Stabilizer of orthogonal group acting on plane

Question

Let $G = O(2) = \{ R \in \mathbb{R}^{ 2 \times 2} \ | \ R^TR=I \}$ act on $\mathbb{R}^2$ by the usual matrix multiplication $R \cdot v = Rv$ for each $v \in \mathbb{R}^2$ and $R \in G$. The stabilizer of $v$ in $G$ is denoted $G_v$ and $$ G_v = \{ R \in O(2) \ | \ Rv = v \} $$ it is not hard to prove $G_v \leq O(2)$. One simple point I can solve via explicit algebra, $$ G_{(0,0)} = \{ R \in O(2) \ | \ R(0,0) = (0,0) \} = O(2) $$ On the other hand, if $v \neq (0,0)$ then the explicit algebraic calculation of $G_v$ is currently escaping my grasp. When I visualize rotations and reflections in the plane then it seems that $G_v = \{ I, R_v \}$ where $R_v$ is the reflection about the line with direction vector $v$.

How can we show $G_v = \{ I, R_v \}$ without resorting to geometric intuition?

I am interested in the algebraic proof $G_v = \{ I, R_v \}$. If someone could provide the argument and/or show me my geometric intuition is off-base then I would be ever so grateful. (I plan to add a 50pt bounty when I am allowed by the site software...)

Answer 1

Here is an easy way to look at it. For any matrix $A$, we consider the subspace $fix(A):=\{u\in \mathbb{R}^2: Au=u\}$. Since $\mathbb{R}^2$ is a two dimensional vector space, then $dim_{\mathbb{R}}fix(A)\in \{0,1,2\}$. Now if $R$ is an isometry that fixes a nonzero $v$, then $v\in fix(R)$. Thus, $fix(R)$ is either 1 dimesnional or two dimensional. If $fix(R)$ is two dimesnional, then $R=I_2$ and we are done. Now lets consider the case when $dim fix(A)=1$:

From properties of the dot product and the fact that the plane is 2D we know that there are exactly two unit vectors that are perpendicular to $v$, these are $w_1,w_2$ (or at least you should know this from your natural geometric intuition about the plane). We have $w_1,w_2\not\in fix(R)$ because $fix(R)$ is 1 dimensional. Thus, $Rw_1\not=w_1$. Since $R$ is orthogonal, therefore $Rw_1$ must be perpendicular unit vector to $v$. Hence, $Rw_1$ can only be $w_2$. So to summarize what we have so far: Any non identity orthogonal matrix that fixes $v$ must satisfy: $Rv=v$ and $Rw_1=w_2$. Since the plane is two dimensional and $v,w_1$ are linearly independent (therefore form a basis for the plane), it follows that there exists a unique such matrix which we will call/define: reflection about v

Answer 2

A $2\times 2$ orthogonal matrix is of the form $$ R=\begin{bmatrix}\cos t&\sin t\\ c\sin t&-c\cos t\end{bmatrix}, $$where $c\in\{-1,1\} $.

We have, then, if $v=\begin{bmatrix}x&y\end{bmatrix}^T$ and $Rv=v$, that $$ \begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix} \cos t&\sin t\\ c\sin t&-c\cos t\end{bmatrix} \begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}x\cos t+y\sin t\\ x c \sin t -y c \cos t\end{bmatrix}. $$ That is, we have a linear system in the unkowns $\cos t $ and $\sin t $ with determinant \begin{align} \begin{vmatrix} x&y\\ -c y&c x\end{vmatrix}=c (x^2+y^2). \end{align} So, as long as $(x,y)\ne (0,0) $ the system has unique solution. We thus get two solutions, one for each choice of $c $.

When $c=-1$, $$\begin{bmatrix}\cos t\\ \sin t\end{bmatrix}=-\frac1{x^2+y^2} \begin{bmatrix} -x& -y\\ -y& x\end{bmatrix}\, \begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}1\\ 0\end{bmatrix}. $$

When $c=1$, $$\begin{bmatrix}\cos t\\ \sin t\end{bmatrix}=\frac1{x^2+y^2} \begin{bmatrix} x& -y\\ y& x\end{bmatrix}\, \begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}\frac {x^2-y^2}{x^2+y^2}\\ \frac {2xy}{x^2+y^2}\end{bmatrix}. $$

In the first case we obtain $R=I $, and in the second one $R=R_v $.