Problem 1.3
To find the equivalent angle/axis of a rotation \({\cal R}\) (so as to express \({\cal R}\) as \({\cal R}_{\alpha,\boldsymbol{\hat{u}}}\)), we first find the vectors which are invariant by the rotation, that is, satisfying \({\cal R}(\boldsymbol{V}) = \boldsymbol{V}\). If this set is of dimension 1, we can identify one of two possible unit vectors \(\pm \boldsymbol{\hat{u}}\): this gives the equivalent axis. To find the equivalent angle \(\alpha\), we use \(2\cos\alpha+ 1 = \text{tr}({\cal R})\). To find \(\alpha\) uniquely, corresponding to \(\boldsymbol{\hat{u}}\), we can use the identity \(\sin\alpha= (\boldsymbol{\hat{v}}, {\cal R}(\boldsymbol{\hat{v}}), \boldsymbol{\hat{u}})\) where \(\boldsymbol{\hat{v}}\) is a unit vector normal to \(\boldsymbol{\hat{u}}\) (finding a candidate for \(\boldsymbol{\hat{v}}\) is straightforward).
Case 1: the matrix is a rotation. The equation \({\cal R}(\boldsymbol{V}) = \boldsymbol{V}\) gives the 3 scalar equations \[ x_3 = x_1, \qquad x_1 = x_2, \qquad x_2 = x_3 \] This represents a 1-dimensional subspace of \(\cal E\) spanned by \(\boldsymbol{\hat{u}}= (\boldsymbol{\hat{e}}_1 + \boldsymbol{\hat{e}}_2 + \boldsymbol{\hat{e}}_3)/\sqrt{3}\). Choosing \(\boldsymbol{\hat{v}}= (\boldsymbol{\hat{e}}_1 - \boldsymbol{\hat{e}}_2)/\sqrt{2}\) we find \[ 2 \cos\alpha+1 = 0, \qquad \sin\alpha= (\boldsymbol{\hat{v}}, {\cal R}(\boldsymbol{\hat{v}}), \boldsymbol{\hat{u}}) = \frac{\sqrt{3}}{2} \] giving \(\boxed{\alpha = \frac{2\pi}{3}}\).
Case 2: the matrix is a rotation. Solving \({\cal R}(\boldsymbol{V}) = \boldsymbol{V}\) gives \(\boldsymbol{\hat{u}}= (3\boldsymbol{\hat{e}}_1 + 4 \boldsymbol{\hat{e}}_2)/5\). Choose \(\boldsymbol{\hat{v}}= \boldsymbol{\hat{e}}_3\) and find \(\cos\alpha= 0\) and \(\sin \alpha=-1\): \(\boxed{\alpha= 3\pi/2}\).
Case 3: Proceed as in the previous cases to find \({\cal R}= {\cal R}_{\alpha, \boldsymbol{\hat{u}}}\) with $ $ and \(\boldsymbol{\hat{u}}= (\boldsymbol{\hat{e}}_1 + \boldsymbol{\hat{e}}_2 + \boldsymbol{\hat{e}}_3)/ \sqrt{3}\).