Proof of Chasles theorem (Kinematics)

Question

I have been trying to prove Chasles theorem using linear algebra. I am especially doubtful about whether the matrix can be inverted in the plane $\Pi$. And does this theorem also hold for infinitesimal transformations?

Theorem. Any rigid-body displacement in $\mathbb{R}^3$ is equivalent to a single screw displacement: a rotation by an angle $\phi$ about a line (the screw axis) combined with a translation $h$ parallel to that line.

Proof. Any proper Euclidean isometry in $\mathbb{R}^3$ can be written as $$\mathbf{r}' = \mathbf{R}\mathbf{r} + \mathbf{d},$$ where $\mathbf{R}\in \text{SO(3)}$ and $\mathbf{R}\neq \mathbf{I}.$ The screw displacement is defined by an axis location $\mathbf{r}_A$ and a parallel pitch $h$: $$\mathbf{r}' = \mathbf{R}(\mathbf{r}-\mathbf{r}_A)+ \mathbf{r}_A + h\mathbf{u},$$ where $\mathbf{u}$ is the unit vector in the direction of the rotation axis whose existence is guaranteed by the Euler rotation theorem. Hence, $\mathbf{Ru}=\mathbf{u}.$ Equating the the two forms we get: $$(\mathbf{R}-\mathbf{I}) \mathbf{r}_A = h\mathbf{u}-\mathbf{d}. \quad (1)$$ Now, our goal is to prove the existence of $\mathbf{r}_A$ and $h$. We decompose $\mathbf{r}_A$ and $\mathbf{d}$ into components perpendicular to and along $\mathbf{u}$: $$\mathbf{d} = \mathbf{d}_{\parallel} + \mathbf{d}_{\perp} \quad \text{where } \mathbf{d}_{\parallel} = (\mathbf{d} \cdot \mathbf{u})\mathbf{u} $$ $$\mathbf{r}_{A} = \mathbf{r}_{A,\parallel} + \mathbf{r}_{A,\perp}$$ Thus, we get $(\mathbf{R}-\mathbf{I})\mathbf{r}_{A,\parallel}=\mathbf{0}.$ Substituting in $(1)$, we have:

$$(\mathbf{R}-\mathbf{I}) \mathbf{r}_{A,\perp} = h\mathbf{u}-\mathbf{d}_\parallel - \mathbf{d}_\perp \quad (2)$$ Now, we take the dot product of the above equation with $\mathbf{u}$. The left hand side becomes: \begin{align*} \mathbf{u} \cdot [(\mathbf{R} - \mathbf{I})\mathbf{r}_{A,\perp}] &= \mathbf{u}\cdot(\mathbf{R}\mathbf{r}_{A,\perp}) - \mathbf{u}\cdot \mathbf{r}_{A,\perp} \\ &= (\mathbf{R}^\mathsf{T}\mathbf{u})\cdot \mathbf{r}_{A,\perp} - {0} \\ &= \mathbf{u}\cdot \mathbf{r}_{A,\perp} = {0}. \end{align*} Thus, we have:$$ 0=h(\mathbf{u}\cdot \mathbf{u})-\mathbf{u} \cdot \mathbf{d}_\parallel - \mathbf{u}\cdot \mathbf{d}_\perp $$ $$\implies h = \mathbf{u} \cdot \mathbf{d}_\parallel = \mathbf{u} \cdot \mathbf{d}.$$ Substituting this back into $(2)$, we are left with the equation in the plane $Π$ perpendicular to $\mathbf{u}$: $$(\mathbf{R}-\mathbf{I}) \mathbf{r}_{A,\perp} = - \mathbf{d}_\perp \quad (3).$$ Now, we show that $(\mathbf{R}-\mathbf{I})$ is invertible in $\Pi$. We represent the transformation $(\mathbf{R}-\mathbf{I})\mid_\Pi$ ​using a $2×2$ matrix $M$ defined by an orthonormal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ in $Π$. The transformation's matrix form is: $$ M = \begin{pmatrix} \cos\phi - 1 & -\sin\phi \\ \sin\phi & \cos\phi - 1 \end{pmatrix} $$

The determinant of $M$ is:

$$ \begin{align*} \det(M) &= (\cos\phi - 1)^2 - (-\sin\phi)(\sin\phi) \\ &= (\cos^2\phi - 2\cos\phi + 1) + \sin^2\phi \\ &= (\cos^2\phi + \sin^2\phi) + 1 - 2\cos\phi \\ &= 1 + 1 - 2\cos\phi \\ &= 2(1 - \cos\phi) \\ &= 4\sin^2\!\left(\frac{\phi}{2}\right) \end{align*} $$ which is $0$ iff $\phi \equiv 0 \pmod{2\pi}$, which is what we have assumed $(\mathbf{R}\neq \mathbf{I}).$ Hence we have: $$ \mathbf{r}_{A,\perp} = -(\mathbf{R} - \mathbf{I})^{-1}\big|_{\Pi} \, \mathbf{d}_{\perp}, $$ Which proves our claim. $\quad \square$

Any help would be appreciated.

Also I couldn't find elementary proofs of Chasles theorem using linear algebra, that is why I am asking here on MSE. Thank you for your time.

Answer 1

Since $R\ne I$, the restriction of $R$ on the plane $\Pi$ is a rotation for an angle $0<\theta<2\pi$. Hence $R\mathbf x\ne\mathbf x$ for every nonzero vector $\mathbf x\in\Pi$, meaning that $(R-I)|_{\Pi}$ is invertible.

Anyway, suppose $R$ is a rotation about the axis $\mathbf u$ by an angle $\theta\in(0,2\pi)$. Then $$ R=Q\pmatrix{1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&\cos\theta}Q^T $$ for some matrix $Q\in SO(3,\mathbb R)$ whose first column is $\mathbf u$. It follows that \begin{align*} \frac{(I-R)(I-R^T)}{\operatorname{tr}(I-R)} =\frac{2I-R-R^T}{2(1-\cos\theta)} =Q\pmatrix{0&0&0\\ 0&1&0\\ 0&0&1}Q^T =I-\mathbf u\mathbf u^T. \end{align*} Let $\mathbf r_A = \dfrac{(I-R^T)}{\operatorname{tr}(I-R)}\mathbf d$. Then $(I-R)\mathbf r_A=(I-\mathbf u\mathbf u^T)\mathbf d$. Let also $h = \mathbf u\cdot \mathbf d$. Then $h\mathbf u = \mathbf u\mathbf u^T\mathbf d$ and \begin{align*} &R(\mathbf r-\mathbf r_A)+\mathbf r_A+h\mathbf u\\ &=R\mathbf r+(I-R)\mathbf r_A+\mathbf u\mathbf u^T\mathbf d\\ &=R\mathbf r+(I-\mathbf u\mathbf u^T)\mathbf d+\mathbf u\mathbf u^T\mathbf d\\ &=R\mathbf r+\mathbf d. \end{align*}