proof of Rodrigues’ rotation formula

Let $[\mathbf{x},\mathbf{y},\mathbf{z}]$ be a frame of right-handed orthonormal vectors in $\mathbb{R}^{3}$ , and let $\mathbf{v}=a\mathbf{x}+b\mathbf{y}+c\mathbf{z}$ (with $a,b,c\in\mathbb{R}$ ) be any vector to be rotated on the $\mathbf{z}$ axis, by an angle $\theta$ counter-clockwise.

The image vector $\mathbf{v}^{\prime}$ is the vector $\mathbf{v}$ with its component in the $\mathbf{x},\mathbf{y}$ plane rotated, so we can write

$\mathbf{v}^{\prime}=a\mathbf{x}^{\prime}+b\mathbf{y}^{\prime}+c\mathbf{z}\,,\\$

where $\mathbf{x}^{\prime}$ and $\mathbf{y}^{\prime}$ are the rotations by angle $\theta$ of the $\mathbf{x}$ and $\mathbf{y}$ vectors in the $\mathbf{x},\mathbf{y}$ plane. By the rotation formula in two dimensions, we have

	$\displaystyle\mathbf{x}^{\prime}$	$\displaystyle=\cos\theta\,\mathbf{x}+\sin\theta\,\mathbf{y}\,,$
	$\displaystyle\mathbf{y}^{\prime}$	$\displaystyle=-\sin\theta\,\mathbf{x}+\cos\theta\,\mathbf{y}\,.$

$\mathbf{v}^{\prime}=\cos\theta(a\mathbf{x}+b\mathbf{y})+\sin\theta(a\mathbf{y}% -b\mathbf{x})+c\mathbf{z}\,.$

The vector $a\mathbf{x}+b\mathbf{y}$ is the projection of $\mathbf{v}$ onto the $\mathbf{x},\mathbf{y}$ plane, and $a\mathbf{y}-b\mathbf{x}$ is its rotation by $90^{\circ}$ . So these two vectors form an orthogonal frame in the $\mathbf{x},\mathbf{y}$ plane, although they are not necessarily unit vectors. Alternate expressions for these vectors are easily derived — especially with the help of the picture:

	$\displaystyle\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z}$	$\displaystyle=\mathbf{v}-c\mathbf{z}=a\mathbf{x}+b\mathbf{y}\,,$
	$\displaystyle\mathbf{z}\times\mathbf{v}$	$\displaystyle=a(\mathbf{z}\times\mathbf{x})+b(\mathbf{z}\times\mathbf{y})+c(% \mathbf{z}\times\mathbf{z})=a\mathbf{y}-b\mathbf{x}\,.$

Substituting these into the expression for $\mathbf{v}^{\prime}$ :

$\mathbf{v}^{\prime}=\cos\theta(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z% })+\sin\theta(\mathbf{z}\times\mathbf{v})+c\mathbf{z}\,,$

which could also have been derived directly if we had first considered the frame $[\mathbf{v}-(\mathbf{v}\cdot\mathbf{z}),\mathbf{z}\times\mathbf{v}]$ instead of $[\mathbf{x},\mathbf{y}]$ .

We attempt to simplify further:

$\mathbf{v}^{\prime}=\mathbf{v}+\sin\theta(\mathbf{z}\times\mathbf{v})+(\cos% \theta-1)(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z})\,.$

Since $\mathbf{z}\times\mathbf{v}$ is linear in $\mathbf{v}$ , this transformation is represented by a linear operator $A$ . Under a right-handed orthonormal basis, the matrix representation of $A$ is directly computed to be

$A\mathbf{v}=\mathbf{z}\times\mathbf{v}=\begin{bmatrix}0&-z_{3}&z_{2}\\ z_{3}&0&-z_{1}\\ -z_{2}&z_{1}&0\end{bmatrix}\,\begin{bmatrix}v_{1}\\ v_{2}\\ v_{3}\end{bmatrix}\,.$

We also have

$\displaystyle-(\mathbf{v}-(\mathbf{v}\cdot\mathbf{z})\mathbf{z})$	$\displaystyle=-a\mathbf{x}-b\mathbf{y}$	(rotate $a\mathbf{x}+b\mathbf{y}$ by $180^{\circ}$ )
	$\displaystyle=\mathbf{z}\times(a\mathbf{y}-b\mathbf{x})$	(rotate $a\mathbf{y}-b\mathbf{x}$ by $90^{\circ}$ )
	$\displaystyle=\mathbf{z}\times(\mathbf{z}\times(a\mathbf{x}+b\mathbf{y}+c% \mathbf{z}))$
	$\displaystyle=A^{2}\,\mathbf{v}\,.$

$\mathbf{v}^{\prime}=I\mathbf{v}+\sin\theta\,A\mathbf{v}+(1-\cos\theta)A^{2}\,% \mathbf{v}\,,$

proving Rodrigues’ rotation formula.

Relation with the matrix exponential

Here is a curious fact. Notice that the matrix¹¹If we want to use coordinate-free , then in this section, “matrix” should be replaced by “linear operator” and transposes should be replaced by the adjoint operation. $A$ is skew-symmetric. This is not a coincidence — for any skew-symmetric matrix $B$ , we have ${(e^{B})}^{\textrm{t}}=e^{{B}^{\textrm{t}}}=e^{-B}=(e^{B})^{-1}$ , and $\det e^{B}=e^{\operatorname{tr}B}=e^{0}=1$ , so $e^{B}$ is always a rotation. It is in fact the case that:

$\displaystyle I+\sin\theta\,A+(1-\cos\theta)A^{2}=e^{\theta A}$

for the matrix $A$ we had above! To prove this, observe that powers of $A$ cycle like so:

$\displaystyle I,A,A^{2},-A,-A^{2},A,A^{2},-A,-A^{2},\ldots$

Then

	$\displaystyle\sin\theta\,A$	$\displaystyle=\sum_{k=0}^{\infty}\frac{(-1)^{k}\theta^{2k+1}A}{(2k+1)!}=\sum_{% k=0}^{\infty}\frac{\theta^{2k+1}A^{2k+1}}{(2k+1)!}=\sum_{k\textrm{ odd}}\frac{% (\theta A)^{k}}{k!}$
	$\displaystyle(1-\cos\theta)\,A^{2}$	$\displaystyle=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\theta^{2k}A^{2}}{(2k)!}=\sum% _{k=1}^{\infty}\frac{\theta^{2k}A^{2k}}{(2k)!}=\sum_{k\geq 2\textrm{ even}}% \frac{(\theta A)^{k}}{k!}\,.$

Adding $\sin\theta\,A$ , $(1-\cos\theta)A^{2}$ and $I$ together, we obtain the power series for $e^{\theta A}$ .

Second proof: If we regard $\theta$ as time, and differentiate the equation $\mathbf{v}^{\prime}=a\mathbf{x}^{\prime}+b\mathbf{y}^{\prime}+c\mathbf{z}$ with respect to $\theta$ , we obtain $d\mathbf{v}^{\prime}/d\theta=a\mathbf{y}^{\prime}-b\mathbf{x}^{\prime}=\mathbf% {z}\times\mathbf{v}^{\prime}=A\mathbf{v}^{\prime}$ , whence the solution (to this linear ODE) is $\mathbf{v}^{\prime}=e^{\theta A}\mathbf{v}$ .

Remark: The operator $e^{\theta A}$ , as $\theta$ ranges over $\mathbb{R}$ , is a one-parameter subgroup of $\mathrm{SO}(3)$ . In higher dimensions $n$ , every rotation in $\mathrm{SO}(n)$ is of the form $e^{A}$ for a skew-symmetric $A$ , and the second proof above can be modified to prove this more general fact.

Title	proof of Rodrigues’ rotation formula
Canonical name	ProofOfRodriguesRotationFormula
Date of creation	2013-03-22 15:23:25
Last modified on	2013-03-22 15:23:25
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	14
Author	stevecheng (10074)
Entry type	Proof
Classification	msc 51-00
Classification	msc 15-00
Related topic	DimensionOfTheSpecialOrthogonalGroup