Proof of the extrinsic to intrinsic rotation transform

Question

Wikipedia states that:

Any extrinsic rotation is equivalent to an intrinsic rotation by the same angles but with inverted order of elemental rotations, and vice-versa. For instance, the intrinsic rotations $x-y’-z''$ by angles $\alpha, \beta,\gamma$ are equivalent to the extrinsic rotations $z-y-x$ by angles $\gamma, \beta, \alpha$.

Is there a simple proof why this is so?

Answer 1

I googled on the Internet, but also found not much material on this topic. The most relevant references are from Wikipedia as @Apoo suggested, and a blog the proof is at bottom of the page. Though the arguments are quite complete, neither of them could convince me. In order to prove the statement, I have to first introduce changing a transformation between coordinate systems. There is a derivation from blender stackexchange, and I excerpt the equation as follows: $$ T_{\text {world}}=S_{\text {world}} \times T_{s} \times S_{\text {world}}^{-1}, \tag{1} $$ where $T_{\text {world}}$ is the transformation matrix in world coordinate system, $S_{\text {world}}$ is the world matrix of the referenced object $s$, $T_{s}$ is the local transformation matrix based on $s $.

Assume that an intrinsic Euler angle can be represented as a product of three rotation matrices $$ R = Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha). $$

The goal is to prove, there exists an extrinsic rotation sequence, s.t.

$$ Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha) = X(\alpha)Y(\beta)Z(\gamma). $$

Consider $Y(\beta)$ as a rotation matrix around $y^\prime$ of angle $\beta$ relative to $x^\prime - y^\prime - z^\prime$ coordinate system, we can obtain the corresponding rotation matrix in $x-y-z$ coordinate system using conversion eq. $(1)$: $$ Y^\prime = X Y X^{-1}, $$ it follows that $$ Y = X^{-1} Y^\prime X . \tag{2} $$ Through a similar argument, we can also get the world rotation matrix of angle $\gamma$ around $z^{\prime \prime}$ in $x-y-z$ coordinate system: $$ Z^{\prime \prime} = (Y^\prime X) Z (Y^\prime X)^{-1}, $$ and it follows that $$ Z = (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X). \tag{3} $$

Multiplying $X$ by eq. $(2)$ and eq. $(3)$, we obtain $$ \begin{align} X Y Z &= X X^{-1} Y^\prime X (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X) \\ &= Z^{\prime \prime} Y^{\prime} X , \end{align} $$ which completes the proof.

Answer 2

Hey I am not an expert in rotation matrix transforms but while going through few things for my thesis, I came across the following link : https://en.wikipedia.org/wiki/Davenport_chained_rotations If you scroll down the page to the section - The proof of the conversion in the pre-multiply case. I think this is what you are looking for. I was kind of searching for an answer to the same stuff. Hope it is helpful...good luck with your work!!

Answer 3

The statement is valid not only for rotations. According to relativity, there is no preferred frame of reference / coordinate system. Therefore, in kinematics, all (times and) positions and motions are relative. (Dynamics - with forces - is another matter).
Consider a two-dimensional example, as has been found in a book about Computer Graphics (to be precise: J.D. Foley, A. van Dam, Fundamentals of Interactive Computer Graphics, 1982). There are two frames of reference, one attached to the observer (world), one attached to an object (chair):

With a transformation of coordinates, the only thing that is important is the relative position of the object with respect ot the observer. This means that the end result of a coordinate transformation can be achieved in at least two ways. As is displayed in the example:

• Extrinsic.  Rotate $\,R\,$ the chair in the world coordinate system over an angle of $45^o$
  and then Translate $\,T\,$ it over a distance $(4,10)$. Thus resulting in a transformation $\,TR$ .
• Intrinsic.  Translate $\,T^{-1}\,$ the observer in the chair coordinate system over a distance $(-4,-10)$ and then rotate $\,R^{-1}\,$ the world coordinate system over an angle $-45^o$. Thus resulting in a transformation $\,R^{-1}T^{-1}$ .

With this simple example, we can see immediately that the transformations are the inverse of each other: $\,R^{-1}T^{-1}= (TR)^{-1}$ .
Hope you get the idea. Generalizing this to three dimensions is expected to be a matter of filling in the (somewhat more involved) technicalities.

Update. Hmm, "not quite a proof". Then perhaps this.
Let the coordinate system of the object be called $O$ and the world coordinate system be called $W$. Both coordinate systems are coincident in the beginning.
The first step is to apply a transformation $R$ to $O$   (just as in the example, but in general now).
The second step is to make $W$ coincident again with $O$, which is done by applying the same transformation $R$ to $W$ as has been done in the first step with $O$.
Then effectively nothing has changed and we have the original configuration again: the product of step (1) and step (2) is the identity.
It is thus obvious that the first step could also have been accomplished by applying the inverse $R^{-1}$ transformation to $W$ instead of $O$.
Common properties of inverse operations like $(AB)^{-1}= B^{-1}A^{-1}$ are assumed throughout to be well known. This completes the proof.