proof of Rodrigues’ rotation formula
The image vector is the vector with its component in the plane rotated, so we can write
The vector is the projection of onto the plane, and is its rotation by . So these two vectors form an orthogonal frame in the plane, although they are not necessarily unit vectors. Alternate expressions for these vectors are easily derived — especially with the help of the picture:
Substituting these into the expression for :
which could also have been derived directly if we had first considered the frame instead of .
We attempt to simplify further:
We also have
|(rotate by )|
|(rotate by )|
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. is skew-symmetric. This is not a coincidence — for any skew-symmetric matrix , we have , and , so is always a rotation. It is in fact the case that:
for the matrix we had above! To prove this, observe that powers of cycle like so:
Adding , and together, we obtain the power series for .
Second proof: If we regard as time, and differentiate the equation with respect to , we obtain , whence the solution (to this linear ODE) is .
Remark: The operator , as ranges over , is a one-parameter subgroup of . In higher dimensions , every rotation in is of the form for a skew-symmetric , and the second proof above can be modified to prove this more general fact.
|Title||proof of Rodrigues’ rotation formula|
|Date of creation||2013-03-22 15:23:25|
|Last modified on||2013-03-22 15:23:25|
|Last modified by||stevecheng (10074)|