Question about change of basis and reflection about the line $y=2x$

Question

In Friedberg et al. Linear Algebra there is an example that asks to find the matrix of the linear operator in the standard basis which reflects vectors of $\mathbb{R}^2$ across the line $y=2x$.

Identifying that we can reflect across any line $y=mx$ by $(x,m)\mapsto(-m,x)$ we may choose the basis $\mathcal{B'}=\{\binom{1}{2},\binom{-2}{1}\}.$

The text next claims that the operator $T$ in this basis is given by $$[T]_{\mathcal{B'}}=\begin{pmatrix}1&0\\0 &-1\end{pmatrix}.$$

However, I'm not understanding why the matrix in the basis $\mathcal{B}'$ is the matrix above. Could anyone shed some light on my point of confusion?

Answer 1

$u_1=\begin{pmatrix}{1}\\{2}\end{pmatrix}$ is a vector in the direction of the line and $u_2=\begin{pmatrix}{-2}\\{1}\end{pmatrix}$ is perpendicular, so

$$\begin{cases} T(u_1)=u_1\\T(u_2)=-u_2\end{cases}\underbrace{\Rightarrow}_{\text{transposing}\\\text{coefficients}} [T]_{\mathcal{B'}}=\begin{pmatrix}1&0\\0 &-1\end{pmatrix}.$$

Answer 2

The line $y=2x$ reflects the point $(1,2)$ to itself i.e.
$T(1,2)=(1,2)={\color{green}1}(1,2)+{\color{green}0}(-2,1)$
The line $y=2x$ reflects the point $(-2,1)$ to the point $(2,-1)$ (Use coordinate geometry to see this!) i.e.
$T(-2,1)=(2,-1)={\color{red}0}(1,2)+{\color{red}-}{\color{red}1}(-2,1)$
Arrange the coefficients in color columnwise to get the required transformation.
$T|_{B'}=\begin{pmatrix}{\color{green}1}&{\color{red}0}\\{\color{green}0}&{\color{red}-}{\color{red}1}\end{pmatrix}$