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\lecture{10}{Extreme Points}{Swati Goyal}
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%%                          Beginning of Section 1.
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\section{Extreme Points}
\begin{Def}
An extreme point in a convex set is a point which cannot be
represented as a convex combination of two other points of the
set.
\end{Def}
In the last lecture, we proved that in a space of dimension n, an
extreme point is the intersection point of n linearly independent
hyperplanes i.e
\newline
    Take an $x_0$ such that $Ax_0 \leq b$ given by \newline
\begin{equation}
      A^{'}x_{0} = b^{'}
\end{equation}
\begin{equation}
 A^{''}x_{0} < b^{''}
\end{equation}
then $x_0$ is an extreme point iff $A^{'}$ consists of n linearly
independent rows(hyperplanes).
Note that we have assumed  $Ax_0 \leq b$ to be non-degenerate.

\section{Convex Combination of Extreme Points}
Consider a region $Ax \leq b$
\newline
Let $p_{1},p_{2}..p_{n}$ be the extreme points of this region.
Then this region can also be represented as the convex hull of
these extreme points. \newline We will prove this statement in the
following section.
\begin{Thm}
Any point $x$, such that $Ax \leq b$, can be written as convex
combination of the extreme points of this region.
\end{Thm}
\begin{proof}
The proof will use induction on dimension of the region.
\newline
Base Case: Dimension of the region is 0.Then there is just one point and hence the theorem is vacuously true.
We illustrate the main technique by looking at an example in 2 dimension. Consider the region in Fig1. 
\newpage
 \begin{figure}[htp]
\centering
\includegraphics[scale=0.4]{basecase.ps}
\hspace{1.0in}
\includegraphics[scale=0.4]{basecase2.ps}
 \caption{Convex polygon in dimension 2}
\label{figure:convexPolygon in dimension 2}
\end{figure}

Let p be any point in the given region and say $p_{1},p_{2}$ \dots be
the extreme points of the region. Now join $p_{1}$ and p  and
extend it to intersect $p_{3}p_{4}$ at q. Since q lies on the line
joining $p_{3}p_{4}$ it can be written as a convex combination of
$p_{3}$ and $p_{4}$.
\begin{equation}
   q = \Sigma\alpha_{i}p_{i} ,  0\leq\alpha_{i}\leq1,
    \Sigma\alpha_{i}=1 , i = 3,4
    \end{equation}
    \begin{equation}
    p = \beta_{1}p_{1} + \beta_{2}q ,0\leq\beta_{i}\leq1, \Sigma\beta_{i}=1
\end{equation}
and hence
\begin{equation}
    p = \Sigma\gamma_{i}p_{i} , 0\leq\gamma_{i}\leq1, \Sigma\gamma_{i}=1
\end{equation}
Hypothesis: Assume that any $x$, such that $Ax\leq b$ in space of
dimension $n-1$ can be written as convex combination of the
extreme points of the region.
\newline
\newline
Induction: Consider a region $S : Ax \leq b$ in space of dimension
$n$ with $p_{1},p_{2}$ \dots as the extreme points
\newline
Consider a point $p \in S$. Join $p_{1}$ and p and extend the line
segment to meet a point q on the boundary of $S$.At this boundary, at least one of the inequalities in  $Ax\leq b$ has to be equality. For simplicity assume that the first is equality.
\newline
Thus at q,
\begin{equation}
A_{1}q = b_{1}
\end{equation}
\begin{equation}
A^{''}q < b^{''}
\end{equation}
where $q = [q_1, q_2, \dots]$
\newline
Solve (6) for some $q_{i}$ and substitute its value in (7).
\newline
Thus (7) now reduces to another region $S^{'}$ of dimension $n-1$
of the form $Cx \leq D$.
\newline
By assumed hypothesis, q can be written as convex combination of
extreme points of $S^{'}$.
\begin{equation}
    q = \Sigma\alpha_{i}(p^{'})_{i}\hspace{1.5mm}
   \tt{where} \hspace{1.5mm} \Sigma\alpha_{i} = 1 \hspace{1.5mm}\tt{ and } \hspace{1.5mm}0 \leq\alpha_{i}\leq1
\end{equation}
Now p can be written as a convex combination of q and $p_{1}$ i.e.
as a convex combination of $p^{'}_{1} \dots$ and p.But this still
does't prove that p is a convex combination of $p_{i}
\dots$(extreme points of S)\newline For this we would have to
prove that the extreme points of $S^{'}$ are a subset of the
extreme points of S.
\newline
Assume this is not true,that is, $p^{'}$ is an extreme point of $S^{'}$
but not of $S$.
\newline
Since, $p^{'} \in S^{'}$ and as $S^{'}$ is a subset of S, thus
$p^{'} \in S$.
\newline
Also , $\exists p_{1},p_{2} \in S$ such that $p^{'}$ is the convex
combination of $p_{1},p_{2}$(as $p^{'}$ is not an extreme point in
S).
\begin{equation}
p^{'} = \lambda p_{1} + (1-\lambda)p_2 \hspace{1.5mm}
0\leq\lambda\leq1
\end{equation}
Now, for $p^{'}$ to be an extreme point in S,
\begin{equation}
  A_{1}p^{'} = b_{1}
\end{equation}
must be true, from (6), as (7) is already satisfied for all points
in region $S^{'}$.
\newline i.e.
\begin{equation}
A_{1}\lambda p_{1} + A_{1}(1-\lambda)p_2 = b_{1} \hspace{1.5mm}
\tt{from(9) and (10)}
\end{equation}
Since $A_{1}p_{1} \leq b_{1}$ and $A_{1}p_{2} \leq b_{1}$, both of them have to be equal to
$b_{1}$. Thus, $p_{1}$ and $p_{2}$ satisfy (6) and thus belong to
$S^{'}$.
\newline
This leads to a contradiction since $p^{'}$(extreme point) cannot
be expressed as convex combination of two or more points of the
region.
\newline
Hence, the extreme points of $S^{'}$ are a subset of the extreme
points of $S$.
\newline
Thus,
\begin{equation}
q = \Sigma\alpha_{i}p^{'}_{i} = \Sigma\alpha_{i}p_{i}
\end{equation}
\begin{equation}
p = \beta_{1}p_{1} + \beta_{2}q
\end{equation}
\begin{equation}
p = \Sigma\gamma_{i}p_{i} \hspace{1.5mm} \tt{from (12) and (13)}
\end{equation}
Thus, every point in S can be expressed as the convex combination
of the extreme points of S.
\end{proof}
\section{Maximize $c^{T}x$}
Given a linear function $c^{t}x$ and a region $Ax\leq b$, such
that the dimension of the space is $n$ and the number of rows in A is m,then the number of the extreme points
of the region is C(m,n). Each extreme point is an intersection of n linearly independent 
planes, and thus the number of extreme points is at most the number of
possible ways of choosing n planes from m choices.
\newline
But before finding the maximum value for the function in the given
region, we will have to prove a theorem.
\newline
\begin{Thm}
The maximum value of a linear function f(x) occurs at an extreme
point of the region $Ax \leq b$.
\end{Thm}
\begin{proof}
Let f be maximum at point p of the region $Ax \leq b$. \newline
From the previous theorem we have,
\begin{equation}
p = \sum\alpha_{i}p_{i} \hspace{1.5mm} 0\leq\alpha_{i}\leq1
\hspace{1.5mm} \Sigma\alpha_{i} = 1
\end{equation}
where $p_{1} \dots $are the extreme points of S. \newline Thus,
\begin{equation}
f(p) = f(\sum\alpha_{i}p_{i})
\end{equation}
\begin{equation}
f(p) = \sum \alpha_{i}f(p_{i})
\end{equation}
since f is a linear function.
\begin{equation}
\sum\alpha_{i}f(p_(i)) \leq \theta\sum\alpha_{i}
\end{equation}
where $\theta$ is the maximum value among
f($p_{i}$).Thus
\begin{equation}
\theta = f(p) = \sum \alpha_{i}f(p_{i}) \leq \theta\sum\alpha_{i} \leq
\theta
\end{equation}
 Thus there must exist some extreme point$p_{i}$ such that $f(p)=
f(p_{i})$, otherwise the equality cannot be satisfied in (19).
\newline Hence, the maximum value of f occurs at an extreme point.
\end{proof}
\newline
In next lecture, we will look at the simplex algorithm for finding
the extreme points of the region $Ax \leq b$.
\end{document}