Define the matrix representation of the cross product as the following:
Define $so(n)$ :
And finally define the following Lemma:
(Source: Page 26 and 28 of A Mathematical Introduction to Robotic Manipulation by Richard M. Murray, Zexiang Li and S. Shankar Sastry)
Using all of these I want to prove $\omega \times (-\omega \times q) = q$ when $||\omega|| = 1$.
Here I go:
\begin{align*} \omega \times (-\omega \times q) & = \hat{\omega}(-\hat{\omega}q)\\ & = -\hat{\omega}(\hat{\omega}q) \tag*{Pulling a constant out of matrix multiplication}\\ & = -(\hat{\omega}\hat{\omega})q \tag*{Because matrix multiplication is associative}\\ & = -\hat{\omega}^2q \end{align*}
Using Lemma 2.3, $\hat{\omega}^2 = \omega\omega^T - ||\omega||^2I = \omega\omega^T - I$ and so $-\hat{\omega}^2q = (I - \omega\omega^T)q$.
So we need $\omega\omega^T = 0$ to prove that $-\hat{\omega}^2q = q$.
$$\omega\omega^T = \begin{pmatrix}\omega_1^2 & \omega_1\omega_2 & \omega_1\omega_3 \\ \omega_1\omega_2 & \omega_2^2 & \omega_2\omega_3 \\ \omega_1\omega_3 & \omega_2\omega_3 & \omega_3^2\end{pmatrix}$$
However this matrix is not 0 unless all components of $\omega$ are 0, which violates $||\omega|| = 1$ so we have a contradiction. Where is my mistake?