The linear transformation matrix for a reflection across the line $y = mx$ is:
$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$
My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks
Best Answer
You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be:
The line $y = mx$ is parametrised by $t \cdot \begin{pmatrix}1\\m\end{pmatrix}$. The line orthogonal to it is parametrised by $r \cdot \begin{pmatrix}-m\\1\end{pmatrix}$. The line $y = mx$ shall be fixed, the line orthogonal to it shall be reflected, so you want a matrix $R$ with
$$R \begin{pmatrix}1 & -m\\ m & 1\end{pmatrix} = \begin{pmatrix}1 & m\\ m & -1\end{pmatrix},$$
and that means
$$\begin{align} R &= \begin{pmatrix}1 & m\\m&-1\end{pmatrix} \begin{pmatrix}1&-m\\m&1\end{pmatrix}^{-1}\\ & = \begin{pmatrix}1&m\\m&-1\end{pmatrix}\cdot \frac{1}{1+m^2}\begin{pmatrix}1&m\\-m&1\end{pmatrix}\\ &= \frac{1}{1+m^2} \begin{pmatrix}1 - m^2 & 2m\\2m &m^2-1\end{pmatrix}. \end{align}$$