4.2.4 Rotations and Euclidean Motions

A general reflection at the plane $w^\perp$ can be described by first rotating $w$ into some vector $v$, then reflect at $v^\perp$ and now rotate back $v$ to $w$. In fact, let $u:=v+w$. Then this first rotation is given by $S_u\o S_w=R=S_v\o S_u$ and we have

$$S_v\o R=S_v\o (S_v\o S_u) = (S_v)^2\o S_u=S_u=S_u\o (S_w)^2=R\o S_w.$$

How can rotations be described by matrices? Let us first consider 2-dimensional space. The rotation by $90^o$ maps any vector $v$ to a normal vector $v^\perp$ of the same length. There are only two possibilities for $\langle a,b\rangle^\perp$ namely $\pm \langle -b,a\rangle$, and the one with $+$ is rotation in the positive direction, i.e. counterclockwise. The matrix corresponding to this rotation is given by

$$R_{\pi/2} =\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}.$$

Now what about a rotation around 0 by an arbitrary angle $\varphi$? By elementary geometric calculations we see, that the image of $x$ is $\cos{\varphi }\,x+\sin{\varphi }\,y$, and the image of $y$ which is $x$ turned $90^o$ counterclockwise is the vector $-\sin{\varphi }\,x+\cos{\varphi }\,y$. Thus in coordinates this linear mapping is given by the unitarian matrix

$$\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{pmatrix}$$

Note that this can also be paraphrased by ``the reference frame is rotated in the opposite direction by $\varphi$''. In three dimensions a rotation by $\varphi$ around the $z$-axes does not change the $z$-coordinate and rotates the $(x,y)$-coordinates as before and thus is given by the unitarian matrix

$$\begin{pmatrix} \cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

In Pov-Ray: rotate $\varphi$*z.

Similarly rotations around the $x$- and around the $y$-axes are given by

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\varphi & -\sin\varphi \\ 0 & \sin\varphi & \cos\varphi \end{pmatrix}$$    and $$\begin{pmatrix} \cos\varphi & 0 & -\sin\varphi \\ 0 & 1 & 0 \\ \sin\varphi & 0 &\cos\varphi \end{pmatrix}$$

In Pov-Ray: rotate $\varphi$*x and rotate $\varphi$*y.

Note that the composition of two rotations by angles $\varphi$ and $\psi$ around the same center is the rotation by angle $\varphi +\psi$. Expressing this via the corresponding matrices gives the addition laws for $\sin$ and for $\cos$.

All these rotation matrices $M$ (as well as arbitrary compositions of such) satisfy $\det(M)=1$ and $M^t\cdot M=\operatorname{id}$, thus are special orthogonal matrices.

A general rotation around the axis spanned by the unit vector $w$ by the angle $\varphi$ is given by considering the orthogonal frame given by $w$, $v\times w$, $v-\langle v\vert w\rangle w$. The length of these vectors are $1$, $\sin(\angle(v,w))\Vert v\Vert$, $\sin(\angle(v,w))\Vert v\Vert$. The vector $v$ is given in this frame as

$$v=\langle v\vert w\rangle\cdot w + 1\cdot(v-\langle v\vert w\rangle w) + 0\cdot v\times w,$$

so the rotation is

$$v\mapsto \langle v\vert w\rangle\cdot w + \cos(\varphi )\cdot(v-\langle v\vert w\rangle w) + \sin(\varphi )\cdot v\times w$$

Figure: Orthogonal decomposition of a vector $v$

Note that

$$w\times (v \times u)=(u\times v)\times w = \langle u\vert w\rangle v - \langle v\vert w\rangle u$$

$$\qquad v\times (v\times u) = \langle u\vert v\rangle v - \langle v\vert v\rangle u = \langle u\vert v\rangle v - u.$$

See www.martinb.com/.../index.htm.

Let $f$ be an arbitrary length preserving mapping (a so called EUCLIDEAN MOTION), i.e. $\Vert f(v_1)-f(v_2)\Vert=\Vert v_1-v_2\Vert$ for all $v_1,v_2$. Up to the translation by $f(O)$ (i.e. replacing $f$ by $v\mapsto f(v)-f(0)$) it preserves also the origin $O$ and hence $\Vert f(v)\Vert=\Vert v\Vert$ for all $v$. Furthermore by the polarization equality

$$2\langle v\vert w\rangle = \Vert v\Vert^2+\Vert w\Vert^2-\Vert v-w\Vert^2$$

we get

$$2\langle f(v)\vert f(w)\rangle = \Vert f(v)\Vert^2+\Vert f(w)\Vert^2-\Vert f(v)-f(w)\Vert^2$$

$$= \Vert v\Vert^2+\Vert w\Vert^2-\Vert v-w\Vert^2 = 2\langle v\vert w\rangle,$$

i.e. $f$ preserves the inner product and hence maps the standard orthogonal basis $e_1=x$, $e_2=y$ and $e_3=z$ to some other orthonormal basis $f_1:=f(e_1)$, $f_2:=f(e_2)$ and $f_3:=f(e_3)$. For a general vector $v$ we have $v=\sum_k v_k\,e_k$ with $v_k=\langle v\vert e_k\rangle=\langle f(v)\vert f(e_k)\rangle=\langle f(v)\vert f_k\rangle$ and hence

$$f(v)=\sum_k \langle f(v)\vert f_k\rangle f_k=\sum_k v_k\,f_k,$$

i.e. $f$ is linear and, moreover, for the matrix $[f]$ associated to $f$ we have $[f]^t\cdot [f]=\operatorname{id}$, i.e. $f$ is given by an orthogonal matrix: In fact the $jk^{th}$ entry in the matrix $[f]^t\cdot [f]$ is given by

$$([f]^t\cdot [f])_{j,k} = [f^*\o f]_{j,k} = \langle (f^* \o f)e_j\vert e_k\rangle = \langle fe_j\vert fe_k\rangle$$

$$= \langle e_j\vert e_k\rangle =\left\{\begin{array}{ll} 1 &\text{for}j=k, \\ 0 &\text{otherwise}.\end{array}\right.$$

Let now conversely an orthogonal $3\times 3$-matrix $A$ be given. Then $1=\det(\operatorname{id})=\det(A^t\cdot A)=\det(A^t)\cdot \det(A)=\det(A)^2$, i.e. $\det(A)\in\{-1,+1\}$. Let us assume first that $\det(A)=1$.

Let $F:=\operatorname{Fix}(A):=\{x\in E:Ax=x\}$ be the set of its fixed points, i.e. the eigenspace for the eigenvalue 1. This is a linear subspace.

We show next, that $\dim(F)>0$. Let $p(\lambda ):=\det(\lambda -A)$ be the characteristic polynomial of $A$. We have

$$p(\lambda ) = \det(\lambda -A) = \det((\lambda - A)^t) = \det(\lambda -A^t) = \det(\lambda -A^{-1})$$

$$= \det(-\lambda \left(\frac1{\lambda }-A\right)A^{-1}) = \det(-\lambda ) \cdot \det\left(\frac1{\lambda }-A\right)\cdot \det A^{-1}$$

$$= (-\lambda )^3 p(\frac1\lambda ),$$

thus we get for $\lambda =1$ that $2p(1)=0$. Hence $1$ is an eigenvalue.

In case its dimension is 3, we have $A=\operatorname{id}$.

In case its dimension is 2, we find a unit vector $w$ such that $F=w^\perp$. Since $F$ is invariant under $A$ and $A$ is orthogonal the same is true for $F^{\perp}$. Since $w\in F^{\perp}$ and $A$ is an isometry we have $A(w)=-w$ and hence $A$ is the reflection at the plane $F=w^\perp$. In fact, $x-\langle x\vert v\rangle v\in F$ for all $x\in E$, since

$$\langle x-\langle x\vert v\rangle v\vert v\rangle = \langle x\vert v\rangle - \langle x\vert v\rangle \cdot \langle v\vert v\rangle = 0.$$

Thus

$$Ax = A\Bigl( ( x-\langle x\vert v\rangle v) + \langle x\vert v\rangle v \Bigr) = x-\langle x\vert v\rangle v + \langle x\vert v\rangle (-v)$$

$$= x-2 \langle x\vert v\rangle v.$$

Remains to consider the case, where $\dim(F)=1$. Then there exists a unit vector $w$ which spans $F$. The orthogonal plane $w^{\perp}=F^{\perp}$ is also $A$ invariant, so $A\vert _{F^{\perp}}$ is orthogonal on this plane and hence a reflection on a line in this plane or a rotation in this plane. Thus $A$ is a reflection at the plane spanned by $w$ and the reflection line of $A\vert _{F^{\perp}}$ or a rotation with axis $w$.

Therefore any orthogonal mapping on $\protect\mathbb{R}^3$ is the composite of (at most three) reflections. It is a rotation around some axes $v\ne 0$ by some angle $\varphi$ iff it is a composite of two reflections.
In Pov-Ray: $v\mapsto$vaxis_rotate $(v,w,\varphi )$.
Finally the euclidean motions are exactly of the form $v\mapsto R\cdot v + w$, where $R$ is a rotation which is followed by the translation $v\mapsto v+w$.

We show next that any rotation (i.e. special orthogonal matrix) can be obtained by composing 3 of the special rotations discussed above by the so called EULER ANGLES. Consider an airplane or an hang-glider: We have the basis given by the axes of airplane: the direction from the left to the right wing, the vertical direction, and the direction from back to front.

• First we turn the head in the required direction via a rotation $R_1$ around $y$ by angle $\psi$ called heading.
• Next we raise/lower the head by a rotation $R_2$ around the new $x$ by an angle $\th$ called attitude.
• Finally we tilt the airplane by a rotation $R_3$ around the new $z$ by the angle $\varphi$ called bank.

How do the corresponding matrices look like?

• Clearly
  $$[R_1]:=\begin{pmatrix} \cos(\psi ) & 0 & -\sin(\psi ) \\ 0 & 1 & 0 \\ \sin(\psi ) & 0 & \cos(\psi ) \end{pmatrix}$$

• In the new basis $\mathcal B'=(x',y,z')$ we have
  $$[R_2]_{\mathcal B',\mathcal B'}= \begin{pmatrix} 1 & 0 & 0 \\ 0 &\cos(\th ) & -\sin(\th ) \\ 0 & \sin(\th ) & \cos(\th ) \end{pmatrix}$$
  In linear algebra (see [Kri02, 10.28] and [Kri02, 10.20a]) one shows that for the matrix-representation with respect to the original basis $\mathcal B=(x,y,z)$ we have

  $$[R_1]:=[R_1]_{\mathcal B,\mathcal B} =[R_1]_{\mathcal B',\mathcal B'}=[\operatorname{id}]_{\mathcal B',\mathcal B}$$

  $$[R_2]_{\mathcal B,\mathcal B} =[\operatorname{id}]_{\mathcal B',\mathcal B}\, [R_2]_{\mathcal B',\mathcal B'}\, [\operatorname{id}]_{\mathcal B,\mathcal B'}$$

  $$=[R_1] \, [R_2]_{\mathcal B',\mathcal B'}\, [R_1]^{-1}$$

  
  
  The composition $R_2\o R_1$ has thus the following form

  $$[R_2\o R_1] =[R_2]\, [R_1]=[R_1]\, [R_2]_{\mathcal B',\mathcal B'}$$

  $$= \begin{pmatrix}\cos(\psi ) & 0 & -\sin(\psi ) \\ 0 & 1 & 0 \\ \... ...& 0 \\ 0 &\cos(\th ) & -\sin(\th ) \\ 0 & \sin(\th ) & \cos(\th ) \end{pmatrix}$$

  
  

• Similarly
  $$[R_3]_{\mathcal B'',\mathcal B''} = \begin{pmatrix} \cos(\varphi ... ...arphi ) & 0 \\ \sin(\varphi ) & \cos(\varphi ) & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
  and hence

  $$[R_3 \o R_2\o R_1] = [R_3]\,[R_2\o R_1]$$

  $$= [R_2\o R_1]\, [R_3]_{\mathcal B'',\mathcal B''}\,[R_2\o R_1]^{-1}\,[R_2\o R_1]$$

  $$= [R_1]_{\mathcal B,\mathcal B}\, [R_2]_{\mathcal B',\mathcal B'}\, [R_3]_{\mathcal B'',\mathcal B''}$$

  $$= {\scriptsize\begin{pmatrix}\cos(\psi ) & 0 & -\sin(\psi ) \\ 0 ... ...arphi ) & 0 \\ \sin(\varphi ) & \cos(\varphi ) & 0 \\ 0 & 0 & 1 \end{pmatrix} }$$

  $$= {\scriptsize\begin{pmatrix}\cos(\varphi )\cos(\psi )+\sin(\th )... ...i )\sin(\th )-\sin(\varphi )\sin(\psi ) & \cos(\th )\cos(\psi ) \end{pmatrix} }$$

  
  

Figure:                 $R_1$$R_2$                

    

Figure:                 $R_3$$R_1$, $R_2$, $R_3$            

    

Another decomposition into 3 rotations is given via the following Euler-angles:
Let $R$ be a rotation and $f_j:= R(e_j)$ be the images of the standard-basis. We would like to express $R$ as composition of 3 rotations around some coordinate-axes. It suffices to describe the images of these rotations on the first 2 vectors $e_1$ and $e_2$, since $e_3=e_1\times e_2$ is the uniquely determined unit vector normal to $e_1$ and $e_2$ such that $(e_1,e_2,e_3)$ is left oriented.

In order to rotate $e_1$ to $f_1$ we have to keep an axis $k\in \{e_1\}^\perp \cap \{f_1\}^\perp = \langle\{ e_2,e_3\}\rangle \cap \langle\{f_2,f_3\}\rangle$ fixed. In order to rotate afterwards $e_2$ to $f_2$ without destroying the assignment $e_1\mapsto f_1$, we could first rotate $e_2$ to $k$ around $e_1$ and at the end rotate $k$ to $f_2$ around $f_1$.

$$\begin{matrix} e_1 &\mapsto& e_1 &\mapsto& f_1 &\mapsto& f_1 \\ e_2 &\mapsto& k &\mapsto& k &\mapsto& f_2 \\ \end{matrix}$$

Figure:                 Fixed axes $k$$R_1$                

    

Figure:                 $R_2$$R_3$                

    

Let $\varphi _1,\varphi _2,\varphi _3$ be the so-called Euler-angles of $R$, given by

$$\varphi _1 := \measuredangle e_2 k; \quad \varphi _2 := \measuredangle e_1 f_1; \quad \varphi _3 := \measuredangle k f_2.$$

Thus the matrix-representation of the corresponding rotations $R_1$, $R_2$ and $R_3$ are given by:

$$[R_1]_{e_1,e_2} = \begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\varphi _1 & -\sin\varphi _1 \\ 0 & \sin\varphi _1 & \cos\varphi _1 \\ \end{pmatrix}$$

$$[R_2]_{e_1,k} = \begin{pmatrix}\cos\varphi _2 & 0 & -\sin\varphi _2 \\ 0 & 1 & 0 \\ \sin\varphi _2 & 0 & \cos\varphi _2 \\ \end{pmatrix}$$

$$[R_1]_{f_1,k} = \begin{pmatrix}1 & 0 & 0 \\ 0 & \cos\varphi _3 & -\sin\varphi _3 \\ 0 & \sin\varphi _3 & \cos\varphi _3 \\ \end{pmatrix}$$

Hence

$$\begin{multline*}[R_3\o R_2\o R_1]_{e_1,e_2,e_3} = \\ = \begin{pmatrix} 1 & 0 ... ...hi _3 \\ 0 & \sin\varphi _3 & \cos\varphi _3 \\ \end{pmatrix}\end{multline*}$$