Let $Cl(s,t)$ be the Clifford algebra over $\mathbb{R}^{s,t}$ where $(s,t)$ is the signature of the bilinear form $\eta$. Let $Pin(s,t)$ be the associated pin group and define $$R: Pin(s,t) \times \mathbb{R}^{s,t} \rightarrow \mathbb{R}^{s,t}\\ (u,x) \mapsto (-1)^{\text{deg}(u)} u \cdot x \cdot u^{-1}.$$
In Hamilton's Mathematical Gauge Theory, he says to fix a vector $v \in \mathbb{R}^{s,t}$ with $\eta(v, v) = \pm1$. Then $v^{-1} = \mp v$ and $$R_v(x) = -v \cdot x \cdot v^{-1} = \pm v \cdot x \cdot v \tag{1}$$ and furthermore $$R_v(x) = \begin{cases} -x \quad x \parallel v \\ x \quad x \perp v\end{cases} \tag{2}$$ so $R_v(x) \in \mathbb{R}^{s,t}$ and $R_v$ is a reflection in $v^\perp$. In summary, he says
We see that vectors $v \in \mathbb{R}^{s,t}$ with $\eta(v,v) = \pm 1$ act as reflections on $\mathbb{R}^{s,t}$ .
I have three questions about the above.
Why does it suffice to only define $R_v$ for $x \parallel v$ and $x \perp v$? What if $x$ is a linear combination of basis vectors, some of which are parallel and others which are orthogonal to $v$?
I am able to show (2) from (1) if $x \perp v$, but I am having trouble with when $x \parallel v$. For example, using the Clifford relation we have $$ \pm v\cdot x \cdot v = \pm v \cdot \big(-v\cdot x - \eta(x,v)\cdot 1\big) = x \mp \eta(x,v) \cdot v$$ but this is where I am stuck.
Most importantly, how does (2) imply that $v$ acts as a reflection on $v^\perp$? Wouldn't the signs in (2) have to be swapped for this conclusion to hold?
It is a basic fact of linear algebra that if your space $E$ decomposes as $E=U\oplus V$, to determine a linear application $f:E\to F$ it is enough to define its restrictions to $U$ and $V$.
Just write $x=\lambda v$ for some $\lambda\in \mathbb{R}$. Then $v\cdot x\cdot v = \lambda v\cdot v^2 = \eta(v,v)\lambda v = \eta(v,v)x$.
A reflection acts by the identity on a hyperplane and by minus the identity on a line.