# What is the derivative of a skew symmetric matrix?

I'm trying to work out some Jacobians and I ran across a problem. If I have a function of a vector making it a skew symmetric matrix, like below, what is the derivative $f'$?

$$f(\boldsymbol{\omega}) = \lfloor \boldsymbol{\omega} \, \times \rfloor = \left( \begin{array}{ccc} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{array} \right)$$

• Derivatives are usually something you have for a function depending on a single parameter. Perhaps you want the partial derivatives or the differential of the function? That the function is matrix valued is probably the least complication; at the end of the day, the differentiation will occur entrywise. – fuglede May 27 '14 at 10:17

You have a function from $\mathbb R^3$ to $M_{33}$ (the set of $3 \times 3$ matrices. Its three partial derivatives are $$\frac{\partial f}{\partial \omega_1} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \\ \frac{\partial f}{\partial \omega_2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{bmatrix} \\ \frac{\partial f}{\partial \omega_3} = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} .$$
I suppose that I could treat $M_{33}$ as $\mathbb R^9$, and write out a $9 \times 3$ matrix, but would that really be any better?
• Well, maybe a higher-level observation will help: the function $f$ is linear; that means that it is its own derivative. More exactly, $f'(\omega)$ is a constant (transformation-valued) function, whose value at any location $\omega_0$ is just $f(\omega_0)$. Is that more what you were looking for? – John Hughes May 27 '14 at 12:30
Employing the Levi-Civita alternating tensor $\varepsilon$, your function can be written as $$f = \,-\varepsilon\cdot\omega$$ whose derivative is \eqalign{ \frac{\partial f}{\partial \omega} &= \,-\varepsilon\cdot\frac{\partial \omega}{\partial \omega} \cr &= \,-\varepsilon\cdot I \cr &= \,-\varepsilon \cr }