Trying to find a nice basis to realize the quaternion mapping as a rotation matrix Trying to find a nice basis to realize the quaternion mapping as a rotation matrix.
The quaternions are a 4 dimensional division algebra over $\mathbb{R}$ where we label the standard basis vectors as $e_1=$1, $e_2$=i $e_3$=j and $e_4=$k and define the multiplication by ij=k, jk=i, ki=j, ii=jj=kk=-1.
The quaternions were invented by William Rowan Hamilton as a nice way to express rotations in $\mathbb{R}^3$. A rotation in $\mathbb{R}^3$ is completly specified by it's axis of rotation and angle of rotation. Say we want to rotate $\mathbb{R}^3$ an angle of $a$ about the axis of rotation $(b,c,d)$. To achieve this let $r=$a+bi+cj+dk and also associate each point $(x,y,z) \in \mathbb{R}^3$ to a "pure" quaternion xi+yj+zk.
Now, define a linear map:
$R_r: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defind by $R_r(x) = rxr^{-1}$
This map achieves the desired rotation and has many nice properties properties, in particular composition of rotations corresponds to multiplication of quaternions i.e. $R_{rs}=R_rR_s$

I am currently trying to understand (in as many ways as possible) why this map indeed achieves the desired rotation.
For $r=$a+bi+cj+dk, I have verified that $(b,c,d)$ is an eigenvector of $R_r$ with eigenvalue $1$.
Using the basis $\{(b,c,d),(0,1,0)(0,0,1)\}$ I've computd the matrix for $R_r$ as:
$ \begin{bmatrix}
1 & 2(bc-ad) & 2(ac+bd) \\
0 & a^2-b^2+c^2-d^2 & 2(cd-ab) \\
0 & 2(ab+cd) & a^2-b^2-c^2+d^2 
\end{bmatrix}  $
However, to me this is not in any way an obvious rotation matrix. I'd like to find a basis where the matrix has the following form:
$ \begin{bmatrix}
1 & 0 & 0 \\
0 & m & -n \\
0 & n & m 
\end{bmatrix}  $
Furthermore, I'm a scrub who only undestands $2\times2$ rotation matrices, and I'm under the impression that in the standard basis the corresponding matrix is pretty clearly a $3 \times 3$ rotation matrix. However, I want to use the fact that we have this nice eigenvector to express the matrix in a nicer form.

So I guess the point of this post is two things:
$1)$: What is a nice basis for $\mathbb{R}^3$ so that I can express the matrix of $R_r$ in a nice form?
$2)$ What are some other ways to understand that this map is indeed a rotation of $\mathbb{R}^3$?
Thanks!
 A: WLOG we use a unit quaternion, and these have polar forms $\exp(\theta\mathbf{u})=\cos\theta+\sin\theta\,\mathbf{u}$ where $\mathbf{u}$ is a unit vector. (Note the unit vectors are precisely the square roots of $-1$.)
(1) When conjugating by $\exp(\theta\mathbf{u})$, extend to an orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ (oriented appropriately). To calculate the matrix, use the fact $\mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}$ for vectors $\mathbf{a}$ and $\mathbf{b}$. (This follows from the fact multiplication is bilinear, and if we write the multiplication table for $\{\mathbf{i},\mathbf{j},\mathbf{k}\}$ down the scalar and vector components of the entries are the minus dot product and cross product respectively by inspection.)
(2) The fact that the conjugation $pxp^{-1}$ involves two $p$s (where $p=\exp(\theta\mathbf{u})$) and rotates by $2\theta$ suggests both $p$s (on the left and right) contribute a single $\theta$ to the final outcome. This can be seen by looking at the left and right multiplication maps $L_p(x)=px$ and $R_p(x)=x\overline{p}$; each is an isoclinic rotation in a matching pair of planes. To see this, extend to an orthonormal basis $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ and consider the $1\mathbf{u}$ and $\mathbf{vw}$-planes separately. (By Euler's formula for $p$, it suffices to check what happens when $p=1$ and $p=\mathbf{u}$.)
