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${I\!\!P}^{1} \longmapsto {I\!\!P}^{1}$. The general case of a collineation is:

\begin{displaymath}\left( \begin{array}{c} x \\ t \end{array} \right) \longmapst...
...\left( \begin{array}{c} ax + by \\ cx + dy \end{array} \right)

with $ad-cd \neq 0$. Provided $t \neq 0$ and $cx + d \neq 0$, this can be rewritten in inhomogeneous affine coordinates as:

\begin{displaymath}\left( \begin{array}{c} x \\ 1 \end{array} \right) \longmapst...
...begin{array}{c} \frac{ax + b}{cx + d} \\ 1 \end{array} \right)

Property: A translation in affine space corresponds to a collineation leaving each point at infinity invariant.

Proof: The translation $(x_{1}, ..., x_{n}, 1) \longmapsto (x_{1} +
a_{1}, ..., x_{n} + a_{n}, 1)$ can be represented by the matrix:

\begin{displaymath}A = \left(
\begin{array}{ccc\vert c}
1 & \cdots & 0 & a_{1} ...
...& 1 & a_{n} \\
0 & \cdots & 0 & 1
\end{array} \right)

Obviously $A (x_{1}, ..., x_{n}, 0)^\top = (x_{1}, ..., x_{n}, 0) $. $\sqcup \!\!\!\! \sqcap$

More generally, any affine transformation is a collineation, because it can be decomposed into a linear mapping and a translation:

\begin{displaymath}\left( \begin{array}{c} y_{1} \\ \vdots\\
y_{n} \end{array}...
...\begin{array}{c} t_{1} \\ \vdots\\
t_{n} \end{array} \right)

In homogeneous coordinates, this becomes:

\begin{displaymath}\left( \begin{array}{c} y_{1} \\ \vdots\\
y_{n}\\ \hline 1 ...
...y}{c} x_{1} \\ \vdots\\
x_{n}\\ \hline 1 \end{array} \right)

Exercise 2.1   : Prove that a collineation is an affine transformation if and only if it maps the hyperplane at infinity xn+1=0 into itself (i.e. all points at infinity are mapped onto points at infinity).

Bill Triggs