# Geometric transformation classification scheme

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

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

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

$$\left\{ \begin{array}{l} \text{Direct if } \det(A)>0 \left\{ \begin{array}{l} \text{Isometric if } \det(A)=\pm1 \left\{ \begin{array}{l} \text{Translation if } \vec{b}\neq 0 \\ \text{Rotation if } \vec{b}=0 \end{array} \right. \\ \text{Isomorph si} \det(A)=\pm1, \ k>1 \Rightarrow \text{Similiraty} \end{array} \right. \\ \text{Inverse if } \det(A)<0 \Rightarrow \text{Axial Symmetry} \end{array} \right.$$

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.

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

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.

Given the following system $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$

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.

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

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\ -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$

then the determinant of the system is:

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

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