proof for euler-rodrigues formula - matrix form

Question

I need for a matrix representation. Exactly I want to know how to get the Euler-Rodrigues formula in a matrix form like here.

Thanks!

Answer 1

$ \newcommand{\ii}{\mathbf i} \newcommand{\jj}{\mathbf j} \newcommand{\kk}{\mathbf k} \newcommand{\one}{\mathbf 1} \newcommand{\qq}{\mathbf q} \newcommand{\xx}{\mathbf x} $ I'm going to assume you know about quaternions and linear algebra. Let H denote the 4-d vector space of quaternions, with basis $\one, \ii, \jj, \kk$. Let $\qq = a\one + b\ii + c\jj + d\kk$ be a quaternion. Then left multiplication by $\qq$, i.e.,

$$ L_\qq: H \to H : \xx \mapsto \qq \xx $$ is a linear transformation on $H$. Its matrix, wrt the basis above, looks like $$ M_\qq = \begin{bmatrix} a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a \end{bmatrix}. $$ $$\newcommand{\qbar} {\bar{\qq}}$$ Similarly, right multiplication by $\qq$ leads to a matrix $$N_\qq = \begin{bmatrix} a & -b & -c & -d \\ b & a & d & -c \\ c & -d & a & b \\ d & c & -b & a \end{bmatrix}. $$ And right multiplication by $\bar{\qq} = a \one - b \ii - c \jj - d \kk$ therefore leads to $$ N_\qbar= \begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix}. $$.

So conjugation, the map $$ T: H \to H: \xx \mapsto \qq \xx \qbar $$ has matrix $A = N_{\qbar} M_\qq$, which I'll let you compute.

The function $T$ is the identity on the "pure real" subspace, so the first row and column of $A$ are $[1,0,0,0]$. What's the lower right corner of $A$? It's the matrix for the transformation on the "pure vector" subspace of $H$ (the span of $\ii,\jj,\kk$), which can be thought of as $\mathbb R^3$. That matrix, when $\qq$ is a unit quaternion (i.e., $a^2 + b^2 + c^2 + d^2$) is exactly the one shown in the Wikipedia article.

Here's a bit of a reason why:

First, for a unit quaternion $\qq$, the transformation $L_\qq$ turns out to be orthogonal (easy to verify by checking that $M_\qq M_\qq^t = I$) and orientation preserving (easy to show by a continuity argument: the det is $1$ for $a = 1, b = c = d = 0$, and is clearly a continuous function of $a,b,c,d$, which lie on a unit sphere in 3-space; a continuous function from a connected set to a discrete space like $\{-1, 1\}$ is constant). The same goes for right multiplication, and hence for their product. So the matrix $A$ must also be orthogonal and have determinant $+1$. Every such matrix has $1$ as an eigenvalue; it's not too hard to check that $\omega = (b, c, d)$ is actually an eigenvector. So the transformation represented by $A$ is orthogonal, fixes $\omega$, and therefore takes vectors orthogonal to $\omega$ to vectors orthogonal to $\omega$, in an angle-preserving way, i.e., $A$ rotates $3$-space, fixing $\omega$, and rotating in the plane perpendicular to $\omega$. By what angle? By $\theta = 2\cos^{-1} (a)$. Thus the quaternion $\qq$ and the quaternion $-\qq$ turn out to produce the same rotation under this construction. This construction is therefore a 2-to-1 cover of $SO(3)$ by $S^3$. This is all just an amplification of a few sentences in Steenrod's book on Fiber Bundles.