Editing Rotation matrix (section)

===Conversion from rotation matrix to axis–angle=== 
Every rotation in three dimensions is defined by its '''axis''' (a vector along this axis is unchanged by the rotation), and its '''angle''' — the amount of rotation about that axis ([[Euler rotation theorem]]).

There are several methods to compute the axis and angle from a rotation matrix (see also [[axis–angle representation]]). Here, we only describe the method based on the computation of the [[eigenvector]]s and [[eigenvalue]]s of the rotation matrix. It is also possible to use the [[Trace (linear algebra)|trace]] of the rotation matrix.

====Determining the axis====
[[File:Rotation decomposition.png|thumb|A rotation {{mvar|R}} around axis {{math|'''u'''}} can be decomposed using 3 endomorphisms {{math|'''P'''}}, {{math|('''I''' − '''P''')}}, and {{math|'''Q'''}} (click to enlarge).]]

Given a {{nowrap|3 × 3}} rotation matrix {{mvar|R}}, a vector {{math|'''u'''}} parallel to the rotation axis must satisfy
:<math>R\mathbf{u} = \mathbf{u},</math>
since the rotation of {{math|'''u'''}} around the rotation axis must result in {{math|'''u'''}}. The equation above may be solved for {{math|'''u'''}} which is  unique up to a scalar factor unless {{mvar|R}} is the [[identity matrix]] {{mvar|I}}.

Further, the equation may be rewritten
:<math>R\mathbf{u} = I \mathbf{u}  \implies  \left(R - I\right) \mathbf{u} = 0,</math>
which shows that {{math|'''u'''}} lies in the [[null space]] of {{math|''R'' − ''I''}}.

Viewed in another way, {{math|'''u'''}} is an [[eigenvector]] of {{mvar|R}} corresponding to the [[eigenvalue]] {{math|''λ'' {{=}} 1}}. Every rotation matrix must have this eigenvalue, the other two eigenvalues being [[complex conjugate]]s of each other. It follows that a general rotation matrix in three dimensions has, up to a multiplicative constant, only one real eigenvector.

One way to determine the rotation axis is by showing that:

:<math>\begin{align}
0 &= R^\mathsf{T} 0 + 0 \\
&= R^\mathsf{T}\left(R - I\right) \mathbf{u} + \left(R - I\right) \mathbf{u} \\
&= \left(R^\mathsf{T}R - R^\mathsf{T} + R - I\right) \mathbf{u} \\
&= \left(I - R^\mathsf{T} + R - I\right) \mathbf{u} \\
&= \left(R - R^\mathsf{T}\right) \mathbf{u}
\end{align}</math>

Since {{math|(''R'' − ''R''<sup>T</sup>)}} is a [[skew-symmetric matrix]], we can choose {{math|'''u'''}} such that
:<math>[\mathbf u]_{\times} = \left(R - R^\mathsf{T}\right).</math>
The matrix–vector product becomes a [[cross product]] of a vector with itself, ensuring that the result is zero:

:<math>\left(R - R^\mathsf{T}\right) \mathbf{u} = [\mathbf u]_{\times} \mathbf{u} = \mathbf{u} \times \mathbf{u} = 0\,</math>

Therefore, if
:<math>R = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix},</math>
then
:<math>\mathbf{u} = \begin{bmatrix} h-f \\ c-g \\ d-b \\ \end{bmatrix}.</math>
The magnitude of {{math|'''u'''}} computed this way is {{math|{{norm|'''u'''}} {{=}} 2 sin ''θ''}}, where {{mvar|θ}} is the angle of rotation.

This '''does not work''' if {{mvar|R}} is symmetric.  Above, if {{math|''R'' − ''R''<sup>T</sup>}} is zero, then all subsequent steps are invalid.  In this case, the angle of rotation is 0° or 180° and any nonzero column of {{math|''I'' + ''R''}} is an eigenvector of {{mvar|R}} with eigenvalue 1 because {{math|''R''(''I'' + ''R'') {{=}} ''R'' + ''R''<sup>2</sup> {{=}} ''R'' + ''RR''<sup>T</sup> {{=}} ''I'' + ''R''}}.<ref>{{Cite journal |last1=Palais |first1=Bob |last2=Palais |first2=Richard |date=2007-12-20 |title=Euler's fixed point theorem: The axis of a rotation |journal=Journal of Fixed Point Theory and Applications |language=en |volume=2 |issue=2 |pages=215–220 |doi=10.1007/s11784-007-0042-5 |issn=1661-7738 |mr=2372984 |doi-access=free}}</ref>

====Determining the angle====
To find the angle of a rotation, once the axis of the rotation is known, select a vector {{math|'''v'''}} perpendicular to the axis. Then the angle of the rotation is the angle between {{math|'''v'''}} and  {{math|''R'''''v'''}}.

A more direct method, however, is to simply calculate the [[Trace (linear algebra)|'''trace''']]: the sum of the diagonal elements of the rotation matrix. Care should be taken to select the right sign for the angle {{mvar|θ}} to match the chosen axis:
:<math>\operatorname{tr} (R) = 1 + 2\cos\theta ,</math>

from which follows that the angle's absolute value is
:<math>|\theta| = \arccos\left(\frac{\operatorname{tr}(R) - 1}{2}\right).</math>

For the rotation axis <math>\mathbf{n}=(n_1,n_2,n_3)</math>, you can get  the correct angle<ref>{{cite arXiv|last=Kuo Kan|first=Liang|date=6 October 2018|title=Efficient conversion from rotating matrix to rotation axis and angle by extending Rodrigues' formula|eprint=1810.02999|class=cs.CG}}</ref> from

<math>\left\{\begin{matrix}
\cos \theta&=&\dfrac{\operatorname{tr}(R) - 1}{2}\\
\sin \theta&=&-\dfrac{\operatorname{tr}(K_n R)}{2}
\end{matrix}\right.
</math>

where

<math>K_n=\begin{bmatrix}
0   & -n_3 & n_2\\
n_3 & 0 & -n_1\\
-n_2 & n_1 & 0\\
\end{bmatrix}
</math>

====Rotation matrix from axis and angle====
The matrix of a proper rotation {{mvar|R}} by angle {{mvar|θ}} around the axis {{math|'''u''' {{=}} (''u<sub>x</sub>'', ''u<sub>y</sub>'', ''u<sub>z</sub>'')}}, a unit vector with {{math|''u''{{su|b=''x''|p=2}} + ''u''{{su|b=''y''|p=2}} + ''u''{{su|b=''z''|p=2}} {{=}} 1}}, is given by:<ref>{{cite journal |last1=Taylor |first1=Camillo J. |last2=Kriegman |first2=David J. |title=Minimization on the Lie Group SO(3) and Related Manifolds |journal=Technical Report No. 9405 |year=1994 |location=Yale University |url=https://www.cis.upenn.edu/~cjtaylor/PUBLICATIONS/pdfs/TaylorTR94b.pdf }}</ref>
<ref>
{{cite journal|first1 = V. | last1= Balakrishnan|title = How is a vector rotated? | journal = Resonance | volume =4 |issue =10|year=1999 | pages=61–68| doi= 10.1007/BF02834260|url=https://www.ias.ac.in/describe/article/reso/004/10/0061-0068}}
</ref>
<ref>
{{cite book|first1=Adam | last1=Morawiec|title=Orientations and Rotations|year=2004 | doi=10.1007/978-3-662-09156-2|publisher=Springer| isbn=978-3-642-07386-1}}
</ref>
<ref>
{{cite journal| first1=A. |last1=Palazzolo|title=Formalism for the rotation matrix of rotations about an arbitrary axis | journal=Am. J. Phys. | volume=44 | issue=1|pages=63–67| year=1976|doi=10.1119/1.10140|bibcode=1976AmJPh..44...63P}}
</ref>
:<math>R = \begin{bmatrix} 
u_x^2 \left(1-\cos \theta\right) + \cos \theta & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ 
u_x u_y \left(1-\cos \theta\right) + u_z \sin \theta & u_y^2\left(1-\cos \theta\right) + \cos \theta & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ 
u_x u_z \left(1-\cos \theta\right) - u_y \sin \theta & u_y u_z \left(1-\cos \theta\right) + u_x \sin \theta & u_z^2\left(1-\cos \theta\right) + \cos \theta
\end{bmatrix}.</math>

A derivation of this matrix from first principles can be found in section 9.2 here.<ref>{{Cite thesis|url=https://dspace.lboro.ac.uk/dspace-jspui/handle/2134/18050|hdl=2134/18050|title=Modelling CPV|date=January 2015|publisher=Loughborough University|type=thesis|last1=Cole|first1=Ian R.}}</ref> The basic idea to derive this matrix is dividing the problem into few known simple steps.

# First rotate the given axis and the point such that the axis lies in one of the coordinate planes ({{mvar|xy}}, {{mvar|yz}} or {{mvar|zx}})
# Then rotate the given axis and the point such that the axis is aligned with one of the two coordinate axes for that particular coordinate plane ({{mvar|x}}, {{mvar|y}} or {{mvar|z}})
# Use one of the fundamental rotation matrices to rotate the point depending on the coordinate axis with which the rotation axis is aligned.
# Reverse rotate the axis-point pair such that it attains the final configuration as that was in step 2 (Undoing step 2)
# Reverse rotate the axis-point pair which was done in step 1 (undoing step 1)

This can be written more concisely as <!--please LEAVE THE NONTRIVIAL COEFFICIENT OF THE IDENTITY ALONE. READ FOOTNOTE 4-->
<ref>
{{cite journal|first1=Jon|last1=Mathews|title= Coordinate-free rotation formalism|journal=Am. J. Phys. | volume =44 | number=12 | pages=121 | doi=10.1119/1.10264|year=1976|bibcode=1976AmJPh..44.1210M}}
</ref>
:<math>R = (\cos\theta)\,I + (\sin\theta)\,[\mathbf u]_{\times} + (1-\cos\theta)\,(\mathbf{u}\otimes\mathbf{u}),</math>
where {{math|['''u''']<sub>×</sub>}} is the [[Cross product#Conversion to matrix multiplication|cross product matrix]] of {{math|'''u'''}}; the expression {{math|'''u''' ⊗ '''u'''}} is the [[outer product]], and {{mvar|I}} is the [[identity matrix]]. Alternatively, the matrix entries are:
:<math>R_{jk}=\begin{cases}
\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}\left(2u_j^2-1\right), \quad &\text{if }j=k\\
2u_ju_k\sin^2\frac{\theta}{2}-\varepsilon_{jkl}u_l\sin\theta, \quad &\text{if }j\neq k
\end{cases}</math>

where {{mvar|ε<sub>jkl</sub>}} is the [[Levi-Civita symbol]] with {{math|''ε''<sub>123</sub> {{=}} 1}}. This is a matrix form of [[Rodrigues' rotation formula]], (or the equivalent, differently parametrized [[Euler–Rodrigues formula#Vector formulation|Euler–Rodrigues formula]]) with<ref group="nb">Note that

:<math> \mathbf{u}\otimes\mathbf{u} = \bigl( [\mathbf u]_{\times}\bigr)^2+{\mathbf I}</math>
so that, in Rodrigues' notation,  equivalently,

:<math> \mathbf{R} = \mathbf{I} + (\sin\theta)  [\mathbf u]_{\times} + (1-\cos\theta)\bigl( [\mathbf u]_{\times}\bigr)^2.</math></ref>

:<math> \mathbf{u}\otimes\mathbf{u}  = \mathbf{u}\mathbf{u}^\mathsf{T} = \begin{bmatrix}
u_x^2   & u_x u_y & u_x u_z \\[3pt]
u_x u_y & u_y^2 & u_y u_z \\[3pt]
u_x u_z & u_y u_z & u_z^2
\end{bmatrix},\qquad  [\mathbf u]_{\times} = \begin{bmatrix}
0  & -u_z & u_y \\[3pt]
u_z & 0 & -u_x \\[3pt]
-u_y & u_x & 0
\end{bmatrix}.</math>

In <math>\mathbb{R}^3</math> the rotation of a vector {{math|'''x'''}} around the axis {{math|'''u'''}} by an angle {{mvar|θ}} can be written as:
:<math>R_{\mathbf{u}}(\theta)\mathbf{x}=\mathbf{u}(\mathbf{u}\cdot\mathbf{x})+\cos\left(\theta\right)(\mathbf{u}\times\mathbf{x})\times\mathbf{u}+\sin\left(\theta\right)(\mathbf{u}\times\mathbf{x})</math>

or equivalently:
:<math>R_{\mathbf{u}}(\theta)\mathbf{x}= \mathbf{x} \cos(\theta) + \mathbf{u}(\mathbf{x} \cdot \mathbf{u})(1- \cos(\theta)) - \mathbf{x} \times \mathbf{u} \sin{\theta}</math>

This can also be written in [[Tensor|tensor notation]] as:<ref>{{cite journal|first1=T. R.|last1=Koehler | first2=S. B. |last2=Trickey|title=Euler vectors and rotations about an arbitrary axis|year=1978|journal=Am. J. Phys.|volume=46|issue=6|page=650|doi=10.1119/1.11223|bibcode=1978AmJPh..46..650K}}</ref>
:<math>(R_{\mathbf{u}}(\theta)\mathbf{x})_i = (R_{\mathbf{u}}(\theta))_{ij} {\mathbf{x}}_{j} \quad \text{with} \quad  (R_{\mathbf{u}}(\theta))_{ij} = \delta_{ij}\cos(\theta) + \mathbf{u}_i\mathbf{u}_j (1- \cos(\theta)) -  \sin{\theta} \varepsilon_{ijk} \mathbf{u}_{k} </math>

If the 3D space is right-handed and {{math|''θ'' > 0}}, this rotation will be counterclockwise when {{math|'''u'''}} points towards the observer ([[Right-hand rule]]). Explicitly, with <math>(\boldsymbol{\alpha}, \boldsymbol{\beta},\mathbf u)</math> a right-handed orthonormal basis,
:<math>
R_{\mathbf{u}}(\theta)\boldsymbol{\alpha}= \cos\left(\theta\right) \boldsymbol{\alpha} + \sin\left(\theta\right) \boldsymbol{\beta}, \quad
R_{\mathbf{u}}(\theta)\boldsymbol{\beta}= - \sin\left(\theta\right) \boldsymbol{\alpha} + \cos\left(\theta\right) \boldsymbol{\beta}, \quad
R_{\mathbf{u}}(\theta)\mathbf{u}=\mathbf{u}.
</math>

Note the striking ''merely apparent differences'' to the ''equivalent'' Lie-algebraic formulation [[#Exponential_map|below]].