proof of Rodrigues’ rotation formula

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

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

$${𝐯}^{\prime }=a{𝐱}^{\prime }+b{𝐲}^{\prime }+c𝐳,$$

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

${𝐱}^{\prime }$$=\cos \theta 𝐱+\sin \theta 𝐲,$

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

So

$${𝐯}^{\prime }=\cos \theta (a𝐱+b𝐲)+\sin \theta (a𝐲-b𝐱)+c𝐳.$$

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

$𝐯-(𝐯\cdot 𝐳)𝐳$$=𝐯-c𝐳=a𝐱+b𝐲,$

$𝐳\times 𝐯$$=a(𝐳\times 𝐱)+b(𝐳\times 𝐲)+c(𝐳\times 𝐳)=a𝐲-b𝐱.$

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

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

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

We attempt to simplify further:

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

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

We also have

$-(𝐯-(𝐯\cdot 𝐳)𝐳)$$=-a𝐱-b𝐲$(rotate $a𝐱+b𝐲$ by ${180}^{\circ }$)

$=𝐳\times (a𝐲-b𝐱)$(rotate $a𝐲-b𝐱$ by ${90}^{\circ }$)

$=𝐳\times (𝐳\times (a𝐱+b𝐲+c𝐳))$

$=A^2𝐯.$

So

$${𝐯}^{\prime }=I𝐯+\sin \theta A𝐯+(1-\cos \theta )A^2𝐯,$$

proving Rodrigues’ rotation formula.

Relation with the matrix exponential

Here is a curious fact. Notice that the matrix $A$ is skew-symmetric.This is not a coincidence — for any skew-symmetric matrix $B$, we have${(e^B)}^{\text{t}}=e^{B^{\text{t}}}=e^{-B}={(e^B)}^{-1}$,and $det{e}^B=e^{\mathrm{tr}B}=e^0=1$,so $e^B$ is always a rotation. It is in fact the case that:

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

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

$I,A,A^2,-A,-A^2,A,A^2,-A,-A^2,\mathrm{\dots }$

Then

$\sin \theta A$$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(-1)}^k{\theta }^{2k+1}A}{(2k+1)!}}={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{{\theta }^{2k+1}{A}^{2k+1}}{(2k+1)!}}={\displaystyle \sum _{k\text{ odd}}}{\displaystyle \frac{{(\theta A)}^k}{k!}}$

$(1-\cos \theta )A^2$$={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{{(-1)}^{k-1}{\theta }^{2k}{A}^2}{(2k)!}}={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{{\theta }^{2k}{A}^{2k}}{(2k)!}}={\displaystyle \sum _{k\ge 2\text{ even}}}{\displaystyle \frac{{(\theta A)}^k}{k!}}.$

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

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

Remark: The operator $e^{\theta A}$, as $\theta$ ranges over $ℝ$,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 abovecan be modified to prove this more general fact.