Talk:Rotation matrix

Mistake in rotation matrix in introduction

The introduction says that the 2x2 matrix it presents is for counterclockwise rotation but I believe the matrix presented there is for clockwise rotation given what is stated later in the article. — Preceding unsigned comment added by 94.12.234.26 (talk) 19:15, 21 February 2013 (UTC)

Agreed, I found that all the matrix transformations in this article were for clockwise rotations. I don't trust myself, though, so if this is thirded, please change it. 24.217.199.106 (talk) 20:34, 15 July 2013 (UTC)

Yeah I agree, the matrix given in the introduction is clearly for a clockwise rotation — Preceding unsigned comment added by 82.32.167.51 (talk) 12:00, 1 May 2014 (UTC)

That was due to an error in the previous edit. I've reverted back to before that edit, as giving the matrix for a counter-clockwise rotation is better: more consistent with the rest of the angle and showing the 'normal' way of rotating in a positive sense.--JohnBlackburne^words_deeds 12:14, 1 May 2014 (UTC)

That mistake is still there....

Introduction query

Don't see why the 2x2 matrix given in the introduction applies to the x-y plane specifically. It applies to any 2D subspace with a defined origin/axis surely.

Incorrect determinant?

One of the examples claims that the determinant of

$Q={\begin{bmatrix}-1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}$

is -1, so it is not a rotation matrix. The determinant is actually 1, which I believe should make it a rotation matrix.

...Agreed.

12.159.28.3 (talk) 03:49, 20 September 2009 (UTC)

A bit late to reply but I only just noticed this message: I noticed this and fixed this a couple of weeks ago.--JohnBlackburne^words_deeds 09:23, 2 February 2010 (UTC)

Rotation tensor

30 March 2007 - Leonid

Hi everyone!

I just added the bit about the Rotation Tensor representation and Rotation Matrix invariance with respect to change of coordinate frame. I think this invariance note is important in "Rotation Matrix" talks. From the other side, dealing with invariant objects is much more convinient especially if you start doing some advanced stuff like Elasticity Theory and Mechanics of Beams and Shells.

Does rotation tensor thing deserve a page on it's own?

There is some similar bit on "Representation of Rotation" page, but the guy whom wrote that page is cleary in love with quaternions. I am in love with quaternions as well the moment some simulation of rigid body dynamics need to be done, but theory of rotation tensors is very old and deserve it's place (I think). The whole Classical Ellasticity Theory is built on it.

Regards, Dr.Leonid Paramonov PDRA, Imperial College London, UK

17:41, 24 August 2006 - Alanic

I did the changes I talked about and I'm ready to defend them if you don't agree. The big matrices in "three dimensions" section may need an update, too. This was my very first contribution to Wikipedia and I didn't read any help pages, so feel free to cancel my changes if I'm not abiding any rules.

13:25, 22 August 2006 - Alanic

${\begin{pmatrix}\cos {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\cos {\alpha }\\\end{pmatrix}}.{\begin{pmatrix}1\\0\\\end{pmatrix}}={\begin{pmatrix}\cos {\alpha }\\-\sin {\alpha }\\\end{pmatrix}}$ ${\begin{pmatrix}\cos {\alpha }&\sin {\alpha }&0\\-\sin {\alpha }&\cos {\alpha }&0\\0&0&1\end{pmatrix}}.{\begin{pmatrix}1\\0\\0\\\end{pmatrix}}={\begin{pmatrix}\cos {\alpha }\\-\sin {\alpha }\\0\\\end{pmatrix}}$

The first equation shows that the matrix rotated the vector clockwise. This conflicts with the definition that rotation matrices rotate vectors counter-clockwise.

According to the right hand rule, the second equation shows that this matrix rotated this vector around the -z axis. This conflicts with the definitions on the page that a rotation matrix is a matrix that when multiplied with a vector it rotates the vector counter clockwise and that this matrix is a rotation around the z axis. I think all matrices in this page need to be transposed.

67.122.123.121 21:19, 21 August 2006 (UTC) - Geoff Dolan

• Hopefully not stepping on any feet, but I added a minus sign on the 21 term of the three dimensional rotation matrix. Besides disagreeing with my text (Sidi, Spacecraft Dynamics and Control, 1997), it's easy to program the matrix into matlab and see that it isn't a rotation matrix without the minus sign.

I think the signs for the rotation about the z-axis on the 12 and the 21 entries are wrong. Can someone confirm this?

I haven't looked at the other matrices to see if they're wrong too. Thanks

• I don't think they're wrong. They look exactly how I've learnt them on the University a few months ago :)

Torzsmokus 20:36, 3 January 2006 (UTC)

• It depends on how you define your positive angle. Conventions differ.

• I've added a yaw-pitch-roll system today and I've adapted the signs to the current status (out of respect to the authors). However I must say, that at my university the yaw-pitch-roll system was defined like this:

${\mathcal {R}}(\gamma ):={\begin{pmatrix}1&0&0\\0&\cos {\gamma }&\sin {\gamma }\\0&-\sin {\gamma }&\cos {\gamma }\end{pmatrix}}$, ${\mathcal {P}}(\beta ):={\begin{pmatrix}\cos {\beta }&0&-\sin {\beta }\\0&1&0\\\sin {\beta }&0&\cos {\beta }\end{pmatrix}}$, ${\mathcal {Y}}(\alpha ):={\begin{pmatrix}\cos {\alpha }&\sin {\alpha }&0\\-\sin {\alpha }&\cos {\alpha }&0\\0&0&1\end{pmatrix}}$. For the 2-dimenional matrix I agree that it usually is defined like this: ${\begin{pmatrix}\cos {\theta }&-\sin {\theta }\\\sin {\theta }&\cos {\theta }\end{pmatrix}}$ Wedesoft 21 Mar 2006 22:12 BST

• All given matrices are correct. But it should read "with the equivalent counter-clockwise rotation in \mathbb{R}^2". I'll correct this immediately. (AK, 2006-04-05)

• I think the roll and pitch angles are defined in a weird way. I think roll is rotation about the Y axis and pitch is rotation about the X axis. (MTL, 2011-08-16) — Preceding unsigned comment added by 204.11.231.206 (talk) 19:09, 16 August 2011 (UTC)

rewrite by TomViza

I just completed a major rewrite, this article is now much more generalised and also more easy to read. I moved some stuff in from Rotation (mathematics) which is very long.

I also added the formula to find the matrix in terms of the Euler angles.

TomViza 16:41, 21 May 2006 (UTC)

---

The following is the derivation of the second function in the 3D section. When I moved the equation from Rotation (mathematics), I thought that the derivation was not very encyclopdic, but have kept it here for thouroughness. TomViza 16:41, 21 May 2006 (UTC)

Derivation. This matrix is derived from the following vector algebraic equation (see dot product, cross product, and matrix multiplication):

$\mathbf {u'} =(\cos \theta )\mathbf {u} +(1-\cos \theta )\mathbf {v} (\mathbf {v} \cdot \mathbf {u} )+\sin \theta (\mathbf {v} \times \mathbf {u} ),\qquad \qquad (1)$

which in turn is derived from

$\mathbf {u'} =\mathbf {u_{\|}} +(\cos \theta )\mathbf {u_{\perp }} +\sin \theta (\mathbf {v} \times \mathbf {u_{\perp }} ).$

Here

$\mathbf {u_{\|}} =\mathbf {v} (\mathbf {v} \cdot \mathbf {u} ),$

$\mathbf {u_{\perp }} =\mathbf {u} -\mathbf {u_{\|}} ,$

$\mathbf {v} \times \mathbf {u_{\perp }} =\mathbf {v} \times \mathbf {u} ,$

which shows that u is resolved (see Gram-Schmidt process) into a parallel and a perpendicular component (to v). The parallel component does not rotate, only the perpendicular component does rotate. This rotation is similar to a two dimensional rotation, except that instead of x and y axes, there are $\mathbf {u_{\perp }}$ and $\mathbf {v} \times \mathbf {u_{\perp }}$ axes, both of which are perpendicular to v.

Quaternions

It would be nice with a link to quaternions

There is a formula for quaternion -> rotation matrix. This is told to be "left-handed (Post-Multiplied)". This does not make sense for me. "Post-Multiplied" seems to be very OpenGL/Direct3D (computer graphics) oriented, and should not be mentioned. Also, A "Post-Multiplied" matrix M seems to be that the (vector) bases A, B, C for the transformation are placed in each of the rows of the matrix. Hence we must multiply the vector x after the matrix M (xM) which is the opposite of the usual way (Nx, where N is the matrix with bases in each column). (M = N^T). Correct me if this is wrong. — Preceding unsigned comment added by 46.15.111.125 (talk) 11:01, 28 February 2014 (UTC)

Me again... I found this link: http://seanmiddleditch.com/journal/2012/08/matrices-handedness-pre-and-post-multiplication-row-vs-column-major-and-notations/, so I was wrong about Post-Multiplication, Post-Multiplication means "matrix on the left of the vector". Hence Post-Multiplication is the usual way (mathematical notation). False statement above striked out.

However, "Post-Multiplied" in the text should be removed. It does not make sense, since we use the standard mathematical notation when we represent a transformation with a matrix. — Preceding unsigned comment added by 46.15.111.125 (talk) 12:24, 28 February 2014 (UTC)

Agreed that "left-handed (Post-Multiplied)" is quite confusing, especially since the formulas agree with the right-handed basic rotation matrices.--olfactoryScientist (talk) 11:21, 7 May 2014 (UTC)

I'm quite confused with this part. There is often some confusion between pre/post-multiplication and left/right-handeness. This formulation "left-handed (Post-Multiplied)", encourages to think that "Post-Multiplied" is a synonym of "left-handed" just because the matrix is on the left of the multiplied vector. I even have some doubt that the author was doing this confusion himself. Can someone clarify this point ?

I'm also confused about the left handed.Dose this mean a left hand axis?or left handed rule rotations? — Preceding unsigned comment added by 118.9.136.96 (talk) 09:47, 15 May 2016 (UTC)

The convention used in this article is pre-multiplication (matrix on the left), and right-handed axes. Of course, exactly the same results are obtained if you use the unconventional convention of left-handed axes and post-multiplication. Dbfirs 15:56, 15 May 2016 (UTC)

Rotation matrix vs orthogonal matrix

Hi, just came across this article and have a a couple of question.

First, it says in the definition part that a rotation matrix is equivalent to an orthogonal matrix. If this is so, why is there a separate article on rotation matrices when there already is one on orthogonal matrices. What additional information is provided here which does not fit in the article on orthogonal matrices.

Second, is the equivalence of rotation and orthogonal matrices established in the literature? Personally, I would suggest that a rotation matrix is a special case of an orthogonal matrix (for the n-dim case) which only has two eigenvalues not equal to one [my correction]. Such a matrix always appears as a generalization of a 2D rotation for the the n-dim case in the sense that it has one well-defined rotation space in which it rotates with one well-defined angle. Also, in the 2D and 3D cases such a matrix is equivalent to an orthogonal matrix, but in 4 and higher dimensions a general orthogonal matrix is the product of two or more such matrices. Don't know if this is an established way of defining rotation matrices. --KYN 18:48, 12 July 2006 (UTC)

No, a rotation matrix is an orthogonal matrix with the additional restriction that the determinant is +1. Thus

${\begin{bmatrix}0&1\\1&0\end{bmatrix}}$

is an orthogonal matrix, but since its determinant is −1 it is not a rotation matrix. The eigenvalues of a rotation matrix are guaranteed to include a single +1 in odd dimensions, but otherwise may have as many repetitions of −1 or complex conjugate pairs with magnitude +1 as the dimension allows. Thus

${\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}$

is a rotation matrix with four eigenvalues, none of which are +1. We can state the definition as

• An n×n matrix M is a rotation matrix if M^TM = I and if det(M) = +1.

Planar rotations are a very special case, and only 3D rotations have a rotation axis.

We can derive the algebraic conditions from the geometric statement that a rotation is a direct isometry leaving one point fixed. Take the fixed point to be the origin so we are working with a Euclidean inner product space.

• Then isometry means preservation of distances (and by implication, angles), which is equivalent to preservation of the inner product. In vector form the inner product of a vector with itself is v^Tv. Therefore an isometry satisfies (Mv)^T(Mv) = v^Tv, for all vectors v. Rewrite the left-hand side as v^T(M^TM)v, and rewrite the right-hand side as v^TIv; then to obtain equality for all v we must have M^TM = I, as stated.
• For an isometry to be direct it must not reverse "handedness". The identity transformation is obviously direct, and the identity matrix is a (null) rotation, with determinant +1. Furthermore, in the Lie group of Euclidean isometries there must be a connected path from the identity to any direct isometry. Thus in the orthogonal group of n×n matrices there are two disjoint connected components: the special orthogonal group, which contains the identity and all of whose members have determinant +1; and the remaining component (which is only a coset, not a group), all of whose members have determinant −1.

Apparently the article could use some work. --KSmrq^T 13:10, 21 August 2006 (UTC)

This is all very well, but it does not justify that a direct (or special) orthogonal matrices are called rotation matrices in dimensions greater than 3. The rotation article only talks about dimensions 2 and 3, and says that a rotation is about a point or an axis (which means only two eigenvalues not 1). Also this article talks almost exclusively about dimensions 2,3. I don't think the usage (and meaning) of the term rotation in dimensions 2 and 3 is well established, and in any case it is not sourced here. Also, as per WP:LEAD the lead should summarize the article, so in particular not introduce generality that is not discussed in the article. So I will remove mention of dimension greater than 3 from the lead. Marc van Leeuwen (talk) 06:32, 21 August 2011 (UTC)

Use of homogenous Coords?

Hello,

This may be confusing for people who have to implement this with translation compounded, as there is no section on homogenous coords, should this be added? 129.78.208.4 02:51, 7 November 2006 (UTC)

Added generators, reformat

I have added a description of the generators of the group, and reorganized things so that the roll, pitch, and yaw matrices come first. That way, the other representations are more easily understood as compositions of these basic rotations. I also added expressions for the generators, except the Euler angle matrix, because I don't have an expression for that generator. Nothing has been deleted, but some things have been rephrased. PAR 20:52, 5 January 2007 (UTC)

How to find the generator A?

I'm reading the section about the properties of the rotation matrix...if I have M, but I want to find the generator A, what is the best way to do this? --HappyCamper 22:27, 26 April 2007 (UTC)

Transpose as inverse

Hmm...it doesn't seem as obvious to me that the transpose of a rotation matrix is its inverse as it perhaps was to the writer of this article. Might this section perhaps be altered? I'm not actually certain of the best clear way of showing this - perhaps showing that under the appropriate orthogonal basis, the leading diagonal is 1's except for 2 Cos(theta)s, and everywhere else in 0 except the sin thetas, then simply subbing in -theta for theta? Wrayal 15:17, 13 May 2007 (UTC)

My way of showing it would be to point out that the length of vectors shouldn't change under rotation, so that

$\Vert {\underline {x}}\Vert =\Vert R{\underline {x}}\Vert$

${\underline {x}}^{T}{\underline {x}}=\left(R{\underline {x}}\right)^{T}\left(R{\underline {x}}\right)={\underline {x}}^{T}R^{T}R{\underline {x}}$

$R^{T}R=I$

Using this property you can also show why the determinant must be (plus or minus) one:

$det(A)=1/det(A^{-1})$

$det(R)=det(R^{T})=det(R^{-1})=1/det(R)$

I think the above section on "Rotation matrix vs orthogonal matrix" explains why it should be +1 but it's a little beyond me. A possible argument could be that the identity matrix is a rotation, and swapping a row = changing the handedness, so that all rotations must have a determinant the same sign as the identity matrix.

To me it'd make much more sense to introduce rotation matrices as transformations that leave vector lengths, angles between vectors and handedness unaffected and show from there why the properties of its transpose and determinant hold.Michael.Clerx (talk) 15:22, 19 July 2009 (UTC)

Generator Question

The rotation matrices about the x, y, and z axis do not seem to be equate to the "generator" written to their right (the rotation matrices with the matrix exponentials). It seems that one is the transpose of the other. Can someone please explain?

Thanks- James

Major rewrite

Against my better judgment I have attempted to make this a solid article. Experience strongly suggests that before long an endless stream of editors who barely understand anything about the topic will attempt to "improve" it. I do hope that someone besides me who does understand and who cares will keep an eye out and revert the damage. Lacking the patience to babysit, I leave it to its fate. (But I am open to serious questions on my talk page.)

This, by the way, is a trimmed down version of an earlier draft. I have (modestly?) added a B rating; feel free to adjust up or down. Enjoy. --KSmrq^T 08:35, 30 August 2007 (UTC)

What do you want, a medal? Jeez. —Preceding unsigned comment added by 71.111.251.229 (talk) 17:50, 16 February 2008 (UTC)

The equations used to express the rotation are for angles opposite in sign to their definition in the descriptions. For example, consider a test vector having coordinates of x=+1, y=0, z=0 in the old, unprimed coordinate system. Rotate this vector positively (counterclockwise) by 90 degrees about the z axis. After the rotation, the z' axis is identical with the previous z axis, the x' axis is where the previous y axis was, and the y' axis is pointed in the direction which was the negative direction along the old x axis. The vector itself, of course, does not move, meaning that its coordinates in the new, prime coordinate system are x'=0, y'=−1, z'=0. But that's not what you get from the equations describing the rotation. Those equations erroneously return x'=0, y'=+1, z'=0, which are the primed coordinates that the test vector would have if you had rotated a right-handed coordinate system negatively (clockwise) by 90 degrees about the z axis. So you've got the rotation equations wrong by the sign of the angle of rotation. That's probably true for all the other rotations as well. Jenab6 (talk) 01:47, 7 May 2010 (UTC)

"Obviously", the transformation indicated by a rotation of the coordinate system is the inverse (and hence corresponding to a negated angle) indicated by a rotation of the vectors. It appears that you are rotating the coordinate system, while most people rotate the vectors. — Arthur Rubin (talk) 03:47, 7 May 2010 (UTC)

Deficiency in the article?

It seems to me that there's a major deficiency in this article, which is that it's not made clear what the connection is between a rotation matrix and the geometric operation of rotation -- this connection being that if you premultiply a rotation matrix by a column vector, then the result is another column vector that is just a rotation of the first.

Okay, it is stated eventually, but not until the geometry section, and there only in the abstract language of linear algebra, putting the main point of rotation matrices out of reach of anyone except those who should know it already.

I'm hesitating to jump in and revise it, as I'm kind of new to linear algebra, but will probably go ahead and take a crack at it eventually if no one with more expertise wants to step up.

168.156.89.237 (talk) 23:51, 27 November 2007 (UTC)

That's true. I already moved the remark "this works for column vectors" from deep inside the dimension 3 section to top of "dimension 2 and 3" section, but it is an important fact that probably applies to the whole article.

Also, rotations work on row vectors too (but then the matrices need be transposed), that would be worth a word too. MathsPoetry (talk) 10:59, 1 June 2009 (UTC)

Knowledge Gap

As interesting as Lie algebras, spin groups, quaternions, or the intricacies and many properties of SO(3) group theory are, many editors are forgetting that the article is about Rotation Matrices and not about Algebra, Physics, Peculiar 19th century math, or theoretical mathematics. There clearly is a huge disconnect here between what should be here (and is missing) and that which is optional (which is abundant, even overshadowing).

Glimpsing over the article I'd swear it was about quaternions, until the caption told me it should have been about Rotation Matrices, the simple representation between 2 rotated coordinate systems in an orthonormal basis nowhere to be found, not even the very essential composition of rotation matrices about the three cartesian axes are mentioned!

With all due respect to the various authors, 90% of the material is better off in other articles where it doesn't bumb the signal to noise ratio off the scale. 82.72.87.196 (talk) 17:20, 18 February 2008 (UTC)

You may not be aware of this, but all Wikipedia content is contributed by unpaid volunteers, and anyone can edit the articles if they think they can improve them. So, be bold. If you add new content, the best is to follow a published accessible treatment from a good textbook, which can be cited as a source. I do not see how the article would be improved by removing more advanced content (which, unlike you, I think is highly relevant and important information).  --Lambiam 09:31, 20 February 2008 (UTC)

Ambiguities

It's good that ambiguities are listed, but they seem not to be resolved in article itself. E.g. whether presented matrices multiply rows or columns? Roman Cheplyaka (talk) 11:49, 27 June 2008 (UTC)

sandwich character

At the end of section 6.5 "spin group" there is the "sandwich" expression which uses the code ∗ instead of the ordinary asterix (*). The lowast character does not read on my terminal and perhaps on other readers too. Thus I made an edit to better express the sandwich; said edit being un-done. I suggest using $v\to qvq^{-1}$ to better express the required sandwich map. In the present case q* = q^-1 so there is no difference.Rgdboer (talk) 22:49, 7 July 2008 (UTC)

Total re-write?

It has been proposed that the Eigenvector slew article should be merged with the "Rotation matrix" article although this is basically a spacecraft article, only secondary a mathematics article. Attached my proposal for a new mathematical "Rotation matrix" article.

This would then be the reference for a very short spacecraft article.

If people also want to keep the old text a solution has to be found

PS:

I would also add some stuff about quaternions as this just is a slight change in format of the "canonical form". And Quaternions are used a lot to specify spacecraft attitude, there should be some suitable stuff for this!

Stamcose (talk) 16:52, 29 July 2008 (UTC)

Has been implemented

This new article should better correspond to what the user needs/expects! And it contains the material of Eigenvector slew as a mathematical (linear algebra) article what fits better! I.e. a "merge" has been done!

Stamcose (talk) 11:41, 31 July 2008 (UTC)

The new draft article proposed

Let

${\hat {e}}_{1}\ ,\ {\hat {e}}_{2}\ ,\ {\hat {e}}_{3}$

be an orthogonal positively oriented base vector system in $R^{3}$

The linear operator

"Rotation with the angle $\theta$ around the axis defined by ${\hat {e}}_{3}$"

has the matrix representation

${\begin{bmatrix}Y_{1}\\Y_{2}\\Y_{3}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}X_{1}\\X_{2}\\X_{3}\end{bmatrix}}$

relative this basevector system

This then means that a vector

${\bar {x}}={\begin{bmatrix}{\hat {e}}_{1}&{\hat {e}}_{2}&{\hat {e}}_{3}\end{bmatrix}}{\begin{bmatrix}X_{1}\\X_{2}\\X_{3}\end{bmatrix}}$

is rotated to the vector

${\bar {y}}={\begin{bmatrix}{\hat {e}}_{1}&{\hat {e}}_{2}&{\hat {e}}_{3}\end{bmatrix}}{\begin{bmatrix}Y_{1}\\Y_{2}\\Y_{3}\end{bmatrix}}$

by the linear operator

The determinant of this matrix is

$det{\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}=1$

and the characteristic polynomial is

${\begin{aligned}det{\begin{bmatrix}\cos \theta -\lambda &-\sin \theta &0\\\sin \theta &\cos \theta -\lambda &0\\0&0&1-\lambda \end{bmatrix}}&={\big (}{(\cos \theta -\lambda )}^{2}+{\sin \theta }^{2}{\big )}(1-\lambda )\\&=-\lambda ^{3}+(2\ \cos \theta \ +\ 1)\ \lambda ^{2}-(2\ \cos \theta \ +\ 1)\ \lambda +1\\\end{aligned}}$

The matrix is symmetric if and only if $\sin \theta =0$, i.e. for $\theta =0$ and for $\theta =\pi$

The case $\theta =0$ is the trivial case of an identity operator

For the case $\theta =\pi$ the characteristic polynomial is

$-(\lambda -1){(\lambda +1)}^{2}$

i.e. the rotation operator has the eigenvalues

$\lambda =1\quad \lambda =-1$

The eigenspace corresponding to $\lambda =1$ is all vectors on the rotation axis, i.e. all vectors

${\bar {x}}=\alpha \ {\hat {e}}_{3}\quad -\infty <\alpha <\infty$

The eigenspace corresponding to $\lambda =-1$ consists of all vectors orthogonal to the rotation axis, i.e. all vectors

${\bar {x}}=\alpha \ {\hat {e}}_{1}+\beta \ {\hat {e}}_{2}\quad -\infty <\alpha <\infty \quad -\infty <\beta <\infty$

For all other values of $\theta$ the matrix is un-symmetric and as ${\sin \theta }^{2}>0$ there is only the eigenvalue $\lambda =1$ with the one-dimensional eigenspace of the vectors on the rotation axis:

${\bar {x}}=\alpha \ {\hat {e}}_{3}\quad -\infty <\alpha <\infty$

The "rotation operator" is an orthogonal mapping and its matrix relative any base vector system is therefore an orthogonal matrix with determinant 1. A non trivial fact is the opposite, i.e. that for any orthogonal linear mapping in $R^{3}$ having determinant = 1 there exist base vectors

${\hat {e}}_{1}\ ,\ {\hat {e}}_{2}\ ,\ {\hat {e}}_{3}$

such that the matrix takes the "canonical form"

${\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}$

for some value of $\theta$.

In fact, if a linear operator has the orthogonal matrix

${\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}}$

relative some base vector system

${\hat {f}}_{1}\ ,\ {\hat {f}}_{2}\ ,\ {\hat {f}}_{3}$

and this matrix is symmetric the "Symmetric operator theorem" valid in $R^{n}$ (any dimension) applies saying

that it has n orthogonal eigenvectors. This means for the 3 dimensional case that there exists a coordinate system

${\hat {e}}_{1}\ ,\ {\hat {e}}_{2}\ ,\ {\hat {e}}_{3}$

such that the matrix takes the form

${\begin{bmatrix}B_{11}&0&0\\0&B_{22}&0\\0&0&B_{33}\end{bmatrix}}$

As it is an orthogonal matrix these diagonal elements $B_{ii}$ are either 1 or -1. As the determinant is 1 these elements are either all 1 or one of the elements is 1 and the other two are -1. In the first case it is the trivial identity operator corresponding to $\theta =0$. In the second case it is a rotation with $\theta =\pi$ around the eigenvector having 1 as eigenvalue.

If the matrix is un-symmetric the vector

${\bar {E}}=\alpha _{1}\ {\hat {f}}_{1}+\alpha _{2}\ {\hat {f}}_{2}+\alpha _{3}\ {\hat {f}}_{3}$

where

$\alpha _{1}=A_{23}-A_{32}$

$\alpha _{2}=A_{31}-A_{13}$

$\alpha _{3}=A_{12}-A_{21}$

is non-zero. This vector is an eigenvector with eigenvalue

$\lambda =1$

Setting

${\hat {e}}_{3}={\frac {\bar {E}}{|{\bar {E}}|}}$

and selecting any two orthogonal unit vectors in the plane orthogonal to ${\hat {e}}_{3}$:

${\hat {e}}_{1}\ ,\ {\hat {e}}_{2}$

such that

${\hat {e}}_{1}\ ,\ {\hat {e}}_{2},\ {\hat {e}}_{3}$

form a positively oriented trippel the operator takes the desired form with

$\cos \alpha ={\frac {A_{11}+A_{22}+A_{33}-1}{2}}$

$\sin \alpha ={\frac {|{\bar {E}}|}{2}}$

Stamcose (talk) 16:52, 29 July 2008 (UTC)

Error in rotation matrix around Y

In the text we have the following:

$Q_{y}={\begin{pmatrix}\cos \theta &0&-\sin \theta \\0&1&0\\\sin \theta &0&\cos \theta \end{pmatrix}}$

I think this is wrong. I just changed it to this.

$Q_{y}={\begin{pmatrix}\cos \theta &0&\sin \theta \\0&1&0\\-\sin \theta &0&\cos \theta \end{pmatrix}}$

Can someone please confirm.

Thank you DaffyDuck1981

Gvozdimirka Rucović (talk) 02:50, 13 October 2008 (UTC)

Dude, this is now wrong, it was right before! I checked myself and confirmed with other sources. You cost with this mistake Apple millions of dollars, since Llanelly R. O. implemented your matrices in the new iphone software which is causing serious problems. What you wrote is actually a vector rotation matrix. What was written was a coordinate system rotation matrix.

—Preceding unsigned comment added by Gvozdimirka Rucović (talk • contribs) 19:28, 13 July 2010 (UTC)

Nonsense. If $Q_{x}$ has a $-\sin \theta$ in the yz component, then $Q_{y}$ has a $\sin \theta$ in the xz component. — Arthur Rubin (talk) 21:25, 13 July 2010 (UTC)

DaffyDuck is incorrect. The y rotation matrix that DaffyDuck wrote is one that corresponds to a left handed coordinate system. The set of rotation matrices (for x, y, and z) would be correct if the article said it was for a left handed coordinate system, but it does not. It says it's for a right handed coordinate system. I've double checked this with several sources, including wolfram and other wiki articles that list right handed rotation matrices. I've also come across left handed rotation matrices, and the difference is the y rotation matrix, which points down, not up. I thought I had made a mistake in my own derivation of rotation matrices. The handedness of the matrix made sense only after I derived the rotation matrix for a left handed coordinate system. I lost quite a bit of time due to this mistake. Since this was last addressed a year ago, I will fix this error. Melihelibol (talk) 04:46, 7 August 2011 (UTC)

I'm not entirely sure whether the representation is left-handed or right-handed, but the present formulation is, at least, consistent. DaffyDuck's proposal is not consistent. — Arthur Rubin (talk) 05:03, 7 August 2011 (UTC)

The current rotation matrix is left-handed. The version prior to DaffyDuck's correction was neither left handed nor right handed, so a correction was indeed necessary for consistency. Prior to DaffyDuck's correction, the rotation matrix rotated the x and z axes clockwise, while the y axis was rotated counterclockwise. I've made a minor change noting the left handedness of the current rotation matrix. Melihelibol (talk) 19:47, 7 August 2011 (UTC)

I've reverted my changes due to the ALIAS / ALIBI convention discussion in the rotation in three dimensions section.Melihelibol (talk) 20:15, 7 August 2011 (UTC)

The original one is correct. Dude is wrong and I have modified it.

I changed the matrices back to DaffyDuck's version as a few students in my robotics class noticed a inconsistency in the Ry rotation direction (opposite to Rx and Rz).

According to Peter Corke's Robotics, Vision and Control text, particularly p27 (chapter covering pose and coordinate frames), I think the Ry matrix should be "the other way around" in terms of the negative sines, meaning that DaffyDuck is correct. The book comes with a MATLAB toolbox (freely available unlike the book unless you have access to SpringerOnline) which also follows this convention and is used by many roboticists for actual rotation calculations in real world systems. I have used this book and toolbox in by postgraduate research as well as a unit I co-taught recently at 3rd year undergraduate level. The different Ry was a common stumbling block for many students but I am fairly sure the convention in the book (DaffyDuck's) is correct for real world use. Here's a screencap of the book:
http://imgur.com/muYMR

Note also that the toolbox has a tranimate(R) function that will show the rotation.

Edit: Wolfram Mathworld has a similar convention: http://mathworld.wolfram.com/RotationMatrix.html

Edit: The matrices in Wolfram Mathworld are correct, try by yourself, and the ones in the article are "wrong". Or maybe if you use some not usual convention please explain it, but this page at the moment for me is wrong. The most common convention used in mathematics are that the rotations are the rotation of a vector in a counter-clockwise angle around the given and fixed axes.

For the expert in robotics: remember that exists two type of rotation the first one is the rotation of a point respect to an axis the second one is the rotation of an axis system respect to a point or an angle, and in robotics the most common operation is the second one, that can be obtained with the inverted matrices (the ones currently in the article).

So please correct this article or at least explain the two type of rotation (or conventions).

Read this: http://www.cprogramming.com/tutorial/3d/rotationMatrices.html

And: http://en.wikipedia.org/wiki/Right-hand_rule

— Preceding unsigned comment added by 2A02:1205:5044:E470:6E62:6DFF:FE47:4FC (talk) 19:33, 5 November 2012 (UTC)

--WaihoRobot (talk) 10:49, 22 October 2012 (UTC)

These is quite a mess. The matrix in the following section "General rotations" is not consistent with the previous section either, the result is actually different if you multiply them together (I did it symbolically to make sure). Also, as has been pointed out above the individual (basic) rotations are dubious, but looking around a bit it seems everybody defines them slightly differently (the angles are negated or not, causing/canceling a minus sign in front of the sines). This depends on the direction of rotation and the handednesss of the system, but both Wiki and Wolfram claim to be right-handed and yet have different rotations. I'm not sure if I'm missing something here, but this needs to be clarified and I'm adding a dubious tag to warn other people from relying on this alone. 2001:6B0:17:F028:84A1:5579:BF19:1A8A (talk) 15:49, 8 April 2013 (UTC)

Guys, this is ridiculous. This fundamental thing has been "controversial" for some time now. The cprogramming page addresses this fairly explicitely, that the difference between them is that one follows the "right handed" rule, and the other follows the left handed rule. It's a matter of perspective, and both types of matrices should be included in the article along with a brief explanation of the right/left handed rules (and a link to their corresponding wiki pages, if they exist). Somebody hop to it! I'd do it myself but I have a test tomorrow to study for (on this material btw), and a bunch of school projects to work on. This is my effort to modify the page so that I can sleep at night. 128.227.227.248 (talk) 23:04, 18 April 2013 (UTC)

Just tested out rotating a vector using the defined Rx, Ry, and Rz and ended up with an apparently correct LEFT-HANDED answer. However in the following section the multiplied RzRyRx yields the correct RIGHT-HANDED answer. The wolfram page yields the same right-handed answer as the one in this wiki when you multiply its matrices (http://mathworld.wolfram.com/RotationMatrix.html). — Preceding unsigned comment added by 99.62.235.24 (talk) 22:45, 21 April 2013 (UTC)

Significant change

I've added the algebraic forms of 2D and 3D matricies. They currently arn't in this article (although they are referenced somewhat obliquely in the examples) and for real-geometries (i.e. R2 and R3) they really should appear because frankly they're more important than a whole screen-ful of examples. MattTait (talk) 00:44, 18 October 2008 (UTC)

I do not quite see why you would need to add this, since the rotations are already given a little farther below. I fixed them nevertheless, the y matrix was wrong.DaffyDuck1981 (talk) 15:52, 27 October 2008 (UTC)

Ambiguities

Some of these ambiguities are not real:

Positive or negative sense

A positive rotation can mean clockwise or the opposite.

Well no, actually. By definition a rotation by a positive angle is a clockwise rotation. Perhaps the ambiguity lies because rotation by a positive angle means rotating the axes anti-clockwise? Either way, it's a duff-argument. This should be absorbed into the alibi/alias problem.

You are an idiot. —Preceding unsigned comment added by 146.169.6.124 (talk) 19:38, 13 December 2010 (UTC)

Matrices are not left-handed nor right-handed. A rotation matrix does not rotate "counterclockwise" unless you respect the usual conventions for the orientation of the plane or the 3D space. Changing the text accordingly. MathsPoetry (talk) 06:05, 17 May 2009 (UTC)

Row or column vectors

A square matrix can multiply a column vector or a row vector.

So what? That doesn't change the mathematics of a rotation matrix. A rotation matrix only guarrantees to send v $v\in \mathbb {R} ^{n}$ to a rotation of itself about the origin, it says nothing about what it does to $v\in \mathbb {R} ^{1{\text{x}}n}$.

Row- or column-major storage

Matrix elements may be stored in computer memory in either row-major order or column-major order, depending on the programming language and API.

This is certainly not an ambiguity to do with math. This is an ambiguity perhaps for algorithms and how matricies are stored in computers, but it is nothing to do with the math.

Cartesian or homogeneous representation

Homogeneous coordinates carry an extra dimension compared to Cartesian coordinates to allow more flexibility.

A rotation matrix uses cartesian notation. homgeneus representation is a generalization of rotation matricies that allow affine rotations (a rotation + translation).

MattTait (talk) 02:01, 18 October 2008 (UTC)

Converting Matrix to Quaternion

Is there a foolproof way to convert any rotation matrix to a quaternion, such that a fool like myself can use it? I was using

w = 0.5*sqrt(1+Qxx+Qyy+Qzz)
x = copysign(0.5*sqrt(1+Qxx-Qyy-Qzz),Qzy-Qyz)
y = copysign(0.5*sqrt(1-Qxx+Qyy-Qzz),Qxz-Qzx)
z = copysign(0.5*sqrt(1-Qxx-Qyy+Qzz),Qyx-Qxy)

from the article. The matrix

${\begin{bmatrix}0&0&-1\\0&-1&0\\-1&0&0\end{bmatrix}}.$

gives 0 + (root2, 0 root2), but I think the correct result should be 0 + (root2, 0 -root2)

Does this formula not work for 180 degree rotations, or am I missing something else? —Preceding unsigned comment added by 198.99.123.63 (talk) 19:19, 6 November 2008 (UTC)

-1, -1, -1, -1 example

Ok, my correction was wrong, but then the comment after the example is ambiguous. It does not reverse the direction of a 3d vector in homogeneous coordinates. Is this an irrelevant point? --sissyneck (talk) 22:12, 28 December 2008 (UTC)

I thought it was a 4D rotation, rather than a 3D rotation in homogeneous coordinates. — Arthur Rubin (talk) 23:05, 28 December 2008 (UTC)

I understand it as a 4D rotation that inverses 4D vectors. Why would vectors stop existing after dimension 3 ? MathsPoetry (talk) 11:07, 1 June 2009 (UTC)

X mistake

In section Euler angles there are two "z" in first equation I think "x" should be instead of the first of them, but I'm not sure, so I didn't change that. Can someone check that?

No, Z-X-Z is a common order for Euler angles. Remind that, unlike linear coordinates, the axes themselves are moved by previous rotations. So it is possible to have twice a rotation around the "same" axis to gain 3 degrees of freedom. Perharps have a look at illustrations on Euler angles or gimbal lock to understand it if I'm not being clear. MathsPoetry (talk) 11:05, 1 June 2009 (UTC)

Citations

Currently the references to citations in the text are in two formats. The first one is with small numbers like ² or ³, the second one is the "academic style" like (Smith, 2009) or (Jones, 1995).

Wouldn't an homogeneous typography be better ? If yes, which one should we choose? It seems to me the ² or ³ style is more common in wikipedia, and it also generates a small arrow like ↑ in the references list that allows to go back to the text. MathsPoetry (talk) 11:14, 1 June 2009 (UTC)

Sign error?

The direction given for the rotation in two dimensions appears backwards to me. CW should be CCW, etc. Try it. Mikiemike (talk) 18:45, 25 July 2009 (UTC) I agree. I checked its validity by trying to rotate (1,1) counterclockwise by 90 degrees, but the result is (1,-1), which is clearly a clockwise rotation. Anon 07:22 2 February 2010 (UTC) —Preceding unsigned comment added by 24.17.41.38 (talk)

I don't. I get

${\begin{bmatrix}0&-1\\1&0\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}-1\\1\end{bmatrix}}$

as expected: a 90° clockwisecounterclockwise rotation. The others all seem OK in two dimensions too--JohnBlackburne^words_deeds 09:21, 2 February 2010 (UTC)

That doesn't seem right to me. For a 90° clockwise rotation it should be:

${\begin{bmatrix}0&1\\-1&0\end{bmatrix}}{\begin{bmatrix}a_{x}\\a_{y}\end{bmatrix}}={\begin{bmatrix}a_{y}\\-a_{x}\end{bmatrix}}$

because in the new frame, the y-axis is pointing in the negative direction of the x-axis. So a positive a_x is a negative component of the new y-axis. —Preceding unsigned comment added by 130.221.224.5 (talk) 23:52, 3 February 2010 (UTC)

Rereading what I wrote I mistated what the result was. The following

${\begin{bmatrix}0&-1\\1&0\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}-1\\1\end{bmatrix}}$

Is a counterclockwise rotation. It takes a vector in the upper-right quadrant and rotates it into the upper-left quadrant (x < 0, y > 0). The multiplication is just done as described at Matrix (mathematics)#Matrix multiplication, linear equations and linear transformations. The matrix is the first 90° rotation matrix, R(90°). I've corrected it above as well as hopefully explaining it fully here. All the other matrices can be related to this one.--JohnBlackburne^words_deeds 00:30, 4 February 2010 (UTC)

I guess there are two ways to think about it, and the article doesn't really address this. When I think about and use rotation matrices, I do so for the purpose of rotating from one coordinate frame to another. Thinking about it this way, then the sign should be switched on the two "sine" terms in the rotation matrix. However, if the rotation matrix is used to rotate a vector from one direction to a new direction within the same coordinate frame, then the sign of the "sine" terms is correct. So, it is all a matter of reference, whether the rotation matrix is rotating the coordinate frame to a new coordinate frame, or if the rotation matrix is rotating a vector within a fixed coordinate frame, the rotation matrix will be slightly different (opposite signs on the sine terms). —Preceding unsigned comment added by Belegorn (talk • contribs) 18:26, 4 February 2010 (UTC)

I see what you mean. The article is about the matrix used to represent a rotation, as in a rotation (mathematics) of a object described by vectors. That's the generally accepted usage in mathematics, especially as doing it the other way is trivial mathematically. It does touch on what you're thinking of, under ambiguities, but that sections not at all clear - for a start if it's formulated properly nothing should be ambiguous. I think that could be replaced with a better statement of how rotation matrices work when it's the axes not objects that are rotated - I'm not sure the other 'ambiguities' are at all needed.--JohnBlackburne^words_deeds 19:16, 4 February 2010 (UTC)

Three dimensional rotation matrix

It is stated in the article (General rotations section) that a three dimensional rotation matrix is the product of the three rotation matrices:

$R_{x}(\gamma )\,R_{y}(\beta )\,R_{z}(\alpha )\,\!$

I think the order is opposite: first apply rotation around x, then around y and finally around z. Thus:

$R_{z}(\alpha )\,R_{y}(\beta )\,R_{x}(\gamma )\,\!$

In the linked Yaw, pitch and roll article the matrix product is indeed written as the latter. 79.26.188.44 (talk) 14:46, 24 August 2009 (UTC)

xy-plane

Does the 2×2 matrix given in the introduction really rotate all vectors in the xy-plane? Does it, for example, have any meaning to a vector with a component also in the z-direction? I didn't think it did, in which case it should really read that the matrix "rotates two-dimensional vectors expressed in cartesian coordinates counterclockwise..." and not "in the xy-plane". Bigbluefish (talk) 21:46, 6 November 2009 (UTC)

Error in R_y

I don't know how to modify the graphics, but in section 1.2 on three dimensions, I recommend removal of words like "clockwise" -- such things are ambiguous unless you are very careful about specifying the axis and what direction along that axis you're looking at.

Instead, I recommend a statement like: "For example, in the customary right-hand Cartesian coordinate system when theta = pi/2, the rotation matrix R_x takes the unit vector j into k, R_y takes i into k, and R_z takes i into j."

Note that the given rotation matrix R_y is wrong for the previous sentence to be true -- it should be the transpose of the given matrix.

Someonesdad363616 (talk) 19:44, 14 January 2010 (UTC)

I hadn't noticed the error before. A potentially correct statement would be, "For example, in the customary right-hand Cartesian coordinate system when theta = pi/2, the rotation matrix R_x takes the unit vector j into k, R_y takes k into i, and R_z takes i into j," although I'm not sure of the overall handedness. At least those are consistent. — Arthur Rubin (talk) 04:59, 7 August 2011 (UTC)

3D Basic rotations

Just a quick note as to why the matrices in the current version are correct, as there seems to be some confusion based on recent edits. The matrix at the top of the section,

$R(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}},$

rotates in 2D from the x-to the y-axis, as shown in the diagram and described in detail in the maths below it. This generalises straightforwardly to the following in 3D,

$R_{z}(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta &0\\\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}},$

which also rotates the x-axis to the y-axis, as it says in the description, while keeping the z-axis fixed: a rotation about the z axis. Because it's the same rotation in x and y the upper-left 2×2 matrix is the same. The other matrices follow from this by permuting the axes. So the matrices match the descriptions below and match the 2D matrices above, to hopefully make them easier to understand.--JohnBlackburne^words_deeds 17:11, 19 October 2010 (UTC)

As a note, I spend some time before I realized that the 3D rotation matrices are multiplied by a column vector AFTER the matrix. The reason I didn't realize is because in the vector library I'm using a vector is a row, as is the case in most software packages I've seen. Multiplying a row vector on the left of the given matrices yields COUNTER-clockwise rotation, contradicting the text. Thus, I suggest for clarity, we mention the multiplication order explicitly. 79.100.93.24 (talk) 18:42, 4 January 2011 (UTC)

Please check formula for quaternions

The formula given in the article is:

t = Qxx+Qyy+Qzz (trace of Q)
r = sqrt(1+t)
w = 0.5*r
x = copysign(0.5*sqrt(1+Qxx-Qyy-Qzz), Qzy-Qyz)
y = copysign(0.5*sqrt(1-Qxx+Qyy-Qzz), Qxz-Qzx)
z = copysign(0.5*sqrt(1-Qxx-Qyy+Qzz), Qyx-Qxy)

Copysign was defined as follows: where copysign(x,y) is y with the sign of x.

The definition of copysign seemed to be wrong (a sign cannot be obtained from a square root, and multiplying by 0.5 would not change the sign). So, I inverted it. But there's the possibility that the definition of copysign was correct, and the formula was wrong. I am not really interested in quaternions. I just know there was something wrong. Please check.

— Paolo.dL (talk) 17:58, 6 December 2010 (UTC)

I would just remove it altogether. It's unclear what the point of it is: it's not a mathematical formula, nor is it in any programming language that I recognise, though I do a lot of mathematical programming. If it's included at all it should be written in plain English with normal mathematical formulae as at Rotation representation (mathematics), though as the content is there, and linked to from here, there seems no point having it here. --JohnBlackburne^words_deeds 19:31, 6 December 2010 (UTC)

Yes, it is a programming language for sure. It's not MATLAB, it may be some version of Fortran. I have seen the main article Rotation representation (mathematics), but I am not sure it is a good idea to delete this text, as I found here, in another subsection within the same section, useful information (mainly about numerical "robustness", but also different algorithms to solve numerical problems) that is not given in the main article. More exactly, I found very useful the text and the formulas in the subsection about axis-angle, which clearly was written by the same person who wrote the subsection about quaternions, as all the formulas are expressed in the same language. Actually, I found there something that I believe should be added in the main article about using ATAN2 rather than arccos or arcsin. I know little about quaternions, and I have not read that subsection with attention, so I can't judge it. However, there is a request for merging the whole section into the main article. I agree about merging, but cannot do the job. Paolo.dL (talk) 21:11, 6 December 2010 (UTC)

Basic rotation matrices in 3D - direction of rotation

Before editing you should fully understand the ALIAS / ALIBI ambiguity. See this section.

The 3D basic matrices shown in the article produce a counterclockwise rotation of the vectors (relative to the Cartesian axes), and a clockwise rotation of the Cartesian axes (relative to the vectors). The sentences you want to introduce to explain the sense of rotation refer to a rotation of a coordinate system axis. Thus, they suggest an ALIAS rotation, which is clockwise, not counterclockwise as stated.

When these basic matrices are described in the article, the ALIAS / ALIBI ambiguity has not been explained yet to the readers. The intro and previous text refer alsways to rotations of vectors (ALIBI). Even the pictures show vector rotations. So, for the sake of consistency and clarity, we must not use the ALIAS "point of view" yet. Also, we need to assume a right-handed coordinate system, and a pre-multiplication by the matrix, as these are the only conventions used and described in the introduction and in the previous sections. And this consistency is higly desirable.

Even the similarity between the 2-D rotation matrix and the Rz matrix is not casual, but a precise choice of the editors, which is possible because we, thorughout the article, consistently interpret and describe these rotations as rotations of VECTORS, produced by PRE-MULTIPLICATION, in a RIGHT-HANDED coordinate system.

Paolo.dL (talk) 13:08, 7 January 2011 (UTC)

My greatest concern over your edits was that you removed the following "Rx rotates the y-axis towards the z-axis, Ry rotates the z-axis towards the x-axis, and Rz rotates the x-axis towards the y-axis." This is valuable as it relates the 3D matrices to the 2D rotations given immediately above. The way you presented it uses the far more complex observation of the sense of rotation about the axis; complex as it asks the reader to think straight away in 3D and depends on the coordinate system so needs that to be stated.--JohnBlackburne^words_deeds 13:34, 7 January 2011 (UTC)

I do understand your concern. Unfortunately, however, these sentences are wrong! Moreover, they refer to an ALIAS convention which cannot be used in this article. Again, Rx, Ry, Rz rotate the coordinate system clockwise, not conterclockwise (i.e., Rz rotates y towards x, not vice versa!). You do not understand how much the ALIAS/ALIBI ambiguity may confuse readers. A proof is that someone wrote, in the section about 2-D rotations, that the counterclockwise rotation is observed with a LEFT-handed frame! So, if you want to keep reverting, keep doing it, but I won't stop counter-reverting until you understand the reason why I cannot accept your sentences. Paolo.dL (talk) 13:44, 7 January 2011 (UTC)

If you wanted to make your sentences correct, you should write: "Rx would rotate towards the z-axis a vector aligned with the y-axis....". Are you sure that this is clearer than my text? If you want to do it, do not revert. Just add your sentences, as the current edit contains another mistake by the anonymous editor (the axis points toward the observer, not away from the observer), and other parts of my edits are necessary. Paolo.dL (talk) 14:07, 7 January 2011 (UTC)

I added this sentence as an example: "For instance, Rz would rotate towards the y-axis a vector aligned with the x-axis...." (this rotation is similar to the example given in 2D, see also picture and picture caption). Paolo.dL (talk) 14:37, 7 January 2011 (UTC)

I agree with Paolo on Alibi vs. Alias, but his most recent change is not consistent with the current rotation matrix. For the given matrix, a vector aligned with the y-axis would rotate toward the x-axis (Alibi convention). Rz has a sin term for the x component of y, which implies that a vector aligned with the y axis will tend to x, and equal x at PI/2.

As for conventions, it's clear everyone thinks in the Alibi convention, which is consistent with other sources on the web (wolfram etc.) Either the rotation matrix should be rewritten using a right handed coordinate system, or the explanation should describe the characteristics of a left handed coordinate system.Melihelibol (talk) 20:37, 7 August 2011 (UTC)

Rotations in 2D

Just to clarify, the rotations are the same as those in the diagram, a counterclockwise rotation. If for example the angle is 45° the point (1, 0) is rotated to (0.707, 0.707). This is clearly given by

$x'=x\cos \theta -y\sin \theta \,$,

$y'=x\sin \theta +y\cos \theta \,$.

With θ = 45°. Interchanging the signs of the off-diagonal sin will give a different result, (0,707, -0.707), which is a clockwise rotation, the opposite to the sense described here. This is covered in the Non-standard orientation of the coordinate system section.--JohnBlackburne^words_deeds 15:38, 11 April 2011 (UTC)

hi ,i'm 111.255.180.117 in your example,the point (1,0) rotate 45°,then it will become (0.707 ,0.707) in "original coordinate" ,but still remain (1,0) in "new coordinate"

but in equation , (x,y) is the value of point in "original coordinate" and (x',y') is value of the "same point" (which didn't rotate) in "new coordinate" the point is , we want to get the value of "same point" in "different coordinate".

the correct equation is

$x'=x\cos \theta +y\sin \theta \,$,

$y'=-x\sin \theta +y\cos \theta \,$. —Preceding unsigned comment added by 111.255.180.117 (talk) 16:28, 11 April 2011 (UTC)

I'm afraid I don't get your reasoning. In a little more detail here's mine.

Starting with (1, 0), i.e. x = 1, y = 0.

θ = 45°, so sin (θ) = cos (θ) = 0.707 (to 3 decimal places, it's √2/2 exactly)

Put these numbers into the equations, i.e.

$x'=x\cos \theta -y\sin \theta \,$,

$y'=x\sin \theta +y\cos \theta \,$.

gives

x' = 1 × 0.707 - 0 × 0.707 = 0.707

y' = 1 × 0.707 + 0 × 0.707 = 0.707

So the result is x' = 0.707, y' = 0.707, or (0.707, 0.707). This corresponds to the diagram where the point rotated counterclockwise from the x-axis is in the positive x-y quadrant, so must have x and y positive.

Using the equations you give above produces a different result, (0.707, -0.707), in a different quadrant, the result of a clockwise rotation, which does not match the description or the diagram.--JohnBlackburne^words_deeds 16:41, 11 April 2011 (UTC)

In your opinion ,you rotate the "vector" not the "coordinate", likes (talk) says..... x' and y' is "not" the value in original coordinate ,the are the value in "new coordinate". image a apple put on table in front of you ,if you turn left ,where is the apple?

again , rotation matrix represents the relation between two "different" coordinate. —Preceding unsigned comment added by 111.255.180.117 (talk) 16:57, 11 April 2011 (UTC)

That is one way to do it, but not how this article does it, as it describes clearly in the article. It also highlights the difference in the section Ambiguities. The way it is done in the article is the most common way of doing it, with fixed coordinates and objects rotating.--JohnBlackburne^words_deeds 17:08, 11 April 2011 (UTC)

I agree with JohnBlackburne. We previously discussed a similar topic in this talk page. Paolo.dL (talk) 13:33, 12 April 2011 (UTC)

You should be careful in writing about the matrices. You should check the ref (Swokowski, Earl (1979). Calculus with Analytic Geometry (Prindle, Weber, and Schmidt).) better. — Preceding unsigned comment added by 193.255.88.133 (talk) 19:13, 10 April 2014 (UTC)

I don't have a copy of that reference, but I think you will find, if you read it very carefully, that it uses a different convention, thus requiring the transpose of our matrices. Dbfirs 20:05, 10 April 2014 (UTC)

Nope, this part referenced the rotation equations to this book (Swokowski, Earl (1979), I checked it and found that what you keep here is not quite correct. If not convinced, check this: http://math.sci.ccny.cuny.edu/document/show/2685 (page 3)

Swokowski's example has rotating axes. Our article has rotating vectors. Both approaches are common and valid, as discussed above, but rotating vectors is both more common and easier for a beginner to follow (though if you were first taught rotating axes, you will think differently). I have previously made a couple of alterations to the article to clarify this, but you were correct that the reference was inappropriate at that point. I've moved it to a more appropriate place. Dbfirs 17:50, 26 April 2014 (UTC)

CVonline External links

================================================================
Hi - we re-added the CVonline entry to the Rotation_Matrix page to trigger a discussion on the topic below. We were not sure otherwise how to get it to your attention. Our apologies.

================================================================
Thanks for your concern about the addition that we made to Wikipedia. I understand what the potential problem is.

Allow me to explain what we are trying to achieve and then perhaps you can advise us on the correct procedure.

Over the past 10+ years we have developed a resource called CVonline

      http://homepages.inf.ed.ac.uk/rbf/CVonline/

which is a sort of encyclopedia of computer vision. The main structure is a hierarchical tree of concepts in computer vision and image analysis, with the root nodes of the tree pointing to 1-10 content URLs that give further information related to the root node concept. These content URLs are hand selected to exemplify the concept, provide examples, show extensions of the basic concept, etc. The content URLs are presented to the user by a trigger to a cgi script (which may look like a search engine, but is in fact simply retrieving a fixed set of entries from a database of URLs encoded by the concept's database key).

In the case of the Wikipedia term Rotation_matrix, there is no content that is not already in Wikipedia. But other entries, eg. the use of Clifford Algebra for computer vision

      http://homepages.inf.ed.ac.uk/cgi/rbf/CVONLINE/entries.pl?TAG106

has some content not already present in Wikipedia:

      http://en.wikipedia.org/wiki/Clifford_algebra

At the moment, CVonline has about 2000 root node concepts (pointing to about 10K URLS) which intersect with about 500-1000 (estimated) Wikipedia entries. CVonline is used heavily by the image analysis community, with on the order of 50-100K front page accesses a year, and many more via search engines.

Now that Wikipedia is starting to have content that overlaps heavily with CVonline, our hope is to somehow help Wikipedia exploit all the content that we already have collected over the past 10 years. It would reduce duplication, and allow CVonline to gracefully fade away, replacing a 1-ish person effort by a community effort.

Eventually, the community sourced content should supplant the CVonline content, but, in the mean time, CVonline has a lot more.

So, what should we do?

1) What would be ideal is to rewrite all the CVonline content into Wikipedia, but we have neither the estimated 100 person-years of resource, nor the copyright permissions to do this with the existing content.

2) Another possibility is to directly copy the URLs from CVonline into the corresponding Wikipedia pages.

3) What we did that triggered your attention was to instead add the cgi script call to the External Link section of a page.

We felt that #3 achieved all that #2 did, except was faster to implement. If you think #3 would not be acceptable even after our explanation, but #2 is acceptable, then we can do this. What do you recommend?

[As a side note, CVonline provides no commercial benefit to myself or anyone else directly. There might be some indirect benefit via a software and book list that is included in CVonline, neither of which are not expected to be moved into Wikipeia.]

There is another point that I'd like your advice on: at the moment, there is a very useful subject hierarchy in CVonline. For example, look at the "Vision Geometry and Mathematics" page at:

http://homepages.inf.ed.ac.uk/rbf/CVonline/geom.htm

While there are some advantages to the flat Wikipedia structure, some summary hierarchies like these pages help people understand related concepts and technical developments.

It would be good if there were a mechanism to include this hierarchical structure page somewhere into Wikipedia. For example:

(1) If it were appropriately retitled, then the page given above could be rewritten into Wikipedia form, where all the current cgi script hot-links are replaced by Wikipedia links (assuming that they exist).

(2) Another possibility would be for the hierarchy page to remain part of CVonline, but the content links replaced by Wikipedia page URLs (especially if we find an acceptable solution to the problem described in the first part of this message). The disadvantage of this solution is that the hierarchy is not extendable and re-organisable by the community.

Again, what do you recommend?

================
Papadim.G (talk) 17:12, 11 May 2011 (UTC)

The link you just added to this article, [1], links to a page which has no content and only two links, one back to this article and one to the equivalent article at Mathworld. As I noted on your talk page external links to external search sites should not normally be added. In this case a link to Google would be far more useful (it returns those two links and many more) but as a reader can trivially search Google themselves such links aren't provided either. So no, links like that should not be added, and you should not add any more.--JohnBlackburne^words_deeds 17:28, 11 May 2011 (UTC)

May I then ask about the possibility of directly copying the URLs from CVonline into the corresponding Wikipedia pages? I have made some example changes to the external links of the following terms:

  Bhattacharyya Distance
  Relaxation labelling
  Hausdorff distance

what do you think? Thanks in advance for your reply. Papadim.G (talk) 10:59, 12 May 2011 (UTC)

Conversion from and to axis-angle

2012-08-23
I can't read picture "Rotation decomposition.png".
At the first try, I had seen "u->" being to the right from y axis, but later thought it is behind y.
Is "v->" and "(I-P)(v->)" below the i-j surface, or above, or on it ?
The image is from a single point of view, but should be at least from two IMO.
Alex_I — Preceding unsigned comment added by Cantregistermynick (talk • contribs) 08:23, 23 August 2012 (UTC)

Comments on the definition of a rotation matrix

As far as I can tell this article on the rotation matrix was created in 2004 and by 2005 the signs on the off-diagonal terms had already been changed to try to find a consistent definition. The current talk page goes back to 2006 with many editors citing references for the location of the minus sign on both sides of the diagonal. In the past I have been reluctant to wade into this discussion, but the recent flare up has moved me to offer the following.

Maybe it will help to begin by saying both views are correct, which sounds strange but hopefully allows us to focus on what are actually two different ways of viewing a rotation: (i) a transformation from coordinates in one frame to coordinates in another frame, and (ii) a transformation between two sets of basis vectors defining coordinates in the same frame. The two formulations are closely related and perhaps it is no surprise that they result in matrices that are inverses of each other, which for rotation matrices means the transpose of each other so they differ only in the location of the minus sign above and below the diagonal.

(i) Consider the first case, where a reference frame M is rotated counter-clockwise by the angle θ relative to a reference frame F. A vector in M has the coordinate x=xi + yj, where i=(1,0) and j=(0,1) are the natural basis vectors along the coordinate axes of M, so x=(x, y) in this reference frame. Now consider the coordinates X=(X, Y) of the same point but now measured in F. This is easily done by considering the vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) that are the images of i and j of M, but now measured in the frame F, so X=xe_r+ye_r, or

${\begin{Bmatrix}X\\Y\end{Bmatrix}}={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}{\begin{Bmatrix}x\\y\end{Bmatrix}}.$

This version of the rotation matrix has the vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) as column vectors, which means the minus sign is located on the upper right sine term.

(ii) Now consider the second case, where the vectors are measured in the same frame F, so i and j are the natural basis vectors along the x and y axis of F, and we have the unit vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) measured in F that define a pair of orthogonal unit vectors rotated by the angle θ in the counter clockwise direction. Now, the transformation from coordinates relative to the basis vectors i and j of F to the new basis vectors e_r=(cosθ, sinθ) and e_t=(-sinθ, cosθ) in F, is easily defined by the computing i=cosθe_r-sinθe_t j=sinθe_r+cosθe_t. This means a vector X=Xi+Yj in F is transformed to a vector in the rotated basis, by the transformation,

${\begin{Bmatrix}x\\y\end{Bmatrix}}={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}{\begin{Bmatrix}X\\Y\end{Bmatrix}}.$

Now the columns of the rotation matrix are the coordinates of the natural basis vectors i and j as measured in the new basis e_r and e_t. This places the minus sign on the sine term below the diagonal.

Hopefully, it is clear that the two ways to define a rotation matrix are inverses of each other, and they differ by the location of the minus sign. This same situation applies to rotations about each of the coordinate axes in three dimensions. The first case results in rotation matrices that have the minus sign above the diagonal for Rx and Rz and below the diagonal for Ry, while the second case yields the transpose of these matrices, with the minus sign below the diagonal for Rx and Rz and above the diagonal for Ry. I hope this is helpful. Prof McCarthy (talk) 05:06, 19 April 2013 (UTC)

Actually, there are several possible sources of ambiguity that need to be considered (passive/active rotation, column/row vector convention, left/right handedness). Anyway, your terminology is somewhat confusing. It is misleading, in my opinion, to state that, in your second example, "vectors are measured in the same frame F". What do you mean exactly? Do you agree that when you change basis you change frame? Paolo.dL (talk) 08:17, 5 May 2013 (UTC)

There is a lot of room for confusion on this point. The analysis of rigid body systems such as linkage systems and robots generally uses separate reference frames attached to each body in the system, such as outlined in the Denavit-Hartenberg convention. However, traditional dynamics texts formulate the velocity and acceleration equations for Newton's Law or Lagrange's Equations in terms of basis vectors attached to various bodies that are defined in the base inertial frame. These approaches are closely related but not the same. In the first case, each body has a frame with a natural basis and a coordinate transformation between these frames, while for the second case there is only the base frame with a natural basis and all vectors are defined in this frame though often using alternate orthogonal bases for convenience of the formulation. Prof McCarthy (talk) 21:01, 5 May 2013 (UTC)

I came to this article from Cauchy stress tensor. I am trying to keep consistency with the convention for rotation matrix. What I see in this article is that it is only addressing rotation matrix for a vector in a fixed coordinate system. The vector is rotating and one wants to find the new coordinates. However, what this article is missing is the rotation matrix for a rotating coordinate system (in other words, if one wants to find the coordinates of a fixed vector in a rotated coordinate system (primed coordinate system). Or if one wants to to a transformation of a tensor. That is what Prof McCarthy is stating. sanpaz (talk) 00:09, 30 May 2013 (UTC)

In the active transformation (left), a point moves from position P to P' by rotating clockwise by an angle θ about the origin of the coordinate system. In the passive transformation (right), point P does not move, while the coordinate system rotates counterclockwise by an angle θ about its origin.

The difference between active and passive transformations is easy to understand, and clearly explained elsewhere (a summary is also included in this article). A transposition of the matrix suffices to switch from active to passive and vice versa. McCarthy's text makes everything much more complex, raises doubts instead of solving them, and therefore in my opinion is not helpful in this context.
By the way, the difference between active and passive transformations is not the only source of ambiguity in the definition and interpretation of a rotation matrix. See Rotation matrix#Ambiguities for details. Paolo.dL (talk) 13:05, 30 May 2013 (UTC)

Thanks Paolo, I did not know about the active and passive transformation page. It seems that page should be merge into this article. What do you think? sanpaz (talk) 16:07, 30 May 2013 (UTC)

You welcome. Active and passive transformation is not only about rotation matrices, and not only about rotations. Not all transformations are rotations, and transformations are not always represented by matrices. So, that article should not be merged into this one. However, a rotation-specific summary of that article is already included in section Rotation matrix#Ambiguities. Paolo.dL (talk) 18:59, 30 May 2013 (UTC)

Understood. I would make a final suggestion: perhaps the title Ambiguities is to general. Perhaps there is a way to quickly guide the reader to the issue/difference between passive and active transformation. sanpaz (talk) 19:09, 30 May 2013 (UTC)

People are still changing the matrix to its transpose, so I think we need to add a (possibly hidden) note early in the article. It is plain to me that the example in the text is of an active (alibi) rotation, but editors familiar with passive (alias) rotations think that the article is wrong and make changes before they get to the relevant section (and despite the fact that the working is laid out plainly). Dbfirs 19:13, 30 January 2014 (UTC)

Comment on accessibility to linear algebra interpretations

Consider

>>where [\mathbf u]_{\times} is the cross product matrix of u, ⊗ is the tensor product and I is the Identity matrix.

into

>>where [\mathbf u]_{\times} is the cross product matrix of u, ⊗ is the tensor product, in this case the Kronecker product, and I is the Identity matrix.

I'm a user, Not a mathematician. Wanted rotation in 3D. This page was long, but the part I wanted eventually found in all this esoteric stuff. But the wiki page on tensor product, well, dense. 15 minutes later, I am sort of happy that if this page had said "kronecker product" I would have just programmed my rotation about an arbitrary axis in 3D, normal euclidean space & had it operating by now. — Preceding unsigned comment added by 131.203.13.81 (talk) 22:14, 13 August 2013 (UTC)

Inconsistency between rotation matrix convention in axis-angle to rotation matrix

If you use the formula given in Rotation_matrix#Rotation_matrix_from_axis_and_angle for a rotation about the positive z-axis by an angle $\theta$, then one gets,

${\begin{bmatrix}\cos \theta &-\sin \theta &0\\[3pt]\sin \theta &\cos \theta &0\\[3pt]0&0&1\\\end{bmatrix}}$

This is the inverse of the rotation matrix for the alibi convention, and listed in other areas of the page, e.g. Rotation_matrix#Basic_rotations. The 3D rotation matrix given elsewhere for rotation about the z-axis is the transpose of what it should be. This error can be checked by rotating $[1\ 0\ 0]^{T}$ using the 3D rotation matrix $R_{z}(\theta )$ given in the text. Also note the inconsistency between the correct 2D rotation matrix and the incorrect 3D matrix for rotation about the z-axis. — Preceding unsigned comment added by V madhu (talk • contribs) 03:30, 15 October 2013 (UTC)

The latest edits by an anonymous user (claiming to be Kunal Kathuria) and User:Paolo.dL appear to have fixed these issues. -V madhu (talk) 08:48, 20 October 2013 (UTC).

Thank you for your warning. ;-) Paolo.dL (talk) 16:49, 20 October 2013 (UTC)

Error in heading

The section heading reads "Conversion from and to axis-angle" I presume this should say "Conversion from rotation matrix to axis-angle"Soler97 (talk) 01:37, 19 December 2013 (UTC)

Exponential Map

I think this is unclear:

An important practical example is the 3×3 case, where we have seen we can identify every skew-symmetric matrix with a vector ω = θ u, where u = (x,y,z) is a unit magnitude vector. Recall that u is in the null space of the matrix associated with ω; so that, if we use a basis with u as the z axis, the final column and row will be zero. Thus, we know in advance that the exponential matrix must leave u fixed. It is mathematically impossible to supply a straightforward formula for such a basis as a function of u (its existence would violate the hairy ball theorem); but direct exponentiation is possible, and yields

${\begin{aligned}\exp({\tilde {\boldsymbol {\omega }}})&{}=\exp \left({\begin{bmatrix}0&-z\theta &y\theta \\z\theta &0&-x\theta \\-y\theta &x\theta &0\end{bmatrix}}\right)={\boldsymbol {I}}+2cs~{\boldsymbol {{\tilde {u}}\cdot A}}+2s^{2}~({\boldsymbol {{\tilde {u}}\cdot A}})^{2}=\\&{}={\begin{bmatrix}2(x^{2}-1)s^{2}+1&2xys^{2}-2zcs&2xzs^{2}+2ycs\\2xys^{2}+2zcs&2(y^{2}-1)s^{2}+1&2yzs^{2}-2xcs\\2xzs^{2}-2ycs&2yzs^{2}+2xcs&2(z^{2}-1)s^{2}+1\end{bmatrix}},\end{aligned}}$

where c = cos ^θ⁄₂, s = sin ^θ⁄₂. We recognize this as our matrix for a rotation around axis u by the angle θ.

In the above formula what is $\mathrm {A}$ and what is $\cdot$ ? The bottom matrix is

$\left[{\begin{array}{lll}1+0-2s^{2}(z^{2}+y^{2})&0-2csz+2s^{2}yx&0+2csy+2s^{2}zx\\0-2csz+2s^{2}xy&1+0-2s^{2}(z^{2}+x^{2})&0-2csx+2s^{2}zy\\0-2csy+2s^{2}xz&0+2csx+2s^{2}yz&1+0-2s^{2}(x^{2}+y^{2})\end{array}}\right]={\boldsymbol {I}}+2cs{\boldsymbol {\tilde {u}}}+2s^{2}{\boldsymbol {\tilde {u}}}^{2}$

StephenK51 (talk) 19:23, 22 January 2014 (UTC)

Alias and Alibi Transformations

I do not find the discussion of alias and alibi rotations to be very helpful to the reader. The following quotes taken from Birkhoff and Mac Lane and by George Francis show that these terms define the point of view of the person using the transformation. They are not inherent to its mathematical formulation.

Birkhoff and Mac Lane provide a useful discussion of transformations which uses the terms alibi and alias to describe the results of the transformations. They say:

Equation (13) was interpreted above as a transformation of points (vectors), which carried each point X = (xb · · · , x") into a new point Y having coordinates (y., · · · , y") in the same coordinate system. We could equally well have interpreted equation (13) as a change of coordinates. We call the first interpretation an alibi (the point is moved elsewhere) and the second an alias (the point is renamed).

George Francis, University of Illinois, elaborates on the idea of alias and alibi transformations to explain the choices taken by Silicon Graphics in the development of their Iris geometric engine. He says,

The geometers at Silicon Graphics, the company that builds the Iris, do not consider an affine transformation as something that transforms the objects to be displayed. They consider the affine transformation to act on the coordinate system. There is no harm in this attitude, it comes from the realm of movie maker and landscape artist. It is not adapted to the attitude of the industrial designer nor to the mathematician. The two attitudes are referred to as the “alias” and “alibi” approach to coordinate changes. To change coordinates means to give a point a false name. To change its position is to give it an alibi as to its whereabouts.

Prof McCarthy (talk) 23:26, 12 March 2014 (UTC)

Isn't that exactly the distinction described in the article? Dbfirs 09:29, 13 March 2014 (UTC)

I do not view this statement "The coordinates of a point P may change due to either a rotation of the coordinate system CS (alias), or a rotation of the point P (alibi)." to be in line with Birkhoff and Mac Lane or George Francis. The "ambiguity" of alias and alibi resides in the viewpoint of the user not the mathematics of the rotation matrix. A person can choose either viewpoint for any transformation. There is nothing in the rotation matrix itself, particularly the minus sign above or below the diagonal, that distinguishes these points of view. Prof McCarthy (talk) 11:55, 13 March 2014 (UTC)

I don't see the difference. Am I missing something? I agree that nothing in the matrix determines whether an alias or an alibi is intended, or whether the rotation is clockwise or anti-clockwise, but we have to agree on one convention for the article, and the one chosen is the usual one at elementary level (except in computing). Dbfirs 07:48, 14 March 2014 (UTC)

I recommend introducing the convention when discussing an application, and using the convention meaningful to the application. Prof McCarthy (talk) 13:42, 14 March 2014 (UTC)

That is a sound policy. This article is about the rotation of vectors, not axes. Where do you think this should be made clearer? Dbfirs 15:41, 14 March 2014 (UTC)

The article is about the rotation matrix. The rotation of vectors and the rotation of axes are applications of the rotation matrix.Prof McCarthy (talk) 03:07, 15 March 2014 (UTC)

Yes, I see the point that you are making now. The article chooses to describe examples of rotation of vectors because that is conceptually slightly simpler, and this approach is the one that most people will meet first (unless they programme computers with rotating axes at a very early age). Perhaps the lead could make this clearer, with the information that I relegated to a note moved to the lead and clarified. I'll do this. Dbfirs 07:38, 15 March 2014 (UTC)

Thank you for your patience with this. Your edits are helpful, but remember the same matrix can be viewed in either way. Prof McCarthy (talk) 10:17, 15 March 2014 (UTC)

Yes, the example matrices can be either an anticlockwise rotation of the vectors or a clockwise rotation of the axes. I'll add that, too. Dbfirs 13:09, 15 March 2014 (UTC)

Derivation

Rotation of Axes - Stewart Calculus explains where the matrix coefficients come from. It is not about matrix but reading wikipedia I cannot understand where the coefficients come from. They use two angles, of the rotated system and point in that system. The vector with respect to rotated system has coordinates

X = R cos β
Y = R sin β

In standard system, the coordinates are

x = R cos(α+β) = R[cosα cosβ - sinα sinβ] = X cosα - Y sinα
y = R cos(α+β) = R[sinα cosβ + sinβ cosα] = X sinα + Y cosα

The way it is exposed in the article, it seems nonsense to me. --Javalenok (talk) 17:26, 15 May 2014 (UTC)

The confusion possibly arises because the Wikipedia article is about rotation of vectors. Your Stewart Calculus is about rotation of co-ordinate axes, so the derivation will be different. It's easy to show that, for the rotation of vectors, the columns of the matrix are simply the images of the base vectors (unit vectors along the co-ordinate axes). Should we have this as a simple derivation? Dbfirs 07:10, 16 May 2014 (UTC)

I see no difference between rotation of axis and rotation of vector. It is only a matter of reference and The one located at the rotated coordinates observes the rotating vector. If you can provide a simpler derivation it is even greater. --Javalenok (talk) 14:39, 16 May 2014 (UTC)

The difference is that the matrix for one is the transpose of the matrix for the other. Some people prefer the vector convention; others prefer the alternative with rotation of the axes. Dbfirs 16:02, 16 May 2014 (UTC)

Is it covairant vs contravariant? I know how active vs. passive transforms differ. This is only a matter of interpretation. Nowhere I seen that active can be simpler than passive. --Javalenok (talk) 07:10, 20 May 2014 (UTC)

The difference is sometimes described as alias versus alibi transformations, as in the section above. Which people consider simpler usually depends on which they met first. Dbfirs 18:08, 20 May 2014 (UTC)

See my fix in the computation above. The rotation of a axis produces our matrix whereas rotation of the vector produces different matrix, as I demonstrate below. Secondly, the derivation in case of rotating the vectors is absolutely identical to rotation of axis. So, even your subjective assessment (this is simpler because I was exposed to it earlier) does not make sense, regardless of whether it makes sense to claim something "easier" when it is easier by such stupid reason.

If β is again the angle to the vector in the new (X,Y) coordinates but these axis were rotated -α wrt (x,y) then (x,y) coordinates of the vector will be

x = r cos(β-α) = r[cosβ cosα + sinβ sinα] =  X cosα + Y sinα
y = r sin(β-α) = r[sinβ cosα - sinα cosβ] = -X sinα + Y cosα

What if β is angle of the vector in (x,y) and (X,Y) is α-rotated wrt (x,y)? Then, vector coordinates in (X,Y) are (X = R cos(β-α), Y = R sin(β-α)), identical to the previous case. If we specify vector as β angle in (x,y) again and rotate (X,Y) α clockwise then we will receive our article matrix again, I suppose, because vector angle will be β+α in (X,Y).

So, our matrix stems from rotation of axis, not vectors. If you however will rotate vectors in a fixed coordinate system (x,y), say it was initially at angle β and then rotated to β+α, the new coordinates of the vector changed from (x = cos β, y = sin β) to

${\begin{bmatrix}X\\Y\end{bmatrix}}={\begin{bmatrix}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{bmatrix}}\,{\begin{bmatrix}x\\y\end{bmatrix}}$

see above. Wait this is our matrix. Ok, you will need to flip the sign when start from angle β and rotate vector to β-α. So, you see, it does not matter whether we rotate axis or vector, derivation and result is the same. No need to flip anything. I wonder which also stupid excuses you can appeal to to defend your idee fixe. Wait, I feel that I am starting to understand why you associate the matrix with vector rotation and say why it is simpler. I still would like to have derivation. --Javalenok (talk) 11:50, 18 August 2014 (UTC)

I'm not sure what you are saying here, but you seem to have made the simple English in the article less coherent. Dbfirs 20:49, 18 August 2014 (UTC)

What do you mean less coherent? I have related the statement "and the above matrix also represents a rotation of the axes clockwise through an angle" with your discussion of alibi later in the text. I have related the point positions with their representation as vectors and vectors as coordinates in other wikipedia articles. How increased coherence reduces the coherence? Is white = black and vice versa in your Orwell world? --Javalenok (talk) 10:01, 19 August 2014 (UTC)

Why do people hate simple derivation, https://en.wikipedia.org/w/index.php?title=Rotation_matrix&diff=621891816&oldid=621795569? Why the fuck do you hate derivation so much? It seems like you want to keep it in secret. Your 2 dimensions bears no infomration. When I look at the rotation matrix, I go for wikipedia to find out how it is derived. Why the fuck you produced 10 kilotonnes of text, detailing every fine detail related to rotation, all kinds of non-standard axis orientations, but cannot simply permit simple 2D matrix derivation in the text? What the fuck? --Javalenok (talk) 09:48, 19 August 2014 (UTC)

I agree, your addition makes the lead far less clear, its just badly written. As for the derivation it's unnecessary and unhelpful; thinking of the initial point as being rotated about the origin is only one way of thinking of it and not a standard way. It's also badly formatted but fixing that would make it take far too much space. The whole section reads far better without it as it's easier to see how the different matrices relate. Finally please refrain from using profanities. if you cannot express yourself civilly it is best not to do so at all; instead take a break from editing until you you are calmer.--JohnBlackburne^words_deeds 12:03, 19 August 2014 (UTC)

It is nice to hear the proposal to be calmer from menace and to leave from the oppressor. Feel pain? - Shut up, this is what we ultimately want from you. What does "less clear" mean? Can you say more exactly? I have just added a couple of appropriate links in the running text. This can only add to clearity. Derivation unnecessary for whom? For the problem set in your textbook? I have told you that whenever I see the rotation matrix, I want to refresh where it comes from. How is its availability is useless? I can find here everything, from Euler angles and quantum quanetrions in non-standard axes but all I have about 2D matrix is a vector, originally aligned with x-axis rotated at some angle. What is ever point of that example? Why not rotate a zero vector? It is even easier: matrix always map (0,0) to (0,0)! Euler angles and non-standard axis in quanterions spaces is simple and informative but derivation of basic 2D case, which should be understandable by simple, non-math undergraduate is "too complex and confusing". In which world do you live? Which criteria do you use to decide what is simple and informative? How your vector along x axis, which bears 0 information, is more information? You demonstrate how vector, originally aligned with x-axis is rotated. Why arbitrary point rotation is less informative? Are you crazy? Matrix is about arbitrary point rotation, not about the points in the x-axis. I do not understand how you defend that point by

Thinking of the initial point as being rotated about the origin is only one way of thinking of it and not a standard way.

Saying that "understanding the origins of R precludes drawing parallels between different matrices" is as stupid as to say "understanding the operation of diesel will make you harder to see the relationship between different models of combustion engines". Start demonstrating perfect clarity. --Javalenok (talk) 13:55, 19 August 2014 (UTC)

Keep in mind two things that by simply repeating that something is unclear it does not clarify at all what is unclear exactly, what is the the problem. Be specific. Also,

The most interesting facts are those which can be used several times, those which have a chance of recurring... Which, then, are the facts that have a chance of recurring? In the first place, simple facts.

This is (c) Henri Poincaré says considering general case of rotating arbitrary vector/point is immensely more interesting than your single x-aligned vector rotated because it covers infinitely more cases (rotated points). The 2D matrix derivation is also much more important than all those quanterions and Euler angles that you treat later because it is the most basic case, exposed to 1000 times more people (including those who later learn quanterions and euler angles) and everyone has question: how did they get that? Don't they cheat us? Does it apply to every point in the plane? Other cases are less important because they are more rare and interesting to handful of experts in the field and can be derived once you know how 2D rotation is derived. --Javalenok (talk) 14:46, 19 August 2014 (UTC)

Rotation of the base vectors is sufficient to determine the rotation of any other point in the plane. A more complex and badly-expressed derivation is not helpful as a basic introduction. Why do you always get so upset at any criticism? Dbfirs 20:16, 19 August 2014 (UTC)

(I admit that, if I were editing the Russian Wikipedia, the results would be much less coherent than your edits in English.) Dbfirs 20:20, 19 August 2014 (UTC)

You rotate only one of the basis vectors. Therefore, the article has no relationship to what you say me here, not to speak about the derivation -- the matrix just comes out from nothing, without any reason. So, your article does not convey any information, regardless of the language. I can write the same in Russian. It will be similarly non-informative. I will try to remember that "cohesive" means "useless, meaningless and non-informative". --Javalenok (talk) 10:56, 3 September 2014 (UTC)

I disagree with your assessment of the article, and I've no objection to a simple derivation (no need to make it complicated), but the English must be coherent, not a Google translation (and I live in 2014, not 1984) Dbfirs 12:07, 4 September 2014 (UTC)

• I appended the "this is not a forum" template on top of this page: I am noting that the conversation is generating more heat than light. I believe it would be best for all involved to desist from direct edits of the specific contested sections for a month or so. The sections are fine as they stand, and gilding the lily is likely to lead to grief. This is standard textbook material, and not a venue for pedagogical innovation. It would be sad if peremptory measures had to be taken instead of thoughtful restraint. In a month's time, let any proposed improvement be explicitly stated in a new section in this talk page, and consensus vetting of it or not proceed.Cuzkatzimhut (talk) 13:53, 4 September 2014 (UTC)

Which standard textbook material does tell us that derivation is "too complex to be included in wikipedia page because wikipedia is not a forum". Provide a prove link. I do not believe that any sane textbook can teach such nonsense. --Javalenok (talk) 11:57, 9 September 2014 (UTC)

Seek consensus in October. WP:TE & WP:NOTGETTINGIT. To paraphrase Pudd'nhead Wilson, if I had half of that book, I'd burn it. Ambition may well grow on what it feeds on.Cuzkatzimhut (talk) 13:18, 9 September 2014 (UTC)

Erroneous statement

One sentence states:

"As well, there are some irregularities below n = 5; for example, SO(4) is, anomalously, not a simple Lie group, but instead isomorphic to the product of S³ and SO(3)."

The assertion that "SO(4) is . . . isomorphic to the product of S³ and SO(3)" is incorrect. SO(4) as a Lie group is not a direct product.

(As a differentiable manifold, though, SO(4) is indeed diffeomorphic to the cartesian product S³ × P³ (as the quoted statement would imply). This can be seen because every SO(n) is the total space of an SO(n-1) bundle over the sphere S^n-1. Hence SO(4) is an SO(3) bundle over S³. By the clutching construction, the exact bundle in this case is determined by a map f: S² → SO(3). This map must be homotopic to a constant map, and so the bundle is equivalent to, and therefore diffeomorphic to, a trivial bundle.)

(This might be called a Standard Mistake, since so many mathematicians, upon first seeing the diffeomorphism statement, erroneously assume the Lie group isomorphism statement.)Daqu (talk) 02:01, 24 July 2014 (UTC)

I have now fixed this problem.Daqu (talk) 13:34, 24 July 2014 (UTC)

Are 3D rotations under heading Basic rotations incorrect

I couldn't get these transforms to make sense so I referred to another source (http://mathworld.wolfram.com/RotationMatrix.html).

Seems the signs on the 'sine' components are transposed. The Wolfram transformations worked for me.

I am very much a newbie, so I didn't edit the page, but I thought I'd raise a flag.

Dan Dlpalumbo (talk) 14:04, 11 December 2014 (UTC)

Your linked article explains the difference between rotation of a vector and rotation of axes for 2-D but then goes on to show only rotation of axes for 3-D. The Wikipedia article rotates vectors, not axes, hence the difference. Dbfirs 14:14, 11 December 2014 (UTC)

Suggested merge of section Group theory into other articles.

The section Group theory doesn't belong here. It is too far afield, with its spinors, exponential maps and covering groups, from the relatively simple concept of a rotation matrix. It should be split up, rewritten, and merged into Rotation group SO(3) and Rotation group (aka orthogonal group). I'll go about to do this in a couple of days from now unless protests become too violent. YohanN7 (talk) 16:17, 13 December 2014 (UTC)

No violence proffered... It looks like a good idea, and Rotation group SO(3) appears like the optimal target, but there should be a very short residual section with basic formulas, like the last one of 9.3, for example, summarizing basic usable results here, and referring copiously to the departed material in the main article, Rotation group SO(3) . Cuzkatzimhut (talk) 17:17, 13 December 2014 (UTC)

I am in violent agreement. The subsection Baker–Campbell–Hausdorff formula is slightly confused, so the first order of business is to fix that (in place). The last paragraph applies really more to the matrix exponential section than to BCH. YohanN7 (talk) 17:28, 13 December 2014 (UTC)

This page has 100+ watchers and a couple of thousand views per day, so it is best to wait a while with major operations. But, as a piece of motivation for the proposed move/merge; there is no-one reading this article that could possibly be interested in reading the whole article. Those readers struggling with active/passive rotations will not be able to appreciate group theory, while those able to digest the group theory couldn't be bothered with reading the first half or the article. Besides, the article is too big. YohanN7 (talk) 17:46, 13 December 2014 (UTC)

The last paragraph in Baker–Campbell–Hausdorff formula confuses me. What does it refer to? It may refer to the trigonometric formula from the paragraph above. It may not, in which case it is nonsense. There are, AFAIK, no implications between the validity of the BCH formula and simple connectedness. YohanN7 (talk) 18:48, 13 December 2014 (UTC)

Yes, it refers to Z = αA + βB + γ(A × B) , where I think you mean A--> X, and B-->Y. In this paragraph, the coefficients α, β, γ, are computed with some effort in the ref of Engo. However, all structure constants of this Lie algebra are the same for any representation, thus including the doublet, normally associated with SU(2), but of course, it is the spinor rep of SO(3) as well. So the very same coefficients α, β, γ, follow much more directly, indeed by inspection, in the Pauli matrices article, as linked. (The final parenthetical statement is a pedantic aside... Only the excessively curious or enterprising reader would try to reserve the formula in the quartet rep..., etc.. But a fact is a fact and who knows where it would lead.) For the purposes of this multiplication, the topological differences between SO(3) and SU(2) do not enter... Both represent motions continuously connected to the identity. Cuzkatzimhut (talk) 20:15, 13 December 2014 (UTC)

OK, I supplanted A--> X, and B-->Y, and removed the universal cover jazz that might confuse the novice. The statement now amounts to X = iθ n̂ . J where now J are either the 3x3 matrices discussed here, or the three σ/2 which obey the same commutation relations, and so have the same commutators in the BCH expansion. But of course, nobody really sums that, in practice: One just expands the exponential, to a polynomial in the universal Lie algebra, and multiplies two exponentials as polynomials. Now these expansions are not the same in each representation, as they are neither group or algebra identities. The point of the final paragraph is that Pauli matrix multiplication is what every undergraduate is taught, so the answer is available by inspection, and one does not really need Rodrigues formula overkill. But the whole paragraph has a footnote flavor to it. Cuzkatzimhut (talk) 20:42, 13 December 2014 (UTC)

Aha, thanks, now I think see.

${\begin{aligned}e^{ia({\hat {n}}\cdot {\vec {\sigma }})}e^{ib({\hat {m}}\cdot {\vec {\sigma }})}\cdots &=(I\cos {a}+i({\hat {n}}\cdot {\vec {\sigma }})\sin {a})(I\cos {b}+i({\hat {m}}\cdot {\vec {\sigma }})\sin {b})\cdots \\&={\text{ (trigonometry + Pauli matrix relations)...}}\\&=I\cos A(a,b,\ldots ,{\hat {n}},{\hat {m}},\ldots )+i({\hat {N}}(a,b,\ldots ,{\hat {n}},{\hat {m}},\ldots )\cdot {\vec {\sigma }})\sin A(a,b,\ldots ,{\hat {n}},{\hat {m}},\ldots )\\&=e^{iA({\hat {N}}\cdot {\vec {\sigma }})}?\end{aligned}}$

I am a bit slow on the uptake. Never went to school. YohanN7 (talk) 22:02, 13 December 2014 (UTC)

Yes, absolutely. The coefficient of the identity is the cosine of the new angle, which amounts to the spherical law of cosines, and the coefficient of the sigmas has to be the new axes times the sine of same. Frankly, we were asked to do this as an exercise at school, but formal books (such as Biedenharn and Louck, etc.) are too snobbish to provide it... the multiplication law of the simplest nonabelian Lie group at that! Cuzkatzimhut (talk) 22:22, 13 December 2014 (UTC)

While we are at it, there is also the article Axis–angle representation which appears to be begging to be incorporated into this future rearrangement. Cuzkatzimhut (talk) 22:50, 13 December 2014 (UTC)

At long last I have begun. Copy and edit into target first, then think about deletion here is probably best. Lie algebra part now in Rotation group SO(3). YohanN7 (talk) 20:29, 29 December 2014 (UTC)

Just a note in support of what you are doing. As you will know, this was suggested nearly seven years ago. Dbfirs 09:21, 30 December 2014 (UTC)

The end is near...

... of the merge of stuff from group theory into Rotation group SO(3). The original group theory section was 5+ pages, now it is 3- as my screen measures it. My original plan was to "move away" rather than to "trim down". The present version of the section has a lot more the flavor of an overview, with some rudimentary basics, some key results, but little motivation or derivation. I kind of like it this way so I wouldn't want to delete it altogether. While the end is near, it isn't here (there is stuff to do in the SU(2) article teased from the BCH section here, see formulae just above), but it is best I let it rest for a little while to hopefully gather more comments on what remains to be done and what should possibly be undone. YohanN7 (talk) 22:49, 1 January 2015 (UTC)

Looks great. I took the liberty to update the "Main article" headers, since the interested reader is invited to go there for substance. Possibly what is there can be further condensed by a 5% or so...As for comparing the triplet BCH (Engo) to the doublet above and in Pauli matrix, it is just a change of variables, albeit one from hell! If you wish, I could change some Names in the Pauli article and above to avoid predictable confusions of notation. Note that, for the normalization of the generators L to agree with the Pauli matrices, you need L ↔ −iσ/2 , so that, e.g. , calling n̂=u, we identify ia u.σ = −2a u.L , hence θ= −2a (a Pauli, not Engo!). It would be trivial for me go to the Pauli article, and supplant n̂ → û , n̂' → v̂ and n̂' ' → n̂, and use different symbols for a,b,c, (but probably not −θ/2, −φ/2 !) to allow for an easy change of variables form Engo to Pauli. But I would only do it if you needed it in explicit connection. My instinct is to leave it as "an exercise for the student", as long as we don't overwhelm them with ambiguous a,b.c,s designed to confuse them! Cuzkatzimhut (talk) 01:15, 2 January 2015 (UTC)

A better idea... if you let me, change the notation and fill in the missing expression right above here, in the previous section, so n --> u, m --> v , N--> n , a-->A, b-->B, A-->C, and provide the answers right there, so you don't have to go back to the Pauli article, to be left alone for a while. Cuzkatzimhut (talk) 01:39, 2 January 2015 (UTC)

Right now my head is spinning (no pun, S_z = 3⁄2), and I need to sleep before I can think clearly (if ever). I suspected that it might not be that easy to come up with the Pauli version. Not only for algebraic manipulations from hell, but also, the Lie algebra map is one-to-one, but at the group level, while (I think so at least, drawing from what I know from the irreducible Lorentz reps that are all faithful, though some are projective) it can be arranged to be one-to-one locally, we do have a projective representation (1:2) of SO(3), and that has worried me a little. Do I make any sense? I don't doubt the expression is true for SU(2), but what bothers me is that I can't see why it must be true. Anything you can fill in here to make life easier is highly appreciated.

B t w, Engo's paper is available to the public for online reading (see updated ref in Rotation group SO(3). I think the link is legal. YohanN7 (talk) 02:32, 2 January 2015 (UTC)

We might leave spin 3/2 aside for a moment. The Pauli version follows by inspection... do you want me to write in the intermediate step above? It is the changes of variables to identify to Engo which are messy. But it does identify, as long as the perturbative series in the exponent of BCH works, but, of course, everything breaks down when the Engo angles get to 2π (inspect his formulas there) where the Pauli ones do not, but pick up a singularity as well. Cuzkatzimhut (talk) 12:21, 2 January 2015 (UTC)

That would be nice. I don't want to burden you. I haven't made any serious attempts, but I would appreciate if you filled it out. At any rate, it would speed things up. I'll try to knit things together in the article(s) after I have fully understood. We should perhaps have a version also for the t-matrices too (see new addition to rotation group SO(3)), which conform to the math convention used in the article whereas the Pauli matrices conform to the physics convention. That way the Pauli version could go into that article (wiki-linked of course), beefing it up. YohanN7 (talk) 14:03, 2 January 2015 (UTC)

OK, I'll do it above, overwriting your own formula, but without inserting the t's. Actually yours are wrong by a - sign! Compare to Pauli matrices, and specifically sigma-2. This is obvious from normalizing the Pauli matrix algebra by dividing by 2i, not multiplying by it. Cuzkatzimhut (talk) 14:16, 2 January 2015 (UTC)

I put the - sign in the isomorphism (making the t's conform with the Hall ref). I'll change that, it is only a change of basis.

On a side note:

$\alpha X+\beta Y+\gamma [X,Y]=X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}[X,[X,Y]]-{\tfrac {1}{12}}[Y,[X,Y]]+\cdots$

is noteworthy in its own right. Not remarkable, but deserving mention. Bracket series certainly do not behave like power series. YohanN7 (talk) 14:23, 2 January 2015 (UTC)

OK, here

${\begin{aligned}e^{iA({\hat {u}}\cdot {\vec {\sigma }})}e^{iB({\hat {v}}\cdot {\vec {\sigma }})}&=(I\cos {A}+i({\hat {u}}\cdot {\vec {\sigma }})\sin {A})(I\cos {B}+i({\hat {v}}\cdot {\vec {\sigma }})\sin {B})\\&=I(\cos A\cos B-{\hat {u}}\cdot {\hat {v}}\sin A\sin B)+i({\hat {u}}\sin A\cos B+{\hat {v}}\sin B\cos A-{\hat {u}}\times {\hat {v}}~\sin A\sin B)\cdot {\vec {\sigma }}\\&=I\cos {C}+i({\hat {n}}\cdot {\vec {\sigma }})\sin {C}=e^{iC\left({\hat {n}}\cdot {\vec {\sigma }}\right)}.\end{aligned}}$

Note I am using the hat to denote unit vectors, not Engo's non-standard usage! The rest is obvious, by inspection: one directly solves for C from the coefficient of I,

$\cos C=\cos A\cos B-{\hat {u}}\cdot {\hat {v}}\sin A\sin B~,$

the spherical law of cosines. Given C, then, from the coefficient of the sigmas,

${\hat {n}}=({\hat {u}}\sin A\cos B+{\hat {v}}\sin B\cos A-{\hat {u}}\times {\hat {v}}\sin A\sin B)/\sin C~.$

Consequently, the composite rotation parameters in the exponent of this group element simply amount to

$C{\hat {n}}={\frac {C}{\sin C}}({\hat {u}}\sin A\cos B+{\hat {v}}\sin B\cos A-{\hat {u}}\times {\hat {v}}~\sin A\sin B)={\frac {C}{\sin C}}\sin A\sin B~({\hat {u}}\cot B+{\hat {v}}\cot A-{\hat {u}}\times {\hat {v}})~.$

This is straightforward, and bypasses the lengthy huffing and puffing of Engo. It is worth exploring for both sets what happens at the dangerous points 2π,4π, etc... where formulas present multivalued options and the full angles versus half-angles make a difference (For example, as A=2π, solving for C yields B, but alternatively 2π—B). Perturbatively in "powers" of the generators, or the angles, the Z exponent of the respective Pauli vector and Engo spin 1 methods agree. For identification, as i mumbled above, iA u.σ = −2A u.t, hence θ = —2A, φ = —2B, u.v is the axes' cosine, etc. For example, looking at the X component of Z, we get α = (C/sinC) (sinA/A) cosB . To quadratic order in small angles A and B this may well be ≈

${\frac {A^{2}+B^{2}+2{\hat {u}}\cdot {\hat {v}}AB}{6}}={\frac {\theta ^{2}+\phi ^{2}+2{\hat {u}}\cdot {\hat {v}}~\theta \phi }{24}}$, I think... Engo keeps some full angles and some half angles, as per my comment above that half angles are often expeditious...

Note that for Pauli σs (so normalized from generators),

$\gamma ={\frac {1}{2}}{\frac {C}{\sin C}}{\frac {\sin A}{A}}{\frac {\sin B}{B}}.$ Consider now the exact expressions for α/β and α/γ in both cases, easily shown to be identical, respectively, once the "culprits" involving d and C are divided out:

${\frac {\alpha }{\beta }}={\frac {\phi }{\theta }}\tan(\theta /2)\cot(\phi /2),\qquad {\frac {\alpha }{\gamma }}=\phi \cot(\phi /2).$

And, yes, equating the α, β, γ closed expression for Z with the BCH series might be worthwhile, as it dramatizes to one how the same structure constants for commutators would lead to the same closed expression for both spin 1 and spin 1/2, and all spins... the essence of a group identity−−versus wildly differing expansions in the universal Lie algebra! People, including myself, have often run back to such (rare) closed expressions for reassurance of what holds and what does not! (Gilmore's books dwell on it more than others'). Cuzkatzimhut (talk) 14:56, 2 January 2015 (UTC)

Tnx. I'll probably not be editing more today, real life duties, but will get on with it tomorrow. B t w I notice now that the Pauli article did have a section on this even before. I totally missed that. YohanN7 (talk) 15:13, 2 January 2015 (UTC)

Do we really need the caveat about structure constants one we agree to stick to either the physics or the math convention (differing by a factor of i in places)? The BCH formula doesn't refer to them and the quantities on the left are sensitive only to how the inner product is defined. Here there is a built in trap, because

$u\cdot v\leftrightarrow {\frac {1}{2}}\operatorname {Tr} Y^{\mathrm {T} }X,$

(Engo notation) which is half the usual Hilbert-Schmidt inner product.

I only mumbled about structure constants to remind myself how the Pauli matrices, much easier to calculate with (above 3-liner), connect to the standard normalizations, and the X's of the formula... I had to work at being careful yesterday to match them up, especially α/γ where the 2-->1 generators occur, and their normalization matters to produce the same result for Engo and Pauli vectors. That was part of the notational transition hell I was complaining about, X = iA u⋅σ . But anything that works should be fine... Cuzkatzimhut (talk) 11:32, 3 January 2015 (UTC)

The end is here?

I gave it a try in the article with a new hidden box. The only thing I haven't verified is that the constant k in the article actually is equal to 1, but the perturbation expansion above seems to show that. Readers (and I) may wonder if the "t-matrix version" (or the Pauli version) is universal, and if not, why not? The expression is simpler than Engo's. YohanN7 (talk) 00:50, 5 January 2015 (UTC)

The kernel of a Lie algebra homomorphism is an ideal, but since so(3) is simple, it has no nontrivial ideals. The result should hold for all representations. YohanN7 (talk) 01:06, 5 January 2015 (UTC)

Looks nice, except Y= is missing in the 2nd formula line of the hidebox, and in hindsight, the expressions to give are α/γ and β/γ, instead of α/β. But, to a new reader, the dual usage of X and S , and Y and T, etc... would look weird/overkill... note the σ's are hermitean, the t's antihermitean, (not real antisymmetric!) and the a, b, c used in the second hidebox are NOT the a,b,c of the first hidebox of the section (Engo) above... Should I try to condense the hidebox by yet another round of notational changes, both consistent with Engo, above it, and the unfortunately defined a,b,c, of Pauli vector, which I propose to use primes for? Cuzkatzimhut (talk) 01:19, 5 January 2015 (UTC)

Yes, please do so. (The reason for the unfortunate naming is that I wanted the capital Greek A, B, Γ for what it is used for now, I forgot about the a,b, c of the first hide-box.) YohanN7 (talk) 01:41, 5 January 2015 (UTC)

OK, will try that, for a little. How about we continue this commentary in the Talkpae of That page, then? Cuzkatzimhut (talk) 01:48, 5 January 2015 (UTC)