Geometric transformation classification scheme

Any geometric transformation can be written like a linear system of matrices or like a system like:

$(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} u \\ v \end{matrix}) \Leftrightarrow \vec{x} = A \cdot \vec{x} + \vec{b}$

Next, we are going to give a scheme of classification that depends on the previous equations system.

{\begin{cases} Direct if det (A) > 0 {\begin{matrix} Isometric if det (A) = \pm 1 {\begin{matrix} Translation if \vec{b} \neq 0 \\ Rotation if \vec{b} = 0 \end{matrix} \\ Isomorph si det (A) = \pm 1, k > 1 \Rightarrow Similiraty \end{matrix} \\ Inverse if det (A) < 0 \Rightarrow Axial Symmetry \end{cases}

In the scheme is interpreted that the central symmetry is a rotation of $180^{\circ}$ .

This classification is only valid when:

In the axial symmetry, the symmetry axis is one of the axes of coordinates.
In the central symmetry or in the rotation, the center is the origin of coordinates.

In the cases that this is not satisfied, the scheme previously given is false. To do a general classification scheme a little more elaborated, then mathematical concepts would be needed.

Example

Let the equations system be $(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} 2 & 1 \\ - 1 & 1 \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix})$

We are going to see which kind of transformation it is. To start, we are going to calculate:

$det (A) = 2 - (- 1) = 2 + 1 = 3 > 0$

Therefore, it is a question of a direct transformation. Besides, since its determinant is different from $1$ , the transformation is a similarity, therefore, the above mentioned transformation trebles the distances between the points and the lengths of the segments of the plane.

Example

Given the following system $(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix})$

say, what type of transformation is it?

Since $det (A) = - 1$ , the transformation is inverse and the only inverse transformation is the axial symmetry. Therefore this is transformation symmetry of the $y$ -coordinate axis.

Example

Finally, we are going to give an example of draft. Consider the following equations system

$(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix})$

then the determinant of the system is:

$det (a) = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 > 0$

Therefore, it is a direct transformation. Also, as $d e t (A) = 1$ , is an isometric transformation and finally, as we have no term $b$ , it might be a rotation.