Talk:Euler's rotation theorem

message: Why is this article more about quaternions than it is about Euler rotation?
One reference to quaternion algebra is adequate.
The topic has been obfuscated by a writer who wants to steer readers.
65.175.151.8 15:08, 24 October 2006 (UTC)
replies: text: they are intimately connected, same thing in different disguises. IMHO, the section on quaternions could use some elaboration and clarification. Mct mht 01:42, 8 February 2007 (UTC)
thread_title: quaternions

message: * Technically one needs only 3 numbers to represent an arbitrary rotation, since the rotation axis can be normalized, with its magnitude representing the angle (I suspect that this closer to what Euler's rotation theorem states)
:yes, you're right. maybe that'll get corrected. Mct mht 01:42, 8 February 2007 (UTC)
replies: text: right there also. current version can certainly be improved. rotations corresponds to unit quaternions, in a "1 to 2" way, meaning unit quaternions is a double cover of SO(3). Mct mht 01:42, 8 February 2007 (UTC)
thread_title:  unclear about number of parameters and quaternion significance 

message: The statement 
An m×m matrix A has m orthogonal eigenvectors if and only if A is normal.
 is false. The statement should read
An m×m matrix A can be diagonalised if and only if A is normal. An orthogonal set of m eigenvectors can be obtained if A is hermitian. The eigenvectors of a rotation matrix are not in general mutually orthogonal.
:Really? "a matrix is normal if and only if its eigenspaces span Cn and are pairwise orthogonal with respect to the standard inner product of Cn" (Normal matrix#Consequences). Boris Tsirelson (talk) 10:33, 19 April 2010 (UTC)
thread_title:  Excursion into matrix theory 

message: I feel this is an important theorem that can be presented in a somewhat simpler way.  I will developed this in the near future and hope that you find it to be of value.Prof McCarthy (talk) 01:17, 23 June 2011 (UTC)
thread_title:  Proposed edits to this article 

message: I agree with the observations made by 190.188.3.11 in " The matrix proof is wrong ". It is at least incomplete, that is one should indeed prove that any two cartesian coordinate systems (with same orientation) are related by a rotation.
I will try to prove that with " geometric algebra " (I know, some people here will make a grimace ... ).
Let us consider a unit sphere which can slide on itself. After some displacement the extremities of an (e_1,e_2,e_3) reference frame will be situated at (f_1,f_2,f_3). We must be more precise : that means we define arbitrarily two new points (f_1,f_2) with the only condition  \mathbf(f_1). \mathbf(f_2) =0, which assures that we respect conservation of distances and angles. Of course f_3 is determined by {f_1} \times {f_2} = {f_3}.
Each point of the sphere can of course be defined by its (invariant) coordinates relative to both frames, by definition of the displacement of a rigid body.
We know that a rotation in (R^3) can be characterized by a rotor like :
(1) \qquad  R=\cos(\theta /2)+ I a \sin(\theta /2) \qquad \qquad R^{\dagger}=\cos(\theta /2)- I a \sin(\theta /2)
where  a  is the unit vector of the rotation axis, and  \theta  the rotation angle.
What we have to do now is to verify that the knowledge of the couples (e_1,f_1)  and (e_2,f_2) is necessary and sufficient to determine uniquely the rotor and thus the axis and rotation angle.
We must have :
(2) \qquad  f_1=R^{\dagger} e_1 R \qquad \qquad  f_2=R^{\dagger} e_2 R \qquad \qquad f_1.f_2=0 
We can write successively :
(3) \qquad  R f_1=e_1 R
(4) \qquad (\cos(\theta /2)+ I a \sin(\theta /2))f_1=e_1(\cos(\theta /2)+ I a \sin(\theta /2))
(5) \qquad \cos(\theta /2)f_1+\sin(\theta /2) I (a.f_1+a \wedge f_1)=\cos(\theta /2)e_1+\sin(\theta /2) I (a.e_1-a \wedge e_1)
(6) \qquad \cos(\theta /2)(f_1-e_1)+ \sin(\theta /2)I a \wedge (f_1+e_1)+\sin(\theta /2)I a . (f_1-e_1)=0
and :
(7) \qquad \cos(\theta /2)(f_2-e_2)+ \sin(\theta /2)I a \wedge (f_2+e_2)+\sin(\theta /2)I a . (f_2-e_2)=0
By splitting in pseudoscalar and vectorial parts we get:
(6 bis) \qquad a.(f_1-e_1)=0 \qquad \qquad \cos(\theta /2)(f_1-e_1)+\sin(\theta /2)I a \wedge (f_1+e_1)=0
(7 bis) \qquad a.(f_2-e_2)=0 \qquad \qquad \cos(\theta /2)(f_2-e_2)+\sin(\theta /2)I a \wedge (f_2+e_2)=0
As a consequence of the scalar relations we can write :
(8)\qquad a=\lambda I (f_1-e_1)\wedge(f_2-e_2)
where \lambda is the inverse of the norm of (f_1-e_1)\wedge(f_2-e_2).
We may calculate : \lambda=[(f_1-e_1)^2(f_2-e_2)^2-(f_1-e_1).(f_2-e_2)]^{(-1/2)}
As for the vectoriel relations, they should be coherent. After replacing a by its value and simplifying in (6bis) and (7bis) we get :
(6ter) \qquad \cos(\theta /2)-\lambda \sin(\theta /2)(f_2-e_2).(f_1+e_1)=0
(7ter) \qquad \cos(\theta /2)+\lambda \sin(\theta /2)(f_1-e_1).(f_2+e_2)=0
These equations coincide if :
(9)\qquad -(f_2-e_2).(f_1+e_1)=(f_1-e_1).(f_2+e_2)
That is true if and only if the condition \mathbf (f_1) . \mathbf (f_2) =0  is verified.
Thus we get:
(10) \qquad \cot(\theta /2)=\lambda (f_2-e_2).(f_1+e_1)=\lambda (f_2 .e_1-f_1 .e_2) 
and the rotation is unambiguously determined.
Obviously the same demonstration can be made starting with non-orthogonal points and the condition f_1.f_2=e_1.e_2 .
Chessfan (talk) 20:02, 20 January 2012 (UTC)
thread_title:  Demonstration with geometric algebra. 

message: As far as I know, the single rotation axis is typically called "Euler axis". I guess that the "Euler pole" is not the Euler axis, but just the fixed point around which the body rotates. See this image, for instance: http://www.earth.northwestern.edu/people/seth/202/new_2004/euler_pole.html Paolo.dL (talk) 16:30, 2 May 2013 (UTC)
replies: text: Bad guess — it's rather the point where the axis intersects the sphere surface. From what I can understand, Euler pole is mainly a term in geology, even if the basis for it is Euler's rotation theorem. Euler pole used to be a separate article, but got merged into this one, apparently without preserving enough context, so subsequent edits of this math article removed all traces of an explanation of the term. Today I changed the redirect to one actually giving an explanation. 78.73.97.76 (talk) 18:41, 11 August 2016 (UTC)
thread_title:  Euler pole 

message: The introduction states 1775, while the first heading states 1776 as year of proof. Since I can only find 1776 in other sources, is 1775 incorrect? Or is it some difference between year of proof and publication date? Anyone with sources on that?
2003:62:4D10:E301:350D:6E9D:7F67:A333 (talk) 13:58, 14 January 2015 (UTC)
thread_title:  Year of proof 

message: Note 2 explains that A^\dagger indicates an adjoint (conjugate transpose). This is not standard notation. Typically the adjoint of a matrix/operator is denoted A^*. A dagger usually indicates the Moore–Penrose pseudoinverse.
I'm not really in this Wikipedia editing universe, so I don't want to change something that might make somebody howl for a reason I don't understand. Otherwise, I would fix it myself.  — Preceding unsigned comment added by 2601:1C0:C901:569C:DAA2:5EFF:FE96:5FDD (talk) 05:46, 28 June 2015 (UTC)
replies: text: One sees numerous notations for the Hermitian adjoint in the literature. A^* is probably more common than A^\dagger, but the latter is by no means unheard of, and traditions vary between different fields. The pseudoinverse is more commonly A^+ than A^\dagger, I think. 78.73.97.76 (talk) 19:03, 11 August 2016 (UTC)
thread_title:  Notation for adjoint 

message: Hello. In the first paragraph, "It also means that the composition of two rotations is also a rotation." is not untrue? I was going to insert "of the same center", but a confirmation would be appreciated. Thank you. 81.196.9.170 (talk) 10:38, 3 February 2018 (UTC)
thread_title:  Composing rotations