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I'm familiar with the rotation matrix and its three-angle parametrizations (i.e. Euler's angles). I'm stumbling now on axis-angle representation. In the book I'm reading, it says

One approach is to note that the rotational transformation $R = R_{z,α}R_{y,β}$ will bring the world z-axis into alignment with the vector $k$. Therefore, a rotation about the axis $k$ can be computed using a similarity transformation as

$$ \begin{align*} R_{k,\theta} &= RR_{z,α}R^{-1} \\ &= R_{z,α}R_{y,β} R_{z,\theta} R_{y,-β}R_{z,-α} \end{align*} $$

I'm not able to visually understand why the reverse of the first sequence of rotations is carried out.

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CroCo
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The general process works like this - rotate $k$ to align with the $z$-axis, then rotation around the $z$-axis, rotate $k$ back to its original position. Since you are familiar with Euler angles I'm sure you know that rotation do not commute in three dimensions so it's important to undo them in the right order. You'll often see transformations of the form $XAX^{-1}$ in non-commutative settings such as this.

CyclotomicField
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  • Your description doesn't account for rotation about y-axis. Also, is it possible to visually explain this? – CroCo Dec 10 '21 at 20:06
  • @CroCo I accounted for it when we rotated $k$ to and from the $z$ axis. You must undo the operations in the reverse order you did them. Prove that $(R_1R_2)^{-1}=R_2^{-1}R_1^{-1}$ then convince yourself this is true geometrically. In fact, most things must be undone in the opposite order you did them. Imagine assembling a fine watch then disassembling it. You would start with tiny gears inside but if you tried to take it apart starting with the tiny gears it would be impossible. You have to start where you finished and work backwards. – CyclotomicField Dec 10 '21 at 20:15
  • Is it the idea here to transform $k$ axis so that it is treated as a principle axis then rotate about it in the usual way and returns it back to its original position? If so, it means I can transform the $k$ axis in the $x$ principle axis or $y$ principle axis, ending up with different representations, right? – CroCo Dec 10 '21 at 20:19
  • @CroCo you can choose any vector as your principle axis it's just convention to use the $z$-axis since it reduces the problem to the two dimensional case by putting the rotation in the $x,y$-plane. Note that I didn't use the fact that they were rotation explicitly, just that the process is invertible at each step so I could retrace my steps. This means you can apply similar reasoning in cases where the transformations are more general than this. – CyclotomicField Dec 10 '21 at 20:29
  • @CyclotomicField I don't know if my understanding is correct but I have always thought similar transformations as transformations in a different language (basis). Say we apply some transformation $T$ (e.g. scale) on a vector $x$. In basis $a$, we can transform our point as $x'_a = T_a \cdot x_a$. For basis $b$, we have to translate back to basis $a$, do the transformation with $T_a$ and finally, transform back to basis $b$. That is: $x'_b = T_b x_b = P^{-1} T_a P \cdot x_b$. – user599310 Mar 17 '24 at 17:10
  • @CyclotomicField In the rotation example, I can understand that is easier to perform the rotation around the principle axis. However, I am struggling to see how it is a change of basis Wiki. In a change of basis, we just produce a new coordinate vector for the same vector. But here, during the alignment of $k$ with the $z$-axis, we have also changed our vector. We undo the first rotation, do the new rotation and finally we reapply the first rotation. Why this order of operations give the "correct" orientation if rotations don't commute? – user599310 Mar 17 '24 at 17:39
  • @user599310 To see how it's a change of basis instead of considering moving $k$ to the $z$-axis we go in the other direction and move the entire coordinate frame so that the $z$-axis aligns with $k$. Then we rotate the $x,y$-plane around the $z$-axis and finally undo the original rotation on the entire frame. Now we can see that the $z$-axis returns to its original position as does the $x,y$-plane. However the basis vectors in the $x,y$-plane have rotated so $k$ now has a new coordinate representation. It's basically how if you spin clockwise the world spins anticlockwise. – CyclotomicField Mar 17 '24 at 23:22