TITLE: negative quaternion
QUESTION [6 upvotes]: I often see claims that the negative of a quaternion represents the same rotation, just that the axis and angle have both been reversed. However, if I look at the axis-angle representation of the quaternion $q=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$ and then reverse the rotation angle $\theta=-\theta$ and rotation axis $\vec{n}=-\vec{n}$ and plug it in I get 
$$q(-\theta,-\vec{n})=[\cos(-\theta/2),-\vec{n}\sin(-\theta/2)]=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$$ but this is not the same as $-q=[-\cos(\theta/2),-\vec{n}\sin(\theta/2)]$. I must be missing something obvious but cannot figure it out. To me, negating the axis and angle gives $q$ not $-q$, so where am I going wrong?

REPLY [4 votes]: In addition to what Ted said that statement may mean the following. The rotation $R$ related to quaternion $q$ sends a vector $\vec{v}$ to the vector $$R(\vec{v})=q\,\vec{v}\,\overline{q}.$$ Because $\overline{-q}=-\overline{q}$, the rotation $R'$ related to $-q$ sends the vector $\vec{v}$ to
$$
\begin{aligned}
R'(\vec{v})&=(-q)\,\vec{v}\,\overline{-q}\\
&=(-1)^2q\,\vec{v}\,\overline{q}\\
&=q\,\vec{v}\,\overline{q}\\
&=R(\vec{v}).
\end{aligned}
$$
The multiplication of quaternions is bilinear, so we can move those $-1$ factors to the front when they cancel each other.
Anyway, the calculation shows that $R$ and $R'$ are the same rotation.