Homogeneous coordinates

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In this article three-dimensional space is regarded because it is usually used in robotics. Rotation of three-dimensional coordinates can be described by an 3-by-3 matrix. All the components of the matrix are multiplied with one of the three 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 vectors, the additional fourth component is always 1. So a three-dimensional vector $\vec{\mathbf{q}}$ in homogeneous coordinates looks a sfollows:

$\vec{\mathbf{q}}= \left[\begin{array}{c} q_x\\ q_y\\ q_z\\ 1 \end{array}\right]$

A homogeneous transformation matrix $\mathbf{T}$ for three-dimensional space 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:

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

Let the components of the rotation matrix $\mathbf{R}$ and the translation vector $\vec{\mathbf{p}}$ be denoted as follows:

$\mathbf{R}= \left[\begin{array}{ccc} r_{11} & r_{12} & r_{13}\\ r_{21} & r_{22} & r_{23}\\ r_{31} & r_{32} & r_{33} \end{array}\right], \qquad \vec{\mathbf{p}}= \left[\begin{array}{c} p_x\\ p_y\\ p_z \end{array}\right]$

So the multiplication of a homogeneous transformation matrix $\mathbf{T}$ with a vector $\vec{\mathbf{q}}_0$ in homogeneous coordinates leads to:

$\vec{\mathbf{q}}_1= \left[\begin{array}{c} x_1\\ y_1\\ z_1\\ 1 \end{array}\right]= \mathbf{T} \cdot \vec{\mathbf{q}}_0 = \left[\begin{array}{cccc} r_{11} & r_{12} & r_{13} & p_x\\ r_{21} & r_{22} & r_{23} & p_y\\ r_{31} & r_{32} & r_{33} & p_z\\ 0 & 0 & 0 & 1 \end{array}\right] \cdot \left[\begin{array}{c} x_0\\ y_0\\ z_0\\ 1 \end{array}\right]= \left[\begin{array}{ccc} t_{11} & t_{12} & t_{13}\\ t_{21} & t_{22} & t_{23}\\ 0 & 0 & 1 \end{array}\right] \cdot \left[\begin{array}{c} x_0\\ y_0\\ 1 \end{array}\right]= \left[\begin{array}{c} t_{11}\cdot x_0+t_{12}\cdot y_0+t_{13}\cdot 1\\ t_{21}\cdot x_0+t_{22}\cdot y_0+t_{23}\cdot 1\\ 0\cdot x_0+0\cdot y_0+1\cdot1 \end{array}\right]= \left[\begin{array}{c} t_{11}\cdot x_0+t_{12}\cdot y_0+{\color{Green}\mathbf{t_{13}}}\\ t_{21}\cdot x_0+t_{22}\cdot y_0+{\color{Green}\mathbf{t_{23}}}\\ 1 \end{array}\right]$

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.