Collineations

A collineation of is defined as a mapping from the plane to itself such that the collinearity of any set of points is preserved. Such a mapping can be achieved with matrix multiplication by a $3 \times 3$ matrix T. Each point $\ensuremath{{\bf p}}$ is transformed into a point $\ensuremath{{\bf p}} '$:

$$\ensuremath{{\bf p}} ' = T \ensuremath{{\bf p}} .$$

We will use the terms transformation and collineation interchangeably. Since scaling is unimportant, only eight elements of T are independent. Therefore, since each point contains two independent values, four pairs of corresponding points are necessary to determine T.

To transform a line $\ensuremath{{\bf u}}$ into a line $\ensuremath{{\bf u}} '$, we note that collinearity must be preserved, that is, if a point $\ensuremath{{\bf p}}$ lies on the line $\ensuremath{{\bf u}}$, then the corresponding point $\ensuremath{{\bf p}} '$ must lie on the corresponding line $\ensuremath{{\bf u}} '$. Therefore,

$$\ensuremath{{\bf p}} ^T\ensuremath{{\bf u}} = 0 = (T^{-1}\ens... ...u}} = (\ensuremath{{\bf p}} ')^T(T^{-T}\ensuremath{{\bf u}} ),$$

which indicates that

$$\ensuremath{{\bf u}} ' = T^{-T}\ensuremath{{\bf u}} .$$

From these results, it is not hard to show that a point conic C transforms to T^-TCT^-1, and a line conic |C|C^-1 transforms to T|C|C^-1T^T.

Regarding transformations, recall that projective $\supset$ affine $\supset$ similarity $\supset$ Euclidean. Let's study the matrix T to uncover the relationships between these various geometries. First we will write out the elements of T, for reference:

$$T_{projective} = \left[\matrix{t_{11} & t_{12} & t_{13} \cr t_{21} & t_{22} & t_{23} \cr t_{31} & t_{32} & t_{33}}\right].$$

The affine plane is just the projective plane minus the ideal line. Therefore, affine transformations must preserve the ideal line and the ideal points, that is, any point [X, Y, 0]^T must be transformed into $[\ensuremath{\alpha} X, \ensuremath{\alpha} Y, 0]^T$ for some arbitrary scaling $\ensuremath{\alpha}$:

$$\ensuremath{\alpha}\left[\matrix{X \cr Y \cr 0}\right] = T \left[\matrix{X \cr Y \cr 0}\right],$$

which implies that t₃₁ = t₃₂ = 0. The matrix for affine transformation, then, is

$$T_{af\!fine} = \left[\matrix{t_{11} & t_{12} & t_{13} \cr t_{21} & t_{22} & t_{23} \cr 0 & 0 & t_{33}}\right],$$

where once again only six of these parameters are independent, since scale is unimportant.

Unlike affine transformations, similarity transformations preserve angles and ratios of lengths. Delaying the derivation for a moment, we simply state the result:

$$T_{similarity} = \left[\matrix{\cos \theta & \sin \theta & t_... ...sin \theta & \cos \theta & t_{23} \cr 0 & 0 & t_{33}}\right],$$ (3)

where $\theta$ is an arbitrary angle.

Under Euclidean transformation, scale is important, and therefore the point $\ensuremath{{\bf p}}$ must first be converted to Euclidean coordinates by dividing by its third element. The transformation then is

$$\left[\matrix{x' \cr y'}\right] = \left[\matrix{\cos \theta ... ...t[\matrix{x \cr y}\right] + \left[\matrix{t_x \cr t_y}\right].$$

In closing this section, we offer one final proposition, along with its proof:

Proposition 1   A transformation is a similarity transformation if and only if it preserves the absolute points, $[1, \pm i, 0]$.

The ``only if'' is rather easy to see: The absolute point $[1, \pm i, 0]^T$is transformed through equation (3) to the point $e^{\pm i\theta}[1, \pm i, 0]^T$, which is equivalent because the scale factor is ignored.

The ``if'' is a little more complicated, but still rather straightforward. Starting with the unrestricted equation for T,

$$T = \left[\matrix{t_{11} & t_{12} & t_{13} \cr t_{21} & t_{22} & t_{23} \cr t_{31} & t_{32} & t_{33}}\right],$$

the fact that [1, i, 0]^T is preserved yields the following two equations:

$$\begin{eqnarray*}\frac{t_{11} + it_{12}}{t_{21} + it_{22}} & = & \frac{1}{i} \\ \\ t_{31} + it_{32} & = & 0. \end{eqnarray*}$$

Since the elements of T are constrained to be real, this leads to the following three constraints on the elements of T:

$$\begin{eqnarray*}t_{11} & = & t_{22} \\ t_{12} & = & -t_{21} \\ t_{31} & = & t_{32} \ \ = \ \ 0. \end{eqnarray*}$$

So the matrix of T looks like this:

$$T = \left[\matrix{t_{11} & t_{12} & t_{13} \cr -t_{12} & t_{11} & t_{23} \cr 0 & 0 & t_{33}}\right].$$

Given two arbitrary numbers t₁₁ and t₁₂, we can always reparameterize them as $t_{11} = k\cos \theta$ and $t_{12} = k\sin \theta$, where $\theta$ is an angle and k is a scalar. Multiplying the previous equation by 1/k (which is legal because we are working in homogeneous coordinates), we then get

$$T = \left[\matrix{\cos \theta & \sin \theta & t_{13}/k \cr -... ...\theta & \cos \theta & t_{23}/k \cr 0 & 0 & t_{33}/k}\right],$$

which is seen to be the equation of a similarity transformation when compared with equation (3). (NOTE: Using the point [1,-i,0]^T yields the same result.)