Given the poseo of point 1 between $t_k$ and $t_{k+1}$ \begin{equation} \mathbf{P}_{1}\left(t_{k}, t_{k+1}\right)=\left[\begin{array}{cc} \mathbf{R}_{1}\left(t_{k}, t_{k+1}\right) & \mathbf{T}_{1}\left(t_{k}, t_{k+1}\right) \\ 0 & 1 \end{array}\right] . \end{equation} and the relative pose between point 1 and 2, $\mathbf{P_{12}}$ why is the relative pose between $t_k$ and $t_{k+1}$ in point 2 given as:

\begin{equation} \mathbf{P}_{2}\left(t_{k}, t_{k+1}\right)=\mathbf{P}_{12} \mathbf{P}_{1}\left(t_{k}, t_{k+1}\right) \mathbf{P}_{12}^{-1} \end{equation}.

Why is this expression true?

My attempt: I have tried multiplying moth sides by $\mathbf{P}_{12}^{-1} = \mathbf{P}_{21}$.

I get:

$ \mathbf{P}_{21}\mathbf{P}_{2}\left(t_{k}, t_{k+1}\right)$ which in my opinion equals $\mathbf{P}_{1}\left(t_{k}, t_{k+1}\right)$ but instead according to the expression it equals $\mathbf{P}_{1}\left(t_{k}, t_{k+1}\right) \mathbf{P}_{12}^{-1}$.


1 Answer 1


Consider the graph below:

enter image description here

Starting from the point $2k+1$, we can count the transformations anticlockwise, ending up again at the same point; hence, the aggregate transformation shall be the identity matrix $I$.

In formula:

$$ P_{12}^{-1} \cdot P_1^{-1} \cdot P_{12} \cdot P_2 = I, $$

from which you can easily derive the original equation.


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