Question:
Let $\phi \in \mathbb{R}$, and consider the matrix:
\begin{equation} A = \begin{pmatrix} \cos \phi & \sin \phi\\ \sin \phi & - \cos \phi \end{pmatrix} \end{equation}
Prove that:
- $L_A$ is a reflection
- Find the axis in $\mathbb{R}^2$ about which $L_A$ reflects
My attempt:
So I did solve this question, but only after having found what the line of reflection in $\mathbb{R}^2$ is online.
The line is given by:
\begin{equation*} L = \{t \begin{pmatrix} \cos \phi + 1\\ \sin \phi\\ \end{pmatrix} : t \in \mathbb{R}\} \end{equation*}
So I worked backwards and said that given the orthonormal basis :
\begin{equation*} \beta_N = \left\{\frac{1}{c_1} \begin{pmatrix} \sin \phi\\ - \cos \phi - 1\\ \end{pmatrix} ,\frac{1}{c_2} \begin{pmatrix} \cos \phi + 1\\ \sin \phi\\ \end{pmatrix} \right\} \end{equation*}
(for some constants $c_1, c_2$ that make these vectors unit length), we can show that
$[L_A]_{\beta_N} = \begin{pmatrix} -1 & 0\\ 0 & 1\\ \end{pmatrix}$
which by definition means that $L_A$ is a reflection. And then in $\mathbb{R}^2$, the line of reflection is the span of the second vector (i.e. $L$)
However, I was only able to solve this because I knew what the line of reflection was before hand. Without this, how would one solve this question?
Initially, I thought we could use that if $\mathbb{F} = \mathbb{R}$, then we just need to show that det$(A) = -1$. However, nowhere does it say that the field is necessarily real.
So how would we find a suitable orthonormal basis from scratch, or otherwise solve these questions?
Thank you!
Pick a point, any point, in the plane (except the origin), call it $P$. Work out where $A$ takes $P$, call the answer $Q$. Then the line of reflection (assuming $A$ really is a reflection) is just the perpendicular bisector of the line segment joining $P$ and $Q$.