The rotation group is isomorphic to the orthogonal group $SO(3)$. So a rotation matrix can represent all the possible rotation transformations on the euclidean space $R3$ obtainable by the operation of composition.

The axis/angle notation describes any rotation that can be obtained by rotating a solid object around an axis passing on the reference origin by the given angle.

Given a pair $(\mathbf{u}, \theta)$, where $\mathbf{u}$ is a vector and corresponds to the axis rotation and $\theta$ is the roation angle, can this single pair represents all the possible rotations on the euclidean space R3 obtainable by composition?

  • 1
    $\begingroup$ If I understand your question correctly, this is Euler's rotation theorem. $\endgroup$
    – user856
    Feb 9, 2012 at 21:21
  • $\begingroup$ It seems correct! Feel free to make an answer (not a comment) so I can mark it as correct in few days (the time to fully convince myself). $\endgroup$
    – ceztko
    Feb 10, 2012 at 12:25
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    $\begingroup$ Not sure I do understand your question fully. But this might be helpful to you too: For each special orthogonal matrix in SO3, there is at least one axis-angle representation. In other words, the exponential map from tangent space of SO3 (=axis-angle) to SO3 is surjective (=onto), proof is e.g. in [seas.upenn.edu/~jean/diffgeom.pdf, pp.24]. $\endgroup$
    – B0rk4
    Feb 13, 2012 at 22:34
  • $\begingroup$ Hauke, the result "$exp: so(3) -> SO(3)$ is surjective" is actually a direct proof to my question. Euler's rotation theorem is less direct but it's also valid. Feel free to properly answer to the question so I can upvote and mark. $\endgroup$
    – ceztko
    Feb 14, 2012 at 13:37

1 Answer 1


The answer is yes, since the exponential map $\exp: \mathbf{so}(3) \rightarrow \mathbf{SO}(3)$ is surjective (=onto).

Long answer:

Axis-angle can be represented using a $3$-vector $\omega$ while the magnitude $\theta=|\omega|$ is the rotation angle and $\mathbf{u}=^\omega/_\theta$ is the rotation axis. 3-Vectors are closed under the cross product:

$$\omega_1\in \mathbb{R}^3, \omega_2\in \mathbb{R}^3\Rightarrow (\omega_1\times \omega_2)\in\mathbb{R}^3.$$

Each such vector $\omega$ has an equivalent $3\times 3$ matrix representation $\hat{\omega}$ (which is uniquely defined by $\hat{\omega}\cdot \mathbf{a} := \omega\times \mathbf{a}$ for $\mathbf{a}$ being a general 3-vector). The space of matrices of the form $\hat{\omega}$ is called the Lie algebra $\mathbf{so}(3)$. Thus, one can show that matrices of the form $\hat{\omega}$ are closed under the Lie bracket $[A,B]=AB-BA$:

$$\hat{\omega_1}\in \mathbf{so}(3), \hat{\omega_2}\in \mathbf{so}(3)\Rightarrow [\hat{\omega}_1, \hat{\omega}_2]\in\mathbf{so}(3).$$

Now, let us consider the matrix exponential: $\exp(\mathtt{A}) = \sum_{i=0}^\infty \frac{\mathtt{A}^i}{i!} $. Two poperties can be shown:

(1) If $\hat{\omega}\in\mathbf{so}(3)$, then $\exp(\hat{\omega})\in\mathbf{SO}(3)$.

$\mathbf{SO}(3)$ is the special orthogonal group in three dimensions. Thus, it consists of matrices which are orthogonal ($\mathtt{R}\cdot \mathtt{R}^\top=\mathtt{I}$) and the determinant is 1. In other word, it is the group of pure rotations.

(2) The exponential map $\exp: \mathbf{so}(3) \rightarrow \mathbf{SO}(3)$ is surjective.

So, (1) says that every $\exp(\hat{\omega})$ is a rotation matrix. And, (2) says that for each rotation matrix $\mathtt{R}$, there is at least one axis-angle representation $\omega$ so that $\exp(\hat{\omega})=\mathtt{R}$

Proofs of (1) and (2) are in corresponding text books, e.g. [Gallier, page 24].

  • $\begingroup$ Great answer ;) $\endgroup$
    – ceztko
    Feb 17, 2012 at 16:12

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