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Viewing Version 3 of 'Radon's lemma'
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Title of object: Radon's lemma
Canonical Name: RadonsLemma
Type: Theorem

Created on: 2003-09-07 22:44:58
Modified on: 2004-01-25 00:40:00

Creator: bbukh
Modifier: bbukh
Author: bbukh

Classification: msc:52C07, msc:52A20
Keywords: convex hull

Revision comment (for changes between this and next version):

Changes for correction #5659 ('Absolute value of the negative coefficients').

Preamble:

\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}

%\usepackage{xypic}

\makeatletter
\@ifundefined{bibname}{}{\renewcommand{\bibname}{References}}
\makeatother
Content:

Every set $A\subset \mathbb{R}^d$ of $d+2$ or more points can be partitioned into two disjoint sets $A_1$ and $A_2$ such that the convex hulls of $A_1$ and $A_2$ intersect.

\begin{proof}
Without loss of generality we assume that the set $A$ consists of exactly $d+2$ points which we number $a_1, a_2,\dotsc, a_{d+2}$. Denote by $a_{i,j}$ the $j$'th component of $i$'th vector, and write the components in a matrix as
\begin{equation*}
M=\begin{bmatrix}
a_{1,1}&a_{2,1}&\dots&a_{d+2,1}\\
a_{1,2}&a_{2,2}&\dots&a_{d+2,2}\\
\vdots& \vdots&\ddots & \vdots\\
a_{1,d}&a_{2,d}&\dots&a_{d+2,d}\\
1&1&\dots&1
\end{bmatrix}.
\end{equation*}
Since $M$ has fewer rows than columns, there is a non-zero column vector $\mathbf{\lambda}$ such that $M \mathbf{\lambda}=0$ which is equivalent to existence of a solution to the system
%aligned inside equation with a tag does not work with LaTeX2HTML
\begin{align}\label{eqn:sys}
0&=\lambda_1 a_1+\lambda_2 a_2+\dotsb+\lambda_{d+2} a_{d+2}\\
0&=\lambda_1+\lambda_2+\dotsb+\lambda_{d+2}\notag
\end{align}
Let $A_1$ to be the set of those $a_i$ for which $\lambda_i$ is positive, and $A_2$ is the rest. Set $s$ to be the sum of positive $\lambda_i$'s. Then by the system~\eqref{eqn:sys} above
\begin{equation*}
\frac{1}{s}\sum_{a_i\in A_1} \lambda_i a_i=\frac{1}{s}\sum_{a_i\in A_2} \lambda_i a_i
\end{equation*}
is a point of intersection of convex hulls of $A_1$ and $A_2$.
\end{proof}