\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Since rotation tensor is, well, a tensor, it follows that it will also not move any other vectors that are dependent on $\boldsymbol{a}$, i.e., all vectors $\boldsymbol{x}$ of the form $\lambda \boldsymbol{a}$, where $\lambda$ is a real number.

\[\begin{align} \begin{aligned} \boldsymbol{R}\boldsymbol{x} &=\boldsymbol{R}(\lambda \boldsymbol{a})\\ &=\lambda \boldsymbol{R}( \boldsymbol{a})\\ &=\lambda \boldsymbol{a}\\ &= \boldsymbol{x} \end{aligned} \end{align}\]

In general, the rotation tensor will not affect the component of an arbitrary vector that in the directon of $\boldsymbol{a}$.

Previously we showed that given a vector $\boldsymbol{a}$ an arbitrary vector $\boldsymbol{u}$ can be written as $\boldsymbol{u}=\boldsymbol{u}_{\perp}+\boldsymbol{u}_{\parallel}$. From the linearity of $\boldsymbol{R}$ we have that

\[\begin{align} \boldsymbol{R}(\boldsymbol{u}_{\parallel}+\boldsymbol{u}_{\perp})&=\boldsymbol{R}\boldsymbol{u}_{\parallel}+\boldsymbol{R}\boldsymbol{u}_{\perp} \end{align}\]

Since \(\boldsymbol{u}_{\parallel}\) is parallel to $\boldsymbol{a}$ (recall that $\boldsymbol{u}_{\parallel}=(\boldsymbol{u}\cdot \boldsymbol{a})\boldsymbol{a}$), we get that \(\begin{align} \boldsymbol{R}(\boldsymbol{u}_{\parallel}+\boldsymbol{u}_{\perp})&=\boldsymbol{u}_{\parallel}+\boldsymbol{R}\boldsymbol{u}_{\perp} \end{align}\)