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I was astonished by ingenuity of many users who demonstrated reasons for why rotational matrices are not commutative.

However in 3d rotations I'm more puzzled by some other theorem ...

How intuitively to show that

composition of rotations about fixed axes of global frame is equal to composition of the same rotations about their current X,Y,Z axes but made in reverse order.

In the previous question the most interesting to me was example with permutations... Maybe someone also knows such nice examples..

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  • $\begingroup$ I was looking for a reference to the statement you just made. Where did you find this statement? In return, I'll provide you with an answer as to why this is so. $\endgroup$ Commented May 7 at 3:37
  • $\begingroup$ @StephenMontgomery-Smith The statement is used in robotics, most often without detailed explanation.. thank you for the answer(+1). Algebraically your explanation is the simplest., I suppose, from possible ones - but we need also a visualization in a more geometric way.. $\endgroup$
    – Widawensen
    Commented May 7 at 12:37

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Suppose you have two rotations, represented as $R_1$ and $R_2$ in the fixed frame. Then the composition $R_1 R_2$ means (according to the most commonly used convention) - first apply $R_2$, then apply $R_1$.

Now suppose we apply them in the opposite order, expressed by the same matrices, but with respect to the moving frame. We first apply $R_1$, then apply $R_2$.

For $R_1$, the fixed frame and the moving frame are the same, because nothing as happened before the rotation is applied.

For $R_2$, the moving frame is not the fixed frame. If the same rotation is expressed in the fixed frame, then the change of basis formula tells us that the matrix is $R_1 R_2 R_1^{-1}$. If we now compose them in the fixed frame doing $R_1$ first, and then $R_1 R_2 R_1^{-1}$, remembering that we multiply them right to left, we get $$ (R_1 R_2 R_1^{-1} ) R_1 .$$ And after multiplying out, you get $R_1 R_2$, as desired.

To visualize this, I find it helpful to consider a composition of a yaw and a pitch, and see how my hand moves.

This shows you why it is true for two rotations. For more than two rotations, you can use induction.

Note that this works for any linear motions. It is nothing to do with being a rotation. Indeed, it seems to me that with the proper interpretation, it would work for non-linear motions. It would certainly apply to more general rigid motions which include a translation as well as a rotation.

Added later: let's consider this example.

First notation. The axes are $x$ is forward, $y$ is left, $z$ is up. Pitch is about $y$, roll is about $x$, yaw is about $z$.

With my hand stretched out, I first do a pitch forwards of 30 degrees. Then I yaw to the left 30 degrees with respect to the fixed frame.

Or I first yaw to the left 30 degrees. Then I pitch forwards 30 degrees, but in the current frame.

You can see in the second case that, in the fixed frame, the pitch is along an axis which is 30 degrees to the left from the $y$-axis.

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  • $\begingroup$ However there is a problem with notation for such explanation. We should somehow discern between fixed and current frames for notation of rotations. Without it we simply obtain $R_1 R_2 = R_2 R_1$ but nowhere we have that rotations on left and right side are not the same ( i.e. made with different conventions). $\endgroup$
    – Widawensen
    Commented May 7 at 14:35
  • $\begingroup$ I thought I did distinguish between the different frames. (I called it "moving" instead of "current", but I did make a distinguishment.) $\endgroup$ Commented May 8 at 0:13
  • $\begingroup$ Yes, distinction in text for names is clear, but should we also need a distinction in symbols for equations? In text there are the same symbols used for rotation in fixed and moving frames - it would be useful to distinguish them. For example rotations in fixed frames could be with asterisk for example $R_{1*}$, in moving frame without: $R_1$. Then an equation above would be $R_1 R_2=R_{2*} R_{1*}$ which has no ambiguity. $\endgroup$
    – Widawensen
    Commented May 8 at 9:48
  • $\begingroup$ @Widawensen Fair enough. Maybe you could write out a solution here, based upon mine, but using different notation. $\endgroup$ Commented May 8 at 23:54
  • $\begingroup$ Yes, only notation is for elaboration... with this restriction your answer is very good. $\endgroup$
    – Widawensen
    Commented May 9 at 12:12
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We'll use three properties of rotations - they are isometries, conformal, and form a group under composition. Letting this group act on the canonical basis vectors we see that it maps them onto other unit vectors being isometries, and that the vectors remain orthogonal, because the map is conformal and so the image is also an orthonormal and linearly independent set of vectors defining another frame of reference. All that is left is to note that in any group $(xyz)^{-1}=z^{-1}y^{-1}x^{-1}$ and we are done.

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  • $\begingroup$ Interesting argumentation (+1) from the abstract algebra point of view, and elements of it can be inspiring but somehow I don't see (up to now) the whole picture. Rotations indeed transform ortonormal set ot vectors into other ortonormal sets and it can be somehow used and I wonder how many such sets we need to illustrate the thesis, 3 is minimum, but probably 4 would be more appropriate. I should add that after 3 years I did not find convincing illustration for this phenomenon, although I was searching. $\endgroup$
    – Widawensen
    Commented Jul 29, 2020 at 6:52
  • $\begingroup$ I think if you use your hand, and use a pitch and a yaw of around 30 degrees as the rotations, then the abstract argument I gave will become obvious. Let me add that to my answer. $\endgroup$ Commented May 8 at 1:56

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