\[q=a+b \mathbf{i}+c \mathbf{j}+d \mathbf{k}\]
where \[a, \; b, \; c, d \in \mathbb{R}\]
with\[\mathbf{i}^2= \mathbf{j}^2 = \mathbf{k}^2 =-1\]
\[ij=k, \; jk=i, \; ki=j, \; ji=-k, \;kj=-i, \; ik=-j\]
forms a vector space over
\[\mathbb{R}\]
with basis \[1, \; \mathbf{i}, \; \mathbf{j}, \; \mathbf{k}\]
.We can represent a vector in
\[\mathbf{v}= \begin{pmatrix}x\\y\\z\end{pmatrix} \in \mathbb{R}^3\]
as the quaternion \[q=x \mathbf{i}+ y \mathbf{j}+z \mathbf{k}\]
. Let
\[\mathbf{v} \in \mathbb{R}^3\]
and let
\[q=a+b \mathbf{v}\]
be a unit quaternion, so that
\[\sqrt{a^2+b^2}=1\]
.Let
\[\mathbf{w} \in \mathbb{R}^3\]
and consider the mapping
\[T: \mathbf{w}q \mathbf{x} \bar{q}=(a+b \mathbf{v})\mathbf{w}(a-b \mathbf{v})\]
.This is also a vector in
\[\mathbb{R}^3\]
so the mapping above is a linear transformation from \[\mathbb{R}^3\]
to \[\mathbb{R}^3\]
.