Inverse transformation

← Back: Combinations of transformations

Overview: Transformations

Next: ??? →

Let $\mathbf{T}$ be a general homogeneous transformation matrix. The inverse transformation $\mathbf{T}^{-1}$ corresponds to the transformation that reverts the rotation and translation effected by $\mathbf{T}$. If a vector is pre-multiplied by $\mathbf{T}$ and subsequently pre-multiplied by $\mathbf{T}^{-1}$, this results in the original coordinates because $\mathbf{T}^{-1}\mathbf{T}=\mathbf{I}$ and multiplication with the identity matrix does not change anything (see transformations).

The general homogeneous transformation matrix $\mathbf{T}$ for three-dimensional space consists of a 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:

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

As stated in the article about homogeneous coordinates, multiplication with $\mathbf{T}$ is equivalent in cartesian coordinates to applying the rotation matrix $\mathbf{R}$ first and then translating the coordinates by $\vec{\mathbf{p}}$:

$\vec{\mathbf{q}}_1= \mathbf{T} \cdot \vec{\mathbf{q}}_0 \equiv \mathbf{R}\cdot \vec{\mathbf{q}}_0 + \vec{\mathbf{p}}$

The equation above is now solved for $\vec{\mathbf{q}}_1$. Here the fact is used, that the inverse of 3-by-3 rotation matrices equals the transpose of the matrix:

$\begin{align} \vec{\mathbf{q}}_1&=\mathbf{R} \vec{\mathbf{q}}_0 + \vec{\mathbf{p}} & & \\ \vec{\mathbf{q}}_1-\vec{\mathbf{p}}&=\mathbf{R} \vec{\mathbf{q}}_0 & & \\ \mathbf{R}^{-1}(\vec{\mathbf{q}}_1-\vec{\mathbf{p}})&=\mathbf{R}^{-1}\mathbf{R} \vec{\mathbf{q}}_0 & & \\ \mathbf{R}^{T}(\vec{\mathbf{q}}_1-\vec{\mathbf{p}})&=\mathbf{R}^{T}\mathbf{R} \vec{\mathbf{q}}_0 & & \\ \mathbf{R}^{T}\vec{\mathbf{q}}_1-\mathbf{R}^{T}\vec{\mathbf{p}}&=\vec{\mathbf{q}}_0 \qquad\quad \rightarrow \qquad\quad \vec{\mathbf{q}}_0=&\underbrace{\mathbf{R}^{T}}&\vec{\mathbf{q}}_1+(\underbrace{-\mathbf{R}^{T}\vec{\mathbf{p}}})\\ & &\mathbf{R}_i& \qquad\qquad \vec{\mathbf{p}}_i\\ \end{align}$