Perron and Frobenius meet Carath\'{e}odory

TL;DR

This paper introduces a novel method leveraging the Perron-Frobenius Theorem to prove Carathéodory-type theorems, potentially opening new avenues for related geometric results.

Contribution

It presents a new approach using matrix theory to prove classical geometric theorems, offering a fresh perspective and potential extensions.

Findings

New proof technique for Carathéodory-type theorems

Potential extension to Colourful Carathéodory Theorem

Open question on broader applicability

Abstract

We present a new approach of proving certain Carath\'{e}odory-type theorems using the Perron-Frobenius Theorem, a classical result in matrix theory describing the largest eigenvalue of a matrix with positive entries. One of the problems left open in this note is whether our approach may be extended to prove similar results in the area, in particular the Colourful Carath\'{e}odory Theorem.

Equations

$$\left\{\sum _{i=1}^n{\lambda }_i{x}_i:{\lambda }_i\ge 0\text{ for all }i\in [n],\sum _{i=1}^n{\lambda }_i=1\right\}.$$

$$0=\mathrm{tr}(H)={\lambda }_1+\cdots +{\lambda }_d+(n-d),\text{and thus, }|{\lambda }_1+\cdots +{\lambda }_d|=n-d\ge d+1.$$

$$\left(\lambda I_n+V^{t}Y\right)x=\rho (H)x,\text{mbox}andthus,V^{t}Yx=(\rho (H)-\lambda )x.$$

$$0=\mathrm{tr}(H)={\lambda }_1+\cdots +{\lambda }_d+(n-d),\text{mbox}andthus,|{\lambda }_1+\cdots +{\lambda }_d|=n-d\ge d+1.$$

$$\left(I_n+V^{t}Y\right)x=\rho (H)x,\text{mbox}andthus,V^{t}Yx=(\rho (H)-1)x.$$

$$\Vert v_{\delta }{\Vert }^2=\Vert (1-\delta )v+\delta v_i{\Vert }^2$$

$$\le \Vert v{\Vert }^2-2\delta \Vert v{\Vert }^2+{\delta }^2(\Vert v{\Vert }^2+\Vert v_i{\Vert }^2).$$

Full text

One of the problems left open in this note is whether our approach may be extended to prove similar results in the area, in particular the Colourful Carathéodory Theorem.

1. Introduction

Carathéodory’s Theorem [Car07] is one of conerstones of combinatorial geometry as Eckhoff [Eck93] called it. The theorem claims that a point in the convex hull of a set $P\subseteq\mathbb{R}^{d}$ is in the convex hull of at most $d+1$ points of $P$; see the definition of convex hull in Section 2. The survey [Eck93] contains a number of generalizations of the theorem. As it usually happens with fundamental statements, Carathéodory’s Theorem is closely connected with many other classical results in convex geometry such as Helly’s and Radon’s Theorems [Hel23, Rad21]. In this note, we provide a new proof of the theorem that shows a connections with a principal theorem in matrix theory, the Perron–Frobenius Theorem.

Our proofs use induction on the number of points (but not on the dimension). Two, most likely related, questions remain open. First, can induction be avoided in the proofs? Second, does our method yield a proof of other Carathéodory-type theorems, specifically, Bárány’s Colourful Carathéodory Theorem [Bár82].

The paper is organized as follows. In Section 2, we introduce notations and the tools that we use. In Section 3, we illustrate the key idea applying it to prove Rankin’s Theorem [Ran55, Theorem 1(iii, iv)]. In Section 4, we use the idea to prove Carathéodory’s and Steinitz’s Theorems [Car07, Ste13].

2. Preliminaries

2.1. Notations

We write $[n]=\{1,\dots,n\}$, for a positive integer $n$. The convex hull of a finite set of points $\{x_{1},\dots,x_{n}\}\subset\mathbb{R}^{d}$, written $\operatorname{conv}\{x_{1},\dots,x_{n}\}$, is the set

[TABLE]

A point $x$ of a set $P\subseteq\mathbb{R}^{d}$ is an interior point of $P$, if the set $P$ contains some open ball with center in $x$.

The spectral radius of a square matrix $A$, written $\rho(A)$, is the largest absolute value of its eigenvalues.

2.2. Tools

Since we do not need the most general form of the Perron–Frobenius Theorem [Per07, Fro12], we state only two of its corollaries.

Lemma 1** (Perron’s Theorem).**

For an $n$-by-$n$ matrix with positive entries, the spectral radius is an eigenvalue of multiplicity $1$, such that its eigenvector has positive entries.

Lemma 2** (Frobenius’s Theorem).**

For an $n$-by-$n$ matrix with non-negative entries, the spectral radius is an eigenvalue such that one of its eigenvectors has non-negative entries.

Also, we need two finite-dimensional versions of the Hahn–Banach Theorem in Section 4.

Theorem 3**.**

If the origin $o$ of $\mathbb{R}^{d}$ does not lie in the convex hull of points $x_{1},\dots,x_{n}\in\mathbb{R}^{d}$, then there exists a vector $y\in\mathbb{R}^{d}$ such that $\langle y,x_{i}\rangle>0$ for all $i\in[n]$.

Theorem 4**.**

If the origin $o$ of $\mathbb{R}^{d}$ is not an interior point of the convex hull of points $x_{1},\dots,x_{n}\in\mathbb{R}^{d}$, then there exists a vector $y\in\mathbb{R}^{d}$ such that $\langle y,x_{i}\rangle\geq 0$ for all $i\in[n]$.

3. Proof of Rankin’s Theorem

We state the two parts of Rankin’s Theorem as separate theorems.

Theorem 5** (Rankin).**

If $\{v_{1},\dots,v_{n}\}$ is a set of non-zero vectors in $\mathbb{R}^{d}$ such that the angle between any two of them is larger than $\frac{\pi}{2}$, then $n\leq d+1$.

Proof.

Suppose to the contrary that $n\geq d+2$. Let $G$ be the Gram matrix of the vectors $v_{1},\dots,v_{n}$, that is, $G=(\langle v_{i},v_{j}\rangle)_{i,j=1}^{n}$. Choose a positive $\lambda$ such that $\lambda>\langle v_{i},v_{i}\rangle$ for all $i\in[n]$, and set $H=\lambda I_{n}-G$. The hypothesis of the theorem implies that all entries of $H$ are positive. By Lemma 1, the spectral radius $\rho(H)$ is the largest eigenvalue of $H$ of multiplicity one.

Obviously, $\operatorname{rank}G\leq d$ because $G$ is the Gram matrix of $d$-dimensional vectors. Hence [math] is an eigenvalue of $G$ of multiplicity at least two, that is, $\lambda$ is an eigenvalue of $H$ of multiplicity at least two. Since the Gram matrix $G$ is positive semidefinite, and thus, all its eigenvalues are non-negative, $\lambda$ must be the largest eigenvalue of $H$. Therefore, $\lambda=\rho(H)$, contradicting the fact that the multiplicity of the largest eigenvalue is one. ∎

Theorem 6** (Rankin).**

If $\{v_{1},\dots,v_{n}\}$ is a set of non-zero vectors in $\mathbb{R}^{d}$ such that the angle between any two of them is at least $\frac{\pi}{2}$, then $n\leq 2d$.

Proof.

Suppose to the contrary that $n\geq 2d+1$. Without loss of generality we can assume that the vectors $v_{1},\dots,v_{n}$ are of unit length. Let $G$ be the Gram matrix of the vectors $v_{1},\dots,v_{n}$. Set $H=I_{n}-G$. The hypothesis of the theorem implies that all entries of $H$ are non-negative. By Lemma 2, the spectral radius $\rho(H)$ is the largest eigenvalue of $H$.

Obviously, $\operatorname{rank}G\leq d$ because $G$ is the Gram matrix of $d$-dimensional vectors. Hence, [math] is an eigenvalue of $G$ of multiplicity at least $n-d$, and thus, $1$ is an eigenvalue of $H$ of multiplicity at least $n-d$. Since the Gram matrix $G$ is positive semidefinite, that is, [math] is its smallest eigenvalue, $1$ is the largest eigenvalue of $H$. It means that $\rho(H)=1$. Let $\lambda_{1},\dots,\lambda_{d},1,\dots,1$ be the eigenvalues of $H$, indexed in non-decreasing order. Thus we have

[TABLE]

However, the last inequality contradicts $|\lambda_{i}|\leq 1=\rho(H)$ for all $i\in[d]$. ∎

4. Proofs of Caratheodory’s and Steinitz’s Theorems

Theorem 7** (Carathéodory’s Theorem).**

If the origin $o\in\mathbb{R}^{d}$ lies in the convex hull of points $v_{1},\dots,v_{n}\in\mathbb{R}^{d}$, then there is a set $J\subseteq[n]$ of size at most $d+1$ such that $o$ lies in the convex hull of $\{v_{j}\;:\;j\in J\}$.

Proof.

We use induction on $n$. The base case, $n=d+1$, being trivial, it is sufficient to show that if $n>d+1$ and $o\not\in\operatorname{conv}\left\{v_{i}:i\in[n]\setminus\{j\}\right\}$ for all $j\in[n]$, then $o\not\in\operatorname{conv}\left\{v_{i}:i\in[n]\right\}$.

By Theorem 3, for all $j\in[n]$, there is a vector $y_{j}$ such that $\langle x_{i},y_{j}\rangle>0$ for all $i\in[n]\setminus\{j\}$. Choose a positive $\lambda$ such that $\lambda+\langle v_{i},y_{i}\rangle>0$ for all $i\in[n]$ and set $H=\lambda I_{n}+V^{t}Y$, where $V=[v_{1},\dots,v_{n}]$ and $Y=[y_{1},\dots,y_{n}]$. By Lemma 1, the spectral radius $\rho(H)$ is the largest eigenvalue of $H$ of multiplicity one.

Obviously, $\operatorname{rank}V^{t}Y\leq d$ because $V$ and $Y$ are $n$-by-$d$ matrices. Hence, [math] is an eigenvalue of $G$ of multiplicity at least two, that is, $\lambda$ is an eigenvalue of $H$ of multiplicity at least two. It follows that $\rho(H)\neq\lambda$, so $\rho(H)>\lambda$, because the spectral radius cannot be less than a positive eigenvalue. Consider the eigenvector $x$ of the eigenvalue $\rho(H)$. By Lemma 1, its entries are positive. Therefore, we obtain

[TABLE]

Hence, $Yx$ is a $d$-dimensional vector such that $\langle v_{i},Yx\rangle>0$ for all $i\in[n]$, because all entries of the vector $(\rho(H)-\lambda)x$ are positive. So the points $v_{1},\dots,v_{n}$ lie in an open half-space bounded by a hyperplane passing through the origin. Thus, their convex hull does not contain the origin. ∎

Theorem 8** (Steinitz’s Theorem).**

If the origin $o$ of $\mathbb{R}^{d}$ is an interior point of the convex hull of points $v_{1},\dots,v_{n}\in\mathbb{R}^{d}$, then there is a set $J\subseteq[n]$ of size at most $2d$ such that the point $o$ is interior of convex hull of $\{v_{j}\;:\;j\in J\}$.

Proof.

Again, we use induction on $n$. The base case, $n=2d$, being trivial, it is sufficient to show that if $n>2d$ and, for all $j\in[n]$, the origin $o$ is not an interior point of $\operatorname{conv}\left\{v_{i}:i\in[n]\setminus\{j\}\right\}$, then $o$ is not an interior point of $\operatorname{conv}\left\{v_{i}:i\in[n]\right\}$.

By Theorem 4, for all $j\in[n]$, there is a non-zero vector $y_{j}$ such that $\langle v_{i},y_{j}\rangle\geq 0$ for all $i\in[n]\setminus\{j\}$. If $\langle v_{i},y_{i}\rangle\geq 0$ for some $i\in[n]$, then the origin is not an interior point of the convex hull of $v_{1},\dots,v_{n}$. So, without loss of generality, we may assume that $\langle v_{i},y_{i}\rangle=-1$ for all $i\in[n]$. Set $H=I_{n}+V^{t}Y$, where $V=[v_{1},\dots,v_{n}]$ and $Y=[y_{1},\dots,y_{n}]$. By Lemma 2, the spectral radius $\rho(H)$ is an eigenvalue of $H$.

Obviously, $\operatorname{rank}V^{t}Y\leq d$ because $V$ and $Y$ are $n$-by-$d$ matrices, and hence, [math] is an eigenvalue of $V^{t}Y$ of multiplicity at least $(n-d)$, that is, $1$ is an eigenvalue of $H$ of multiplicity at least $(n-d)$. Suppose that $\rho(H)=1$. Let $\lambda_{1},\dots,\lambda_{d},1,\dots,1$ be the eigenvalues of $H$, indexed in non-decreasing order. Hence we have

[TABLE]

But the last inequality contradicts $|\lambda_{i}|\leq 1=\rho(H)$ for all $i\in[d]$. Hence $\rho(H)>1$ because the spectral radius cannot be less than a positive eigenvalue. Consider the eigenvector $x$ of the eigenvalue $\rho(H)$. By Lemma 2, its entries are non-negative entries. Therefore, we get

[TABLE]

Hence, $\langle v_{i},Yx\rangle\geq 0$ for all $i\in[n]$. Moreover, $Yx$ is a non-zero vector, because among non-negative entries of $(\rho(H)-1)x$ there is at least one positive. So the points $v_{1},\dots,v_{n}$ lie in a closed half-space bounded by a hyperplane passing through the origin, that is, the point $o$ is not interior of their convex hull. ∎

Appendix A Proofs of Separtion Theorems

Theorems 3 and 4 are fundamental in the study of convex sets. For completeness, we include their proofs.

A.1. Proof of Theorem 3

By compactness of the set $\operatorname{conv}\{v_{1},\dots,v_{n}\}$, it has a point $v\in\operatorname{conv}\{v_{1},\dots,v_{n}\}$ closest to the origin. Suppose that $\langle v,v_{i}\rangle\leq 0$. Write $v_{\delta}:=(1-\delta)v+\delta v_{i}$. Thus we have

[TABLE]

Since the coefficient for $\delta$ in (2) is negative, $\|v_{\delta}\|^{2}>\|v\|^{2}$ for sufficiently small positive $\delta$. Points $v$ and $v_{i}$ lie in $\operatorname{conv}\{v_{1},\dots,v_{n}\}$, so $v_{\delta}\in\operatorname{conv}\{v_{1},\dots,v_{n}\}$, contradicting the choice of $v$. Therefore, $\langle v,v_{i}\rangle>0$ for all $i\in[n]$.

A.2. Proof of Theorem 4

Set $K=\operatorname{conv}\{v_{1},\dots,v_{n}\}$. Denote the ray emanating from a point $x$ and passing through a point $y$ by $[x,y)$.

We claim that there exists a ray $[o,t)$ such that $[o,t)\cap K=\{o\}$. Otherwise, one can choose $d+1$ points $t_{1},\dots,t_{d+1}\in K$ such that the point $o$ is an interior point of $\operatorname{conv}\{t_{1},\dots,t_{d+1}\}\subseteq K$.

Consider a sequence $o_{1},\dots,o_{n},\dots$ of points distinct from $o$ on the ray $[o,t)$ and converging to $o$. By Theorem 3, there exists a sequence of unit vectors $y_{1},\dots,y_{n},\dots$ in $\mathbb{R}^{d}$ such that $\langle y_{i},v_{j}-o_{i}\rangle>0$ for all $j\in[n]$ and all positive integer $i$. Since the unit sphere is compact, there exists a subsequence of unit vectors $y_{i_{1}},\dots,y_{i_{n}},\dots$ converging to a unit vector $y\in\mathbb{R}^{d}$. Therefore, we obtain $\langle y,v_{j}\rangle\geq\langle y,o\rangle=0$.

Acknowledgments

The work was done while the second named author was an academic visitor at EPFL.

The authors are grateful to János Pach and Imre Bárány for fruitful discussions.