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

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.

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