Let $A$ be $2\times 2$ Orthogonal matrix such that det$A= -1$. Show that $A$ represents reflection about the line in $R^2$.

Question

Every matrix is corresponding to a linear transformation. Reflection is a linear transformation called $T$. So $T^2=I$. Also modulus of eigen values of $A$ is $1$.Since it has determinant $-1$ so odd no eigen values are $-1$. I am stucking here. I can't understand how to proceed further

Answer 1

The general form for such a matrix looks like this

$$M = \begin{bmatrix}\cos \theta &\sin \theta \\ \sin \theta &-\cos \theta\end{bmatrix}$$

This matrix is orthogonal, has determinant -1. Now what does it do to a point $(x,y)$?

$$M(x,y) = (x\cos\theta + y\sin\theta, x\sin\theta-y\cos\theta)$$

This point is actually the reflection of point$(x,y)$ about the line making angle $\theta/2$ with x axis. I got to this conclusion by plugging in values of $\theta$, but you can try proving it by proving them equivalent. This is just a hint on how to proceed

Answer 2

If the matrix is $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ you have the conditions $$ \begin{cases} a^2+b^2=1\\ ac+bd=0\\ c^2+d^2=1\\ ad-bc=-1 \end{cases} $$ Therefore, there exist angles $\alpha,\beta\in[0,\pi)$ such that $$ a=\cos2\alpha,\quad b=\sin2\alpha,\quad c=\cos2\beta,\quad=\sin2\beta $$ Hence $\cos2\alpha\cos2\beta+\sin2\alpha\sin2\beta=0$, that is, $\cos(2\alpha-2\beta)=0$. Moreover $\cos2\alpha\sin2\beta-\sin2\alpha\cos2\beta=-1$, that is, $\sin(2\alpha-2\beta)=1$.

The conditions imply that $2\alpha=2\beta+\pi/2+2k\pi$, so we can say that the matrix is $$ \begin{bmatrix} \cos2\alpha & \sin2\alpha \\ \sin2\alpha & -\cos2\alpha \end{bmatrix} $$ Now, suppose you want to find the equations for the axial symmetry with respect to a line through the origin, forming an angle $\alpha$ with the positive $x$-semiaxis, with $0\le\alpha<\pi$..

Complex numbers are very handy here. Take $u=\cos\alpha+i\sin\alpha$. In order to find the symmetric of the point $z$, you can

1. rotate by the angle $-\alpha$;
2. find the symmetric with respect to the $x$-axis;
3. rotate by the angle $\alpha$.

The series of operations on $z$ are, algebraically, $$ z\mapsto zu^{-1}=z\bar{u}\mapsto \overline{z\bar{u}}=\bar{z}u\mapsto\bar{z}u^2 $$ If now $z=x+yi$, we get that its symmetric point is $$ (x-yi)(\cos2\alpha+i\sin2\alpha) $$ In matrix form, $$ \begin{bmatrix} x\cos2\alpha+y\sin2\alpha \\ x\sin2\alpha-y\cos2\alpha \end{bmatrix}= \begin{bmatrix} \cos2\alpha & \sin2\alpha \\ \sin2\alpha & -\cos2\alpha \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$

Answer 3

Writing $A=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$ you find by ortogonality that $a^2+b^2=1$ and that either $(c,d)=(b,-a)$ or $(c,d)=(-b,a)$. Only the first have determinant -1. Thus, $A=\begin{pmatrix} a & b\\ b & -a\end{pmatrix}$. Now, as $A$ is symmetric you have $A^2=A^T A=I$ so $(A-I)(A+I)=0=(A+I)(A-I)$ which provides enough information: Im$(A+I)$ is a line $\ell_+$ which is invariant under $A$ and Im$(A-I)$ is a line $\ell_-$ reflected by $A$. Furthermore, $\ell_-$ and $\ell_+$ are orthogonal (from the same identity and symmetry). Thus $A$ yields an orthogonal reflection in the line $\ell_+$.