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 $\mathbb{R}^3$?



1 Answer 1


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 $p=\mathbf{u}$.)

  • $\begingroup$ What do you mean by $\exp(\theta\mathbf{u})=\cos\theta+\sin\theta\,\mathbf{u}$? It looks like you are adding a scalar $\cos\theta$ to the vector $\sin\theta\,\mathbf{u}$ to me $\endgroup$
    – user637978
    Commented May 6, 2022 at 15:25
  • 1
    $\begingroup$ @Frogwilldo That's exactly what I'm doing! A quaternion is a sum $a+b{\bf i}+c{\bf j}+d{\bf k}$ of a scalar $a$ and a 3D vector $b{\bf i}+c{\bf j}+d{\bf k}$ which doesn't simplify. (I mean, you could also represent quaternions other ways, but I claim this is best for the context.) $\endgroup$
    – anon
    Commented May 6, 2022 at 15:33
  • $\begingroup$ Oooh okay gotcha,thanks! $\endgroup$
    – user637978
    Commented May 6, 2022 at 15:34

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