# proof of Rodriguesβ 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

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

So

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

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 β 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 $90^{\circ}$) $\displaystyle=\mathbf{z}\times(\mathbf{z}\times(a\mathbf{x}+b\mathbf{y}+c% \mathbf{z}))$ $\displaystyle=A^{2}\,\mathbf{v}\,.$

So

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

proving Rodriguesβ 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, βmatrixβ should be replaced by βlinear operatorβ and transposes should be replaced by the adjoint operation. $A$ is skew-symmetric. This is not a coincidence β 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 Rodriguesβ rotation formula ProofOfRodriguesRotationFormula 2013-03-22 15:23:25 2013-03-22 15:23:25 stevecheng (10074) stevecheng (10074) 14 stevecheng (10074) Proof msc 51-00 msc 15-00 DimensionOfTheSpecialOrthogonalGroup