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Show that $O_2(\mathbb{R})$ contains only rotational and reflective matrices.

I know that rotational and reflective symmetries are part of $O_2(\mathbb{R})$. I want to show that there is no other matrix that satisfies $O_2(\mathbb{R})$. This is what I have and I don't know what to do after this or if I'm going about this in the right way at all.

Suppose there exists a matrix $A \in O_2(\mathbb{R})$ such that is not a rotational or reflexive symmetry.

$A^T = A^{-1}$ by definition

Therefore we get the following simultaneous equations:

$$a = \frac{d}{ad - bc}; d = \frac{a}{ad - bc}; b = \frac{-c}{ad - bc}; c = \frac{-b}{ad - bc}$$

using the first two equations, you get $ad - bc =1$ or $a = -d$ and from the second two $ad - bc = -1$ or $b = -c$. I don't know where to go from here to show that $a,b,c,d$ form either a rotational or reflective matrix.

Rodrigo de Azevedo
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  • Undoubtedly you know that $A$ preserves angles and lengths. If $\det A=-1$, then the usual argument: $$\det(I+A)=\det(AA^T+A)=\det(A)\det(A^T+I)=-\det(A^T+I)=-\det(I+A),$$ implying that $\det(I+A)=0$, tells us that $-1$ is an eigenvalue. The determinant is the product of eigenvalues, so $+1$ is also an eigenvalue. Do you see why this implies that $A$ is the orthogonal reflection w.r.t. the line forming the eigenspace of $\lambda=+1$? – Jyrki Lahtonen Mar 30 '21 at 08:04
  • If $\det A=1$ there are many ways to see that it is a rotation depending on what else do you know. Using preserved handedness is one way. Another could be to multiply $A$ with $$\pmatrix{1&0\cr0&-1\cr}$$ reducing it to the previous case (the product of two reflections is a rotation, see the animation). Other ways can undoubtedly be cooked up. – Jyrki Lahtonen Mar 30 '21 at 08:06

1 Answers1

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By definition $A \in O_2(\mathbb{R})$ iff $AA^T = A^TA = I$.

Let \begin{align} A = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \end{align} then we have that \begin{align} \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} a & c\\ b & d \end{pmatrix} = \begin{pmatrix} a^2+b^2 & ac+bd\\ ac+bd & c^2+d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}. \end{align} Hence, $(a, b)$ and $(c, d)$ are unit vectors. Moreover, $(a, b)$ is orthogonal $(c, d)$ since $ac+bd=0$. Then we can rewrite $A$ as follows \begin{align} A = \begin{pmatrix} a & -b\\ b & a \end{pmatrix} \ \ \text{ or } \ \ A = \begin{pmatrix} a & b\\ b & -a \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} b & -a\\ a & b \end{pmatrix} \end{align}

Since $a^2+b^2=1$, we can reparametrize the matrix by noting that $\cos^2\theta+\sin^2\theta = 1$. Hence, we have that either \begin{align} A = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \ \ \text{ or } \ \ A = \begin{pmatrix} \cos\theta & \sin\theta\\ \sin\theta & -\cos\theta \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{pmatrix}. \end{align}

Jacky Chong
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