Axes and the angle of $3\times 3$ rotation matrix

Question

Given matrix: \begin{pmatrix} 1/3 & 2/3 & 2/3\\ -2/3 & -/3 & 2/3\\ 2/3 & -2/3 & 1/3 \end{pmatrix}

I know, how to find angle by trace, but I'm really confused about revealing axis.

Answer 1

$ \def\bbR#1{{\mathbb R}^{#1}} \def\a{\beta}\def\o{{\tt1}} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\p{\partial}\def\grad#1#2{\frac{\p #1}{\p #2}} \def\cay#1{\operatorname{cay}\LR{#1}} \def\skew#1{\operatorname{skew}\LR{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} $A vector $b\in\bbR 3$ can be used to generate a skew matrix $B\in\bbR{3\times 3}\;$ (and vice versa) $$\eqalign{ b &= \m{b_1\\b_2\\b_3}\qif B &= \m{ 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \\} \\ }$$ $B$ has only one real eigenvalue $(\lambda=0),\,$ whose corresponding eigenvector is $b$ $$B\cdot b = 0$$

For a scalar variable, the Cayley Transform is defined as $$\eqalign{ \cay{\a} &= \fracLR{\o-\a}{\o+\a} \\ }$$ When applied to an orthogonal matrix it produces a skew symmetric matrix (and vice versa).

Applying the above operations to your rotation matrix yields $$\eqalign{ B = \LR{I+R}^{-\o}\LR{I-R} = \m{ 0 & -\o & 0 \\ \o & 0 & -\o \\ 0 & \o & 0 \\} \qiq b=\m{ \o \\ 0 \\ \o} \\ }$$ It is easy to verify that $b$ is indeed the axis of rotation for $R$ $$\eqalign{ \LR{I \pm B}\cdot b &= b \\ R\cdot b &= \LR{I+B}^{-\o}\LR{I-B}\cdot b \\ &= \LR{I+B}^{-\o}\cdot b \\ &= b \\ }$$