Measuring centrality and dispersion in directional datasets: the ellipsoidal cone covering approach

Abstract

Consider a finite collection $\{\xi _k\}_{k=1}^p$ of vectors in the space $\mathbb {R}^n$. The $\xi _k$’s are not to be seen as position points but as directions. This work addresses the problem of computing the ellipsoidal cone of minimal volume that contains all the $\xi _k$’s. The volume of an ellipsoidal cone is defined as the usual n-dimensional volume of a certain truncation of the cone. The central axis of the ellipsoidal cone of minimal volume serves to define the central direction of the datapoints, whereas its volume can be used as measure of dispersion of the datapoints.

Appendix

As mentioned already on a number of occasions, two different matrices $B, C\in \mathbb {GL}(n)$ may produce the same ellipsoidal cone. The statement (d) in the next proposition provides an easily verifiable necessary condition for having the equality

$$\begin{aligned} B^{-1}(\mathbb {L}_n) = C^{-1}(\mathbb {L}_n). \end{aligned}$$

(46)

A necessary and sufficient condition for (46) is given in Corollary 6.4. If P and D are convex cones in $\mathbb {R}^n$, then $Q \circeq P$ indicates that either $Q=P$ or $Q=-P$.

Proposition 6.3

For a pair of matrices $B, C\in \mathbb {GL}(n)$, the following conditions are equivalent:

• (a) $B^{-1}(\mathbb {L}_n) \circeq C^{-1}(\mathbb {L}_n)$.

• (b) $(BC^{-1})(\mathbb {L}_n) \circeq \mathbb {L}_n$.

• (c) $(BC^{-1})^\sharp$ is a positive multiple of $J_n$.

• (d) $B^\sharp$ is a positive multiple of $C^\sharp$.

Proof

Clearly, (a) $\Leftrightarrow$ (b) and (c) $\Leftrightarrow$ (d). To prove the equivalence (b) $\Leftrightarrow$ (c), there is no loss of generality in working with the particular case $C=I_n$. So, we must check that

$$\begin{aligned} B(\mathbb {L}_n) \circeq \mathbb {L}_n\;\; \Leftrightarrow \;\; B^\sharp =\mu J_n \text{ for } \text{ some } \mu >0. \end{aligned}$$

(47)

In fact, the above equivalence is a known result by Loewy and Schneider [14, Theorem 2.4]. We give below an alternative proof of (47) that is based on Proposition 3.3 and the spectral decomposition

$$\begin{aligned} B^\sharp \;= \;U \mathrm{Diag}(\lambda _1,\ldots ,\lambda _n)U^T \,=\; \sum _{i=1}^{n-1}\lambda _i u_iu_i^T+ \lambda _nu_n u_n^T. \end{aligned}$$

Let $B(\mathbb {L}_n) \circeq \mathbb {L}_n$. Hence, the ellipsoidal cone $Q=B^{-1}(\mathbb {L}_n)$ is equal to $\mathbb {L}_n$ or equal to $-\mathbb {L}_n$. In particular, Q admits either $u_n=e_n$ or $u_n=-e_n$ as axial symmetry center. In addition, all the semiaxes lenghts of Q are equal to 1. From Proposition 3.3 we get $\lambda _1= \ldots = \lambda _{n-1} =-\lambda _n$ and, a posteriori,

$$\begin{aligned} B^\sharp = \left[ \begin{array}{cc} -\lambda _n I_{n-1}&{}\quad 0 \\ 0 &{}\quad \lambda _n \\ \end{array} \right] = \lambda _n J_n. \end{aligned}$$

This shows that $B^\sharp$ is a positive multiple of $J_n$. Conversely, suppose that $B^\sharp =\mu J_n$ for some $\mu >0$. In such a case, $u_n\in \{ e_n,-e_n\}$ and

$$\begin{aligned} \lambda _1= \ldots = \lambda _{n-1} =-\mu \,, \quad \lambda _n= \mu . \end{aligned}$$

(48)

Hence, the ellipsoidal cone $Q=B^{-1}(\mathbb {L}_n)$ is symmetric with respect to the line generated by $e_n$ and, thanks to (48) and Proposition 3.3, all the semi-axes lengths of Q are equal to 1. In conclusion, Q is either the Lorentz cone or the opposite of the Lorentz cone. $\square$

Corollary 6.4

Consider a pair $B, C\in \mathbb {M}_n$ of invertible matrices and write

$$\begin{aligned} BC^{-1} = \left[ \begin{array}{cc} M &{}\quad z \\ q^T &{}\quad t \\ \end{array} \right] , \end{aligned}$$

with $t\in \mathbb {R}, M\in \mathbb {M}_{n-1}$, and $z,q\in \mathbb {R}^{n-1}$. Then

$$\begin{aligned} B^{-1}(\mathbb {L}_n) = C^{-1}(\mathbb {L}_n)\quad \Leftrightarrow \quad \left\{ \begin{array}{lll} t>\Vert z\Vert \,,\; M^Tz=t\, q,\; \text{ and } \\ M^TM-qq^T= (t^2- \Vert z\Vert ^2)I_{n-1}. \end{array} \right. \end{aligned}$$

Proof

By using Loewy and Schneider [14, Theorem 2.4] or the equivalence (b) $\Leftrightarrow$ (c) in Proposition 6.3, we get

$$\begin{aligned} BC^{-1}(\mathbb {L}_n)\circeq \mathbb {L}_n\quad \Leftrightarrow \quad \left\{ \begin{array}{lll} t^2- \Vert z\Vert ^2 &{}>&{}0\\ M^Tz-t\, q&{}=&{}0\\ M^TM-qq^T&{}=&{} (t^2- \Vert z\Vert ^2)I_{n-1}. \end{array} \right. \end{aligned}$$

The case $BC^{-1}(\mathbb {L}_n)= \mathbb {L}_n$ occurs when the last entry of $BC^{-1}e_n$ is positive, i.e., when $t>0$. $\square$