What is the cardinality of center of $\Bbb O_2(\Bbb R)$?

Question

What is the cardinality of center of $\Bbb O_2(\Bbb R)$? where $\Bbb O_2(\Bbb R)$ denotes the set of all $2\times 2$ orthogonal matrices over $\Bbb R$.

My try:

Let us try some orthogonal matrices i.e.

$\begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$ ,$\begin{bmatrix} 0& 1\\ -1 & 0\end{bmatrix}$ ,$\begin{bmatrix} 1& -1\\ 1 & 1\end{bmatrix}$ .

Now if $\begin{bmatrix} a_{11}& a_{12}\\ a_{21} & a_{22}\end{bmatrix}$ $\in $ centre of $\Bbb O_2(\Bbb R)$ then

$\begin{bmatrix} a_{11}& a_{12}\\ a_{21} & a_{22}\end{bmatrix}$ $\begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$ =$\begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$ $\begin{bmatrix} a_{11}& a_{12}\\ a_{21} & a_{22}\end{bmatrix}$

$\implies a_{12}=a_{21},a_{11}=a_{22}$

Again,$\begin{bmatrix} a_{11}& a_{12}\\ a_{12} & a_{11}\end{bmatrix}$ $\begin{bmatrix} 1& 1\\ -1 & 1\end{bmatrix}$ =$\begin{bmatrix} 1& 1\\ -1 & 1\end{bmatrix}$ $\begin{bmatrix} a_{11}& a_{12}\\ a_{12} & a_{11}\end{bmatrix}$

$\implies a_{12}=0$

$\begin{bmatrix} a_{11}& a_{12}\\ a_{21} & a_{22}\end{bmatrix}$=$\begin{bmatrix} a_{11}& 0\\ 0 & a_{11}\end{bmatrix}$

So the center has cardinality = cardinality of $\Bbb R$

**But the answer is given to be $2$ .Where am I wrong?Please help

Answer 1

You need your matrix $$\begin{bmatrix}a&0\\0&a\end{bmatrix}$$ to be orthogonal, so you need $|a|=1$. As you are in $\mathbb R$, you have $a=1$ or $a=-1$.

Answer 2

You cannot use $\pmatrix{1&1\cr-1&1}$ in your proof since it is not an element of $O(2,R)$. Here is a proof:

$I$ and $-I$ are in the center of $O(2,R)$. Suppose that $A$ is another element of the center. If $A$ is a rotation $A=\pmatrix{\cos a & \sin a\cr -\sin a & \cos a}$ $A$ commutes with the symmetry $B=\pmatrix{0 & 1\cr 1 & 0}$ so $AB=BA$ implies that $\pmatrix{\cos a & \sin a\cr -\sin a & \cos a}\pmatrix{0 & 1\cr 1 & 0}$=$\pmatrix{0 & 1\cr 1 & 0}\pmatrix{\cos a & \sin a\cr -\sin a & \cos a }$. This implies that $\sin a=-\sin a$, $\sin a=0$ and $A=-I$ or $-I$.

If $A$ is a symmetry and not a rotation, $A$ fixes a vector $u$, $A(v)=-v$, the angle between $u$ and $v$ is $\pi/2$, let $B$ be the rotation of angle $\pi/4$, $BA(u)=B(u)\neq A(B(u))$ since the angle between $u$ and $B(u)$ is $\pi/4$ and the angle between $u$ and $A(B(u))$ is $-\pi/4$. Contradiction.