6
$\begingroup$

I am having difficulty understand how to get from a homogeneous transformation matrix:

$ M = \begin{pmatrix} R&t\\0&1 \end{pmatrix} $

Where $R$ is a rotation matrix (or it could be any linear transformation, I just chose rotation for this example) and $t$ is a translation vector.

To it's inverse definition:

$ M^{-1} = \begin{pmatrix} R^{-1}&-R^{-1}t\\0&1 \end{pmatrix} $

The part that I don't understand is the $-R^{-1}t$ term in the $(0,1)$ entry of $M^{-1}$.

My incorrect expectation was:

$ M^{-1} = \begin{pmatrix} R^{-1}&-t\\0&1 \end{pmatrix} $

So where does the extra multiplication by $R^{-1}$ come from?

Thank you! (:

$\endgroup$

2 Answers 2

8
$\begingroup$

If $y=Rx+t$, then $x=R^{-1}(y-t)=R^{-1}y-R^{-1}t$.

$\endgroup$
7
$\begingroup$

In terms of solid math, the user1551's answer is all you need.

For your intuition: the back-translation has to take into account that you do it in the rotated reference frame. The additional $R^{-1}$ in front of $t$ "cancels" that rotation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .