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Mathematics-Online course: Linear Algebra - Analytic Geometry - Orthogonal Groups

Rotation Axis and Angle

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A rotation $ Q$ in $ \mathbb{R}^3$ has an axis of rotation, i.e. $ Q$ fixes an unit vector $ u$, and corresponds to a plane rotation by an angle $ \vartheta$ in the plane orthogonal to $ u$.

With respect to an orthonormal right-handed coordinate system $ u,v,w$, the matrix representation of $ Q$ is given by:

$\displaystyle \tilde Q = \left(\begin{array}{ccc} 1&0&0 \\ 0&\cos\vartheta & -\sin\vartheta \\ 0&\sin\vartheta & \cos\vartheta \\ \end{array}\right) \,. $

For the angle of rotation

$\displaystyle \cos\vartheta = \frac{1}{2}\left(\operatorname{Spur} Q - 1\right) $

holds.
(Authors: Höllig/Reble/Höfert)
For a rotation matrix $ Q$

$\displaystyle Q^{-1} = Q^{\operatorname t}\,, \quad \vert\operatorname{det}\,Q\vert=1 $

holds. The eigenvalues $ \lambda_i$ of an orthogonal matrix have norm 1 and it is $ \lambda_1\lambda_2\lambda_3= \operatorname{det}\,Q=1$. Therefor at least one eigenvalue is 1, namely adapted numeration gives

$\displaystyle \lambda_1=\overline{\lambda_2}\,, \quad \lambda_1\lambda_2 = 1 $

or

$\displaystyle \lambda_i \in \{-1,1\}\,. $

The normed eigenvector $ u$ corresponding to the eigenvalue 1 determines the axis of rotation.

For a right-handed orthonormal coordinate system $ u,v,w$

$\displaystyle Qu$ $\displaystyle =$ $\displaystyle u$  
$\displaystyle Qv$ $\displaystyle =$ $\displaystyle \alpha v + \beta w \qquad (1)$  
$\displaystyle Qw$ $\displaystyle =$ $\displaystyle \gamma v + \delta w$  

holds. The $ u$-independence of $ Qv, Qw$ results from the isogonality of orthogonal matrices

$\displaystyle x\;\bot\; u \Rightarrow Qx\; \bot \; Qu = u \,. $

The matrix representation of (1)

$\displaystyle Q(\underbrace{u,v,w}_{P})=(u,v,w) \underbrace{\left(\begin{array}... ...0&0\\ 0&\alpha&\gamma\\ 0&\beta&\delta\\ \end{array}\right)}_{\tilde Q} \,, $

together with $ \tilde Q = P^{-1}QP$ implies that

\begin{displaymath} \left( \begin{array}{cc} \alpha&\gamma\\ \beta&\delta\\ \end{array}\right) \end{displaymath}

is a rotation matrix. One also gets

$\displaystyle \operatorname{trace} Q = \operatorname{trace} \tilde Q = 1 + 2 \cos\vartheta \, , $

because of the invariance of the trace.
(Authors: Höllig/Reble/Höfert)
With the rotation-matrix

$\displaystyle Q = \frac{1}{2} \left( \begin{array}{ccc} 1 & -\sqrt{2} & 1 \\ \s... ...& 0 & -\sqrt{2} \\ 1 & \sqrt{2} & 1 \end{array} \right) \in \mathbb{R}^{3,3} $

you get

$\displaystyle Q^{\operatorname t}Q = \frac{1}{4} \left( \begin{array}{ccc} 4 & ... ...\\ 0 & 0 & 4 \end{array} \right) = E \Rightarrow Q^{\operatorname t}= Q^{-1} $

and

$\displaystyle \operatorname{det} Q = \operatorname{det} \frac{1}{2} \left( \beg... ...\ \sqrt{2} & 0 & -\sqrt{2} \\ 1 & \sqrt{2} & 1 \end{array} \right) = + 1\,. $

Compute the rotationaxis, the eigenvector $ v$ to the eigenvalue $ \lambda = 1$:

$\displaystyle 2(Q-1E) = \left( \begin{array}{ccc} -1 & -\sqrt{2} & 1 \\ \sqrt{2} & -2 & -\sqrt{2} \\ 1 & \sqrt{2} & -1 \end{array} \right) $

With 2nd row $ - \sqrt{2} \cdot$ 1st row and 3rd row $ +$ 1st row you get

$\displaystyle \left( \begin{array}{ccc} -1 & -\sqrt{2} & 1 \\ 0 & -4 & 0 \\ 0 & 0 & 0 \end{array} \right) $

and so the eigenvector $ v = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right)$.

To compute $ \varphi$ you compute the angle between an unit vector $ n$ with $ n \bot v$ and the unit vector $ x = Qn$.

$\displaystyle n = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right)\,, x... ...{1}{2} \left( \begin{array}{c} -\sqrt{2} \\ 0 \\ \sqrt{2} \end{array} \right) $

$\displaystyle cos \varphi = n \cdot x = 0 \Rightarrow \varphi = \pm \frac{\pi}{2}\,. $

  automatically generated 4/21/2005