Homogeneous coordinates

← Back: Rotation

Overview: Transformations

Next: Combinations of transformations →

Rotation of n-dimensional coordinates can be described by an n-by-n matrix. All the components of the matrix are multiplied with one of the $n$ coordinates. So the transformation is dependent on the original coordinates. Translation however is actually a vector addition and so independent of the original coordinates (see subarticle about translation).

To be able to also apply a translation by matrix multiplication, an additional dimension is introduced. For three-dimensional space (which is usually regarded in robotics), the homogeneous transformation matrix $\mathbf{T}$ is a 4-by-4 matrix. $\mathbf{T}$ consists of the 3-by-3 rotation matrix $\mathbf{R}$ and a 3-by-1 translation vector $\vec{\mathbf{p}}$ combined with the last row of the identity matrix:

$\mathbf{T}= \left[\begin{array}{ccc|c} & & & \\ & \mathbf{R} & & \vec{\mathbf{p}}\\ & & & \\ \hline 0 & 0 & 0 & 1 \end{array}\right]= \left[\begin{array}{ccc|c} & & & \\ \vec{\mathbf{x}} & \vec{\mathbf{y}} & \vec{\mathbf{z}} & \vec{\mathbf{p}}\\ & & & \\ \hline 0 & 0 & 0 & 1 \end{array}\right]= \left[\begin{array}{ccc|c} & & & \mathbf{p}_x\\ \vec{\mathbf{x}} & \vec{\mathbf{y}} & \vec{\mathbf{z}} & \mathbf{p}_y\\ & & & \mathbf{p}_z\\ \hline 0 & 0 & 0 & 1 \end{array}\right]$

Let the components of the rotation matrix be denoted as follows:

to the standard transformation matrix matrix, where the top left n-by-n submatrix The additional component of vectors is always a 1. So the multiplication for $n=2$ for example results in:

$\left[\begin{array}{c} x_1\\ y_1\\ 1 \end{array}\right]= \mathbf{T} \cdot \left[\begin{array}{c} x_0\\ y_0\\ 1 \end{array}\right]= \left[\begin{array}{ccc} t_{11} & t_{12} & t_{13}\\ t_{21} & t_{22} & t_{23}\\ 0 & 0 & 1 \end{array}\right] \cdot \left[\begin{array}{c} x_0\\ y_0\\ 1 \end{array}\right]= \left[\begin{array}{c} t_{11}\cdot x_0+t_{12}\cdot y_0+t_{13}\cdot 1\\ t_{21}\cdot x_0+t_{22}\cdot y_0+t_{23}\cdot 1\\ 0\cdot x_0+0\cdot y_0+1\cdot1 \end{array}\right]= \left[\begin{array}{c} t_{11}\cdot x_0+t_{12}\cdot y_0+{\color{Green}\mathbf{t_{13}}}\\ t_{21}\cdot x_0+t_{22}\cdot y_0+{\color{Green}\mathbf{t_{23}}}\\ 1 \end{array}\right]$

As can be seen in the resulting vector, the components $t_{13}$ and $t_{23}$ are independent of the original coordinates and just added to the x- or the y-coordinate, respectively.