# Finding the exponential relation between two 4x4 transition matrices

Im alright with matrices, but this question has dumb-struck me. Suppose I have two known and given $4\times4$ transition matrices, representing transitions in three dimensions with the fourth dimension representing a factor (I've forgotten the name of how such matrices are called), $T_1$ and $T_2$, and the relation between them is an exponential:

$$T_1 = e^{\{S\}\phi} * T_2$$

with $\phi$ being a scalar angle or matrix displacement, $S$ being another $4\times4$ Matrix, in the form of ($\omega$ represents a $3\times1$ rotation vector, $v$ is also a $3\times1$ 'displacement' vector) : $$S = \begin{bmatrix} 0 & -\omega_3 & \omega_2 & v_1 \\ \omega_2 & 0 & -\omega_1 & v_2 \\ -\omega_2 & \omega_1 & 0 & v_3 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

The question is to find $\omega$, $v$ and $\phi$.

This question was put in the middle of a more complexer quiz about screw-axis theorem, but here, I am just totally blank. Can someone help me out, and describe me how one should solve this problem, or give me a reference to a more detailed description? Or is it just easier to read these values directly from the difference between $T_1$ and $T_2$ ?

• Are you talking about Affine transformation matrices? – Omnomnomnom Jul 9 '14 at 14:42
• What is $\phi$ in terms of matrices? What exactly do you understand by $e^{\{S\}\phi}$? – TZakrevskiy Jul 9 '14 at 14:43
• Yes, the name I was looking for was Affine transformation matrices. Thanks for that ref... @TZakrevskiy $\phi$ is a scalar, $\{S\}$ is a $4\times4$ matrix. I understand that $e^{\{S\}\phi}$ can also be converted to a $4\times4$ matrix, but that wouldn't completely answer the asked for variables... – Florian Jul 9 '14 at 14:55
• Take $\bar S=\ln (T_1T_2^{-1})$, then renormalise $\bar S$ so that corresponding $\omega$ in renormalised $\bar S$ can be a rotational vector (if this renormalisation is not unique, then we are screwed). Like this obtain a possible value for $\omega,v,\phi$. – TZakrevskiy Jul 9 '14 at 15:03
• @TZakrevskiy I think there is a simpler process; you cant expect someone on a quiz to calculate $T_2^-1$ for $4\times4$ matrix, unless they have matlab which obviously isn't the expected case. I think Im missing something much simpler.. – Florian Jul 9 '14 at 15:07

I admit, I am partly to blame for perhaps not giving enough trivial information, but then, I didn't know myself what was vital information to know.

For $T_{02}=e^{[S]\phi}T_{01}$, $T_{01}$ and $T_{02}$ can be rewritten as:

$T_{01} = \begin{bmatrix} R_{01} & p_{01} \\ 0 & 1 \\ \end{bmatrix}, T_{02} = \begin{bmatrix} R_{02} & p_{02} \\ 0 & 1 \\ \end{bmatrix}$

This means that the homogeneous transformation representation of the screw motion between any two separate transformations $T_{02}=e^{[S]\phi}T_{01}$ equals to:

$e^{[S]\phi} = \begin{bmatrix} I & (I - e^{[\omega]\phi})q+h\phi\omega \\ 0 & 1 \\ \end{bmatrix}$

We therefore also get the relation:

$R_{02}=e^{[\omega]\phi}R_{01}$

With $R=e^{[\omega]\phi}$, this is a $3\times3$ matrix multiplication, and is generally not hard to find. The next step is to find $\phi$, which can be done by taking the trace of $R$ (sum of diagonal elements):

$tr(R)=1+2cos(\phi)$

And thus obtain $\phi$. Substituting this value into the exponential formula for the rotation matrix:

$R=e^{[\omega]\phi}=I+sin(\phi){\omega}+(1-cos(\phi))[\omega]^2$

Which for $\phi = pi$ follows

$R=I+2[\omega]^2$

And thus obtain $\omega$. To obtain h, we have, with $q$ being any arbitrary point on the screw axis that we manually choose:

$p_{02} = Rp_{01}+(I - R)q+h\phi\omega$

...And thus I also obtained finally $h$.