Derivative of square of skew symmetric matrix times a vector wrt the argument of the skew symmetric argument

Question

I've got a funny derivative that is giving me problems. It mostly has to do with the skew-symmetric matrix and trying to take a derivative with respect to what is inside the operator.

I've defined the skew-symmetric operator as

$$ \lfloor \omega \rfloor = \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x &0 \end{bmatrix}$$

where $\omega = \begin{bmatrix} \omega_x & \omega_y & \omega_z \end{bmatrix} ^\top $

I can solve the case below, because of the identity $ \lfloor \omega \rfloor v = -\lfloor v \rfloor \omega $

$$ \frac{\partial}{\partial \omega} \lfloor \omega \rfloor v = \frac{\partial}{\partial \omega} -\lfloor v \rfloor \omega \\ = -\lfloor v \rfloor $$

but what if I have omega skew squared? $$ \frac{\partial}{\partial \omega} \lfloor \omega \rfloor \lfloor \omega \rfloor v $$

I can't seem to manipulate this in a way that I can find the derivative. And it also seems that the chain rule is not doing what I wanted. I've tried the following:

$$ \frac{\partial}{\partial \omega} \lfloor \omega \rfloor \lfloor \omega \rfloor v = \frac{\partial}{\partial \omega} \lfloor \omega \rfloor \lfloor \omega \rfloor v \\ = \frac{\partial}{\partial \lfloor \omega \rfloor v}\left( \lfloor \omega \rfloor \lfloor \omega \rfloor v \right) \frac{\partial}{\partial \omega} \lfloor \omega \rfloor v \\ = -\lfloor \omega \rfloor \lfloor v \rfloor $$

but numerical differentiation doesn't agree.

I think I'm confused about what the skew-symmetric matrix really means. It feels like a vector, pretending to be a matrix and there must be some be rules when dealing with that I'm not following.

Answer 1

I found the answer using the Vector Triple Product.

$$ \frac{\partial}{\partial \omega} \left( \omega \times \omega \times v \right) = \frac{\partial}{\partial \omega} \left( \omega(\omega ^\top v) - v(\omega^\top \omega ) \right) \\ = I(\omega^\top v) + \omega v^\top - 2v\omega^\top $$

I confirmed this with numerical differentiation, and it seems to check out - python script

Answer 2

$ \def\LR#1{\left(#1\right)} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\rc#1{\color{red}{#1}} \def\CLR#1{\rc{\LR{#1}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $For ease of typing, let me use the notation
$$\eqalign{ W &= \lfloor w \rfloor, \qquad V &= \lfloor v \rfloor, \qquad F &= \lfloor f \rfloor \\ }$$ As you noted, the first function is easy to deal with $$\eqalign{ &f= Wv \;=\; \LR{w\times v} \;=\; -\rc Vw \qquad \\ &\grad fw = -\rc V\gradLR ww \;\equiv\; -V \\ }$$ The second function is a bit trickier $$\eqalign{ &g = W^2v \;=\; \rc Wf \;=\; -\rc Fw \qquad \\ &\grad gw = \rc W\gradLR fw - \rc F\gradLR ww \;\equiv\; -\LR{WV + F} \\ }$$