I am working on a function to find the rotation matrix to transform one vector to another, and the Wikipedia article has a notation that differs from a StackOverflow post that has come up in my search.
Does this notation represent the skew-symmetric cross product? (x and y are vectors, I is the identity matrix)
notation in question: $ yx^T – xy^T $
full equation: $R:= I + yx^T – xy^T + \frac{ 1 }{ 1 + <x, y> } (yx^T – xy^T)^2$
https://en.wikipedia.org/wiki/Rotation_matrix#Vector_to_vector_formulation
Related post to perform what I believe to be the same function, but explicitly states skew-symmetric cross product.
equation from post adapted for clarity: $R = I + [v] _{\times} + [v] _{\times}^2 \frac{ 1 }{ 1 + x \cdot y}$
Calculate Rotation Matrix to align Vector A to Vector B in 3d?
Could anyone point to a resource that dives deeper into the derivation of this formula?
Edit: Found a resource that explains this formulation in more detail.
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Rotation_matrix.html#Vector_rotation
And
Möller, T., & Hughes, J. F. (1999). Efficiently Building a Matrix to Rotate One Vector to Another. Journal of Graphics Tools, 4(4), 1–4. doi:10.1080/10867651.1999.10487509
Best Answer
Yup it does, since if
$$v = x \times y = \begin{pmatrix} x_{2}y_{3} - y_{2}x_{3} \\ -x_{1}y_{3} + y_{1}x_{3} \\ x_{1}y_{2} - x_{2}y_{1} \end{pmatrix} = \begin{pmatrix} v_{1} \\ v_{2} \\ v_{3} \end{pmatrix},$$
then
$$ [v]_{\times} := \begin{pmatrix} 0 && -v_{3} && v_{2} \\ v_{3} && 0 && -v_{1} \\ -v_{2} && v_{1} && 0 \end{pmatrix} = \begin{pmatrix} x_{1}y_{1} && x_{2}y_{1} && x_{3}y_{1} \\ x_{1}y_{2} && x_{2}y_{2} && x_{3}y_{2} \\ x_{1}y_{3} && x_{2}y_{3} && x_{3}y_{3} \end{pmatrix} - \begin{pmatrix} x_{1}y_{1} && x_{1}y_{2} && x_{1}y_{3} \\ x_{2}y_{1} && x_{2}y_{2} && x_{2}y_{3} \\ x_{3}y_{1} && x_{3}y_{2} && x_{3}y_{3} \end{pmatrix} = yx^{T} - xy^{T},$$
keeping in mind that $x,y \in \mathbb{R}^{3}$ are column vectors, or $(3 \times 1)$-matrices (so $x^{T}, y^{T}$ are $(1 \times 3)$-matrices). For the more "complicated-looking" 3rd term, it's still just multiplying everything out - the Wikipedia article considers vectors on the sphere however, so $xx^{T} = 1 = yy^{T}$, etc.