The Kronecker product is compatible with the Euclidean metrics on $E^n$ and $E^{n^2}$ in the sense that

$$||x \otimes y||^2 = ||x||^2\,||y||^2.$$

This is easily demonstrated, because the left hand side is defined as the sum of the squares of all the $x_i y_j$ while the right hand side is the product of their sums of squares. Just expand that product:

$$||x \otimes y||^2 = \sum_i\sum_j (x_iy_j)^2 = \left(\sum_i x_i^2\right)\left(\sum_j y_j^2\right) = ||x||^2\,||y||^2.$$

In particular, when both $x$ and $y$ have unit length, $x \otimes y$ has unit length. Therefore

$$S^{n-1}\otimes S^{n-1} \subset S^{n^2-1}.$$

Uniform distributions

The Euclidean spheres inherit a measure from the usual measure on the Euclidean spaces (the Lebesgue measure, ultimately determined by the Euclidean distance). That measure is preserved by any isometry of a sphere, because (by definition) an isometry preserves distances and the measure ultimately is determined by distances. The group of isometries of the unit sphere in $\mathbb{R}^m$ is denoted $O(m)$, the orthogonal group. It is a classical result, and straightforward to show, that it consists of the linear transformations represented by all $m\times m$ matrices $P$ for which $PP^\prime = P^\prime P = \mathbb{I}_m$.

The orthogonal group $O(n)$ acts transitively on $S^{n-1}$. (Here's a proof Euclid might have made: pick any two distinct points $x$ and $y$ on the sphere. Draw the line segment $xy$ between them. It determines a unique hyperplane perpendicular to $xy$ passing through the midpoint of $xy$. The reflection in that hyperplane maps $S^{n-1}$ to itself and preserves all distances, whence it is in $O(n)$. That reflection swaps $x$ and $y$, showing there exists an orthogonal transformation sending $x$ to $y$.)

To say that "vectors are uniformly distributed on $S^{n-1}$" means that the distribution is invariant under a transitive group of isometries like $O(n)$.

Here's the punchline: the "hypertorus" $S^{n-1}\otimes S^{n-1} \subset S^{n^2-1}$ enjoys a transitive group of isometries isomorphic to a quotient of $O(n)\times O(n)$. Indeed, given an arbitrary $x\otimes y\in S^{n-1}$ and another $x_2\otimes y_2\in S^{n-1}$, pick $P,Q\in O(n)$ for which $Px=x_2$ and $Qy=y_2$. Let $Z=(z_{ij})$ be any $n\times n$ matrix and define

$$(P\otimes Q)(Z) = PZQ^\prime.\tag{2}$$

This is an isometry because, using the formula $(1)$,

$$\eqalign{ ||(P \otimes Q) (Z)||^2 &= \operatorname{Tr}\left((PZQ^\prime)(PZQ^\prime)^\prime\right) = \operatorname{Tr}\left(PZ Z^\prime P^\prime\right) = \operatorname{Tr}\left(PP^\prime Z Z^\prime \right) = \operatorname{Tr}\left(ZZ^\prime \right) \\ &= ||Z||^2.}$$

These steps exploited the orthogonality of $P$ and $Q$ and the fact that $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$ for any square matrices $A$ and $B$.

Because the isometries of $S^{n-1}$ induce a transitive group of isometries of $S^{n-1}\otimes S^{n-1}$ via $(2)$, the uniform (that is, group-invariant) distribution on $S^{n-1}$ is mapped to a uniform distribution on $S^{n-1}\otimes S^{n-1}$, QED.

  • $\begingroup$ @ whuber In you proof, there exist some confusions as follows: 1. the right hand of equation (2) is a matrix of $n$-by-$n$, how can we understand the left hand of the equation (2)? $\endgroup$ Sep 10 '16 at 7:17
  • $\begingroup$ 2. Why you need to choose $x_1\otimes y_1$ and $x_2\otimes y_2$? Because in the rest of the process, you do not use these two vectors. $\endgroup$ Sep 10 '16 at 7:24
  • $\begingroup$ @Mao-linChe Equation $(2)$ defines its left hand side. I don't know what you refer to by "$x_1\otimes y_1$" because that does not appear in my answer. $\endgroup$
    – whuber
    Sep 10 '16 at 18:18
  • $\begingroup$ @whuber I am soory. I mean why you need to define $x\otimes y$ and $x_1\otimes y_1$. Does the matrix $Z$ depend on these vectors? $\endgroup$ Sep 12 '16 at 1:41
  • $\begingroup$ @Mao-linChe Those four vectors partially determine the orthogonal matrices $P$ and $Q$, as explained immediately after those vectors are introduced. $Z$ does not depend them: that's what "any" means in the phrase "let $Z$ be any $n\times n$ matrix...". $\endgroup$
    – whuber
    Sep 12 '16 at 14:16

As long as $x$ and $y$ are independent, then what you say is true.

It's easiest to think about angles rather than points in space.

For example, in the 2-D sphere in $\mathbb R^3$, we can generate points uniformly on the surface of that sphere by sampling $\theta_1$ and $\theta_2$ independently from $U[0,2\pi]$ and taking those as zenith and azimuth respectively.

Thus, we can think of $x$ as $(\theta_1,...,\theta_{n-1}) \sim U[0,2\pi]^{n-1}$ and $y$ as $(\phi_1,...,\phi_{n-1}) \sim U[0,2\pi]^{n-1}$. Their KP is then $(\theta_1,...,\theta_{n-1},\phi_1,...,\phi_{n-1})$ and it's clear from the construction that that follows a uniform distribution $U[0,2\pi]^{2n-2}$ as well.

  • 1
    $\begingroup$ This argument is fundamentally flawed. The zenith runs only from $-\pi/2$ to $\pi/2$ and a uniform distribution on the sphere induces a non-uniform distribution of the zenith. (In $\mathbb{R}^3$, its sine has a uniform distribution.) See stats.stackexchange.com/questions/7977 for details. $\endgroup$
    – whuber
    Aug 31 '16 at 14:13
  • $\begingroup$ Are we dealing with the surface of the sphere or the whole volume? I was assuming the former. $\endgroup$
    – JDL
    Aug 31 '16 at 14:21
  • $\begingroup$ That's correct: $S_{n-1}$ is implicitly identified with the set of unit vectors in $\mathbb{R}^n$. $\endgroup$
    – whuber
    Aug 31 '16 at 14:22
  • $\begingroup$ @whuber; you're right! D'oh. I still think the question is more or less trivially true though. The kronecker product of two independent uniform distributions can only ever be uniform on the product set. $\endgroup$
    – JDL
    Aug 31 '16 at 14:25
  • 1
    $\begingroup$ @JDL have you tried hist(runif(5000) %x% runif(5000)) ..? $\endgroup$
    – Tim
    Aug 31 '16 at 14:35

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