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\lecture{8}{Extreme points}{Aditya G Parameswaran}

In this lecture, we define \emph{extreme points} in geometry and try to give an linear algebra perspective for the same.

\begin{Def}
Given points $p_1, p_2, p_3, \ldots, p_n$, the convex hull is the smallest convex set containing these points. 
\end{Def}

To simplify the discussion, we make the following assumptions on the nature of $Ax \leq {\bf b}$ which hold for the rest of the course unless otherwise stated.

\section{Assumptions on the nature of the convex set $Ax \leq {\bf b}$}
\begin{itemize}
\item \emph{Assumption 1:} $Ax \leq {\bf b}$ is bounded. No single
coordinate of $x$ satisfying $Ax \leq {\bf b}$ can increase or decrease without bound.
\item \emph{Assumption 2:}  $Ax \leq {\bf b}$ has no degeneracies. This is equivalent to saying that not more than $n$ hyperplanes pass through a point in 
an $n$-dimensional space. This assumption makes sense because small perturbations of the hyperplanes can remove the degeneracies.
\item \emph{Assumption 3:} $Ax \leq {\bf b}$ should be \emph{full dimensional}. This is equivalent to saying that one should be able to place a $n$-dimensional
sphere, however small, in the region defined  by  $Ax \leq {\bf b}$. For two dimensions, the convex set should have area, and for three dimensions, the convex set
should have volume. In $n$ dimensions, the convex set should have an $n$-dimensional volume.
\end{itemize}

\section{Extreme Points and its Algebraic Interpretation}
\begin{Def}
An extreme point is a point in a convex set that cannot be represented as a convex combination of any two other distinct points in the convex set. 
\end{Def}

Thus, an extreme point does not lie on the segment between two other distinct points of the convex set. 
Intuitively, all segments with the extreme point in the interior ``stick out'' of the convex set.
\\
\begin{Thm}
Given a convex set $Ax \leq {\bf b}$ in n-dimensional space satisfying the assumptions of Sec. 1, the extreme points of a convex set are precisely those points 
in $Ax \leq {\bf b}$ which can be expressed as an intersection of n linearly independent hyperplanes out of the set of hyperplanes that define the convex set
$Ax \leq {\bf b}$. 
\end{Thm}
\begin{proof} \\
\emph{Part 1:} ($\implies$) Any point which can be expressed as an intersection of n linearly independent hyperplanes out of the set of hyperplanes that define the convex set
$Ax \leq {\bf b}$ is an extreme point.

Let $x_0$ be any such point. We split $Ax_0 \leq {\bf b}$ into two parts 
\begin{eqnarray}
A^{'}x_0 &=& {\bf b^{'}} \\ 
A^{''}x_0 &<& {\bf b^{''}}
\end{eqnarray}
(We effectively move the inequalities in  $Ax_0 \leq {\bf b}$ that are equalities into the first set $A^{'}x_0 = {\bf b^{'}}$)

We notice that since $x_0$ is a point on the intersection of n linearly independent hyperplanes in the defining set of hyperplanes, 
$A^{'}$ has n linearly independent rows.
 
Let there be $x_1$ and $x_2$ in the given convex set such that $x_0 = \lambda x_1 + (1 - \lambda) x_2$, $0 < \lambda < 1$. We then have 
\begin{equation}
\lambda A^{'} x_1 + (1 - \lambda) A^{'} x_2 = {\bf b^{'}}
\end{equation} 
But since $x_1$ and $x_2$ are part of the convex set,  $A^{'}x_1 \leq {\bf b^{'}}$ and  $A^{'}x_2 \leq {\bf b^{'}}$.

Neither of these inequalities can be strict (or else Eq. 1 will not hold) and therefore $A^{'}x_1 = {\bf b^{'}}$ and $A^{'}x_2 = {\bf b^{'}}$.

Now, we are given that $x_0$ is a solution of $A^{'}x = {\bf b^{'}}$, which is nothing but the intersection of n linearly independent hyperplanes. Thus, we have $ x_0 = x_1 = x_2 $.

Thus $x_0$ cannot be expressed as a convex combination of two other distinct points in the convex set and is therefore an extreme point.

\emph{Part 2:} ($\Leftarrow$) Let $x_0$ be an extreme point. If we split $Ax_0 \leq {\bf b}$ into two parts 
\begin{eqnarray}
A^{'}x_0 &=& {\bf b^{'}} \\ 
A^{''}x_0 &<& {\bf b^{''}}
\end{eqnarray}
then $A{'}$ has n linearly independent rows.

See next lecture for a proof.

\end{proof}
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