Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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
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
$$\begin{align} R &= \begin{pmatrix}1 & m\\m&-1\end{pmatrix} \begin{pmatrix}1&-m\\m&1\end{pmatrix}^{-1}\\ & = \begin{pmatrix}1&m\\m&-1\end{pmatrix}\cdot \frac{1}{1+m^2}\begin{pmatrix}1&m\\-m&1\end{pmatrix}\\ &= \frac{1}{1+m^2} \begin{pmatrix}1 - m^2 & 2m\\2m &m^2-1\end{pmatrix}. \end{align}$$
Another way. To reflect along a line that forms an angle $\theta$ with the horizontal axis is equivalent to:
Further, $y=mx$ implies $\tan \theta = m$, and $1+m^2 = \frac{1}{\cos^2\theta}$ .
Then, assumming you know about rotation matrices, you can write
$$\begin{align}T&=\begin{pmatrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix} \begin{pmatrix}1&0\\ 0 & -1\end{pmatrix} \begin{pmatrix}\cos \theta & \sin \theta\\ -\sin \theta & \cos \theta\end{pmatrix} \\ &= \begin{pmatrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix} \begin{pmatrix}\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta\end{pmatrix} \\ &= \cos^2 \theta \begin{pmatrix}1 & -\tan \theta\\ \tan \theta & 1\end{pmatrix} \begin{pmatrix}1 & \tan \theta\\ \tan\theta & -1\end{pmatrix} \\ &= \frac{1}{1 + m^2} \begin{pmatrix}1 & -m\\ m & 1\end{pmatrix} \begin{pmatrix}1 & m\\ m & -1\end{pmatrix} \\ &=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix}\end{align}$$
Vectors on the line obey the equation
$$y - mx = 0$$
Let $e_x, e_y$ be Cartesian basis vectors associated with the $x, y$ coordinates, respectively. The above equation implies that any vector $r = x e_x + y e_y$ that lies on the line must satisfy
$$r \cdot n = 0, \quad n = -m e_x + e_y$$
The vector $n$ is the normal vector to the line, perpendicular to the line. The associated unit normal is $\hat n = n/\sqrt{1+m^2}$.
Any vector $a$ can be broken down into a component that is parallel to the line and a component that is perpendicular. This is written $a = a_\parallel + a_\perp$. When the vector is reflected by a reflection map $\underline N$, the perpendicular component changes sign; the parallel component does not. That is,
$$\underline N(a) = a_\parallel - a_\perp = a - 2 a_\perp$$
The perpendicular component $a_\perp$ is given by
$$a_\perp = (a \cdot \hat n) \hat n$$
where $a = a^x e_x + a^y e_y$. You should be able to recognize that this is merely a projection map onto the vector $\hat n$.
Thus, the reflection map is given as
$$\underline N(a) = \underline I(a) - 2(a \cdot \hat n) \hat n$$
where $\underline I$ is the identity map.
From here, one need only evaluate this in terms of basis vectors to find the matrix components.
$$\underline N(e_x) = e_x - 2 (e_x \cdot \hat n) \hat n = e_x - \frac{2(-m)(-m e_x + e_y)}{1 + m^2} = \frac{(1-m^2)e_x + 2m e_y}{1+m^2}$$
and
$$\underline N(e_y) = e_y - 2 (e_y \cdot \hat n) \hat n = e_y - \frac{2(1)(-me_x + e_y)}{1+m^2} = \frac{2m e_x + (m^2 - 1)e_y}{1+m^2}$$
Both of these are columns of the associated matrix representation.
Here is a slightly different take. One can check with a picture that $R=2P-I$, where $P$ is the projection onto the line. Taking $v=(1,m)^T$ a vector along the line, then $$ P\begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 1&m\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}\,\begin{bmatrix} 1\\ m\end{bmatrix} / \begin{bmatrix} 1&m\end{bmatrix} \begin{bmatrix} 1\\ m\end{bmatrix} =\frac1{1+m^2} \begin{bmatrix} x+my\\ mx+m^2y\end{bmatrix}. $$ So $$P=\frac1{1+m^2} \begin{bmatrix} 1&m\\ m&m^2\end{bmatrix}$$ and $$ R=2P-I=\frac1{1+m^2} \begin{bmatrix} 1-m^2&2m\\ 2m&m^2-1\end{bmatrix}. $$
If you made a sketch you will se that $R(x)=2 \Pi_v(x)-x$ where $v=(1,m)$ and $\Pi_v$ is the projection of the vector $x$ over the vector $v$.
It is derived from physics of reflection.
The reflected ray rotates by an amount equal to $2 \theta,$ if the mirror itself rotates by $\theta,$ when we are given
$$ \tan \theta = m$$
Rotation matrix for double angle
$$ \begin{pmatrix}\cos 2 \theta & \sin 2 \theta\\\sin 2 \theta &\cos 2 \theta\end{pmatrix}$$
Using Weierstrass half angle relations
$$ = \frac{1}{1+m^2}\begin{pmatrix}1 - m^2 & 2m\\2m &1-m^2\end{pmatrix},$$
for consistency of rotation direction. Cannot explain the sign.
