Example

Say you are standing on the ground and you pick the direction of gravity to be the negative z direction. Then if you turn to your left, you will rotate -π/2 radians (or -90°) about the -z axis. Viewing the axis-angle representation as an ordered pair, this would be

$${\displaystyle (\mathrm {axis} ,\mathrm {angle} )=\left({\begin{bmatrix}e_{x}\\e_{y}\\e_{z}\end{bmatrix}},\theta \right)=\left({\begin{bmatrix}0\\0\\-1\end{bmatrix}},{\frac {-\pi }{2}}\right).}$$

The above example can be represented as a rotation vector with a magnitude of π/2 pointing in the z direction,

$${\displaystyle {\begin{bmatrix}0\\0\\{\frac {\pi }{2}}\end{bmatrix}}.}$$

Exponential map from 𝔰𝔬(3) to SO(3)

The exponential map effects a transformation from the axis-angle representation of rotations to rotation matrices,

$${\displaystyle \exp \colon {\mathfrak {so}}(3)\to \mathrm {SO} (3)\,.}$$

Essentially, by using a Taylor expansion one derives a closed-form relation between these two representations. Given a unit vector ${\textstyle {\boldsymbol {\omega }}\in {\mathfrak {so}}(3)=\mathbb {R} ^{3}}$ representing the unit rotation axis, and an angle, θ ∈ R, an equivalent rotation matrix R is given as follows, where K is the cross product matrix of ω, that is, Kv = ω × v for all vectors v ∈ R³,

$${\displaystyle R=\exp(\theta \mathbf {K} )=\sum _{k=0}^{\infty }{\frac {(\theta \mathbf {K} )^{k}}{k!}}=I+\theta \mathbf {K} +{\frac {1}{2!}}(\theta \mathbf {K} )^{2}+{\frac {1}{3!}}(\theta \mathbf {K} )^{3}+\cdots }$$

Because K is skew-symmetric, and the sum of the squares of its above-diagonal entries is 1, the characteristic polynomial P(t) of K is P(t) = det(K − tI) = −(t³ + t). Since, by the Cayley–Hamilton theorem, P(K) = 0, this implies that

$${\displaystyle \mathbf {K} ^{3}=-\mathbf {K} \,.}$$

As a result, K⁴ = –K², K⁵ = K, K⁶ = K², K⁷ = –K.

This cyclic pattern continues indefinitely, and so all higher powers of K can be expressed in terms of K and K². Thus, from the above equation, it follows that

$${\displaystyle R=I+\left(\theta -{\frac {\theta ^{3}}{3!}}+{\frac {\theta ^{5}}{5!}}-\cdots \right)\mathbf {K} +\left({\frac {\theta ^{2}}{2!}}-{\frac {\theta ^{4}}{4!}}+{\frac {\theta ^{6}}{6!}}-\cdots \right)\mathbf {K} ^{2}\,,}$$

that is,

$${\displaystyle R=I+(\sin \theta )\mathbf {K} +(1-\cos \theta )\mathbf {K} ^{2}\,,}$$

by the Taylor series formula for trigonometric functions.

This is a Lie-algebraic derivation, in contrast to the geometric one in the article Rodrigues' rotation formula.^[1]

Due to the existence of the above-mentioned exponential map, the unit vector ω representing the rotation axis, and the angle θ are sometimes called the exponential coordinates of the rotation matrix R.

Log map from SO(3) to 𝔰𝔬(3)

Let K continue to denote the 3 × 3 matrix that effects the cross product with the rotation axis ω: K(v) = ω × v for all vectors v in what follows.

To retrieve the axis–angle representation of a rotation matrix, calculate the angle of rotation from the trace of the rotation matrix:

$${\displaystyle \theta =\arccos \left({\frac {\operatorname {Tr} (R)-1}{2}}\right)}$$

and then use that to find the normalized axis,

$${\displaystyle {\boldsymbol {\omega }}={\frac {1}{2\sin \theta }}{\begin{bmatrix}R_{32}-R_{23}\\R_{13}-R_{31}\\R_{21}-R_{12}\end{bmatrix}}~,}$$

where ${\displaystyle R_{ij}}$ is the component of the rotation matrix, ${\displaystyle R}$, in the ${\displaystyle i}$-th row and ${\displaystyle j}$-th column.

The axis-angle representation is not unique since a rotation of ${\displaystyle -\theta }$ about ${\displaystyle -{\boldsymbol {\omega }}}$ is the same as a rotation of ${\displaystyle \theta }$ about ${\displaystyle {\boldsymbol {\omega }}}$.

The above calculation of axis vector ${\displaystyle \omega }$ does not work if R is symmetric. For the general case the ${\displaystyle \omega }$ may be found using null space of R-I, see rotation matrix#Determining the axis.

The matrix logarithm of the rotation matrix R is

$${\displaystyle \log R={\begin{cases}0&{\text{if }}\theta =0\\{\dfrac {\theta }{2\sin \theta }}\left(R-R^{\mathsf {T}}\right)&{\text{if }}\theta \neq 0{\text{ and }}\theta \in (-\pi ,\pi )\end{cases}}}$$

An exception occurs when R has eigenvalues equal to −1. In this case, the log is not unique. However, even in the case where θ = π the Frobenius norm of the log is

$${\displaystyle \|\log(R)\|_{\mathrm {F} }={\sqrt {2}}|\theta |\,.}$$

Given rotation matrices A and B,

$${\displaystyle d_{g}(A,B):=\left\|\log \left(A^{\mathsf {T}}B\right)\right\|_{\mathrm {F} }}$$

is the geodesic distance on the 3D manifold of rotation matrices.

For small rotations, the above computation of θ may be numerically imprecise as the derivative of arccos goes to infinity as θ → 0. In that case, the off-axis terms will actually provide better information about θ since, for small angles, R ≈ I + θK. (This is because these are the first two terms of the Taylor series for exp(θK).)

This formulation also has numerical problems at θ = π, where the off-axis terms do not give information about the rotation axis (which is still defined up to a sign ambiguity). In that case, we must reconsider the above formula.

$${\displaystyle R=I+\mathbf {K} \sin \theta +\mathbf {K} ^{2}(1-\cos \theta )}$$

At θ = π, we have

$${\displaystyle R=I+2\mathbf {K} ^{2}=I+2({\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}-I)=2{\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}-I}$$

and so let

$${\displaystyle B:={\boldsymbol {\omega }}\otimes {\boldsymbol {\omega }}={\frac {1}{2}}(R+I)\,,}$$

so the diagonal terms of B are the squares of the elements of ω and the signs (up to sign ambiguity) can be determined from the signs of the off-axis terms of B.

Unit quaternions

the following expression transforms axis–angle coordinates to versors (unit quaternions):

$${\displaystyle Q=\left(\cos {\tfrac {\theta }{2}},{\boldsymbol {\omega }}\sin {\tfrac {\theta }{2}}\right)}$$

Given a versor q = s + x represented with its scalar s and vector x, the axis–angle coordinates can be extracted using the following:

$${\displaystyle {\begin{aligned}\theta &=2\arccos s\\[8px]{\boldsymbol {\omega }}&={\begin{cases}{\dfrac {\mathbf {x} }{\sin {\tfrac {\theta }{2}}}},&{\text{if }}\theta \neq 0\\0,&{\text{otherwise}}.\end{cases}}\end{aligned}}}$$

A more numerically stable expression of the rotation angle uses the atan2 function:

$${\displaystyle \theta =2\operatorname {atan2} (|\mathbf {x} |,s)\,,}$$

where |x| is the Euclidean norm of the 3-vector x.