# How do I get a matrix from a reflection?

I have the following question:

$S : \Bbb{R}^2 \to \Bbb{R}^2$ is the reflection at the line $y = 2$.

How do I get the matrix of $S$ at homogeneous coordinates? And to be honest, I don't get this question at all. I have no idea of what I am supposed to do here or how. I do really want to learn this and hope that someone can explain it to me.

I appreciate your help.

• from reading others' answers below, looks like you have four unknowns. Find some equations, that is, from mappings of points. – Brady Trainor Mar 25 '13 at 9:03

## 2 Answers

The reflection S is an affine transformation of the plane $\mathbb{R}^2$. An affine transformation is simply a map sending a vector $\vec{a}$ to the vector $S(\vec{a}) = T(\vec{a}) + \vec{k}$ where $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is a linear transformation and $\vec{k}\in\mathbb{R}^2$ is some constant translation vector.

In homogeneous co-ordinates, we can represent an affine transformation as a linear transformation in the following way. If $A$ is the matrix of the linear transformation $T$, then we have the correspondence

$$A\vec{a} + \vec{k} \hspace{10pt}\leftrightarrow\hspace{10pt} \left[\begin{array}{c|c}A &\vec{k}\\\hline\begin{array}{cc}0&0\end{array}&1\end{array}\right]\left[\begin{array}{c}\vec{a}\\1\end{array}\right]$$ (where the vector on the right is obtained by simply carrying out the matrix multiplication).

Thus in the case of the reflection across the line $y = 2$, we see that $$S\left(\left[\begin{array}{c}x\\y\end{array}\right]\right) = \left[\begin{array}{c}x\\4-y\end{array}\right] = \left[\begin{array}{c}x\\-y\end{array}\right] + \left[\begin{array}{c}0\\4\end{array}\right]$$

So we want to find a matrix $A = \left[\begin{array}{cc}a_1& a_2\\a_3 & a_4\end{array}\right]$ such that $\left[\begin{array}{ccc}a_1 & a_2 & 0\\a_3 & a_4 & 4\\0 & 0 & 1\end{array}\right]\left[\begin{array}{c}x\\y\\1\end{array}\right] = \left[\begin{array}{c}x\\4-y\\1\end{array}\right]$.

i.e. we want, in particular, to have $A$ such that $\left[\begin{array}{cc}a_1 & a_2\\a_3 &a_4\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}x\\-y\end{array}\right]$. From here it is not hard to see that $A$ must be the matrix $\left[\begin{array}{cc}1 & 0\\0 & -1\end{array}\right]$ and thus the matrix of $S$ in homogeneous coordinates is: $$\left[\begin{array}{ccc}1 & 0 & 0\\0 & -1 & 4\\0 & 0 & 1\end{array}\right]$$

• You state that $S\left([x,y]\right)=[x,4-y]$. How do you come to this conclusion? – Daryl Mar 25 '13 at 6:34
• @john thank you for you help john. Where did you get the first matrix from and what is it? Sorry for the questions but it's not a long time ago that I started linear algebra. – n00b1990 Mar 28 '13 at 21:07
• No, the matrix cannot be 3x3 if the codomain and domain are only 2 dimensional. – Goldname Feb 10 '18 at 17:24

$\renewcommand{\vec}[1]{\mathbf{#1}}$ Define $(x,y)$ as the old point and $(x^*,y^*)$ as the reflected point.

Intuitively, rather than a "formula", note that the line $y=2$ is horizontal. This means that the $x$ coordinate won't change. Also, the distance a point is below the line $y=2$ (I.e. $2-y$ where a negative below is taken to mean above) will be reflected to the same distance above the line. These can be written mathematically as \begin{align}x^*&=x,\\y^*&=2+(2-y)=4-y.\end{align}

Edit

The matrix which reflects about the line $\vec{n}\cdot\vec{r}=0$, where $\vec{r}=[x,y]^T$, is $$R_\vec{n}=I-2\vec{n}\vec{n}^T.$$ This is able to represent a reflection about any line through the origin. In $2D$, with $\vec{n}=[n_1,n_2]^T$, this is $$R=\begin{bmatrix}1-2n_1^2&-2n_1n_2\\-2n_1n_2&1-2n_2^2\end{bmatrix}.$$

Now, to perform a reflection about a line which passes through the point $\vec{p}$ rather than the origin, the point $\vec{p}$ is first translated to the origin, the standard reflection is performed, then the point is translated back. This is where the homogeneous coordinates comes into this problem. The matrix which translates $\vec{p}$ to the origin is $$T_p=\begin{bmatrix}I&-\vec{p}\\\vec{0}&1\end{bmatrix}.$$ In our $2D$ case, this is $$T=\begin{bmatrix}1&0&-p_x\\0&1&-p_y\\0&0&1\end{bmatrix},$$ where $\vec{p}=[p_x,p_y]^T$. In homogeneous coordinates, the reflection can be written as $$R_H=\begin{bmatrix}1-2n_1^2&-2n_1n_2&0\\-2n_1n_2&1-2n_2^2&0\\0&0&1\end{bmatrix}.$$ The matrix which performs the transformation is then $$M=T^{-1}R_HT.$$

For your problem, $\vec{n}=[0,1]^T$, $\vec{p}=[0,2]^T$. The matrices can then be written as \begin{align} R_H&=\begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix},\\ T&=\begin{bmatrix}1&0&0\\0&1&-2\\0&0&1\end{bmatrix}, \end{align} from which we can calculate the transformation matrix $$M=\begin{bmatrix}1&0&0\\0&-1&4\\0&0&1\end{bmatrix}$$ which reflects about the line $y=2$.

You would then have $S(\vec{p})=A\vec{p}+\vec{b}$, where $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $\vec{b}=\begin{bmatrix}0\\4\end{bmatrix}$.

• thank you for your help, I appreciate it. But a few things I don't understand. In the beginning you say that the x coordinate won't change, why not? And why does 2-y represent the difference of the x coordinate and y=2? – n00b1990 Mar 28 '13 at 21:01
• @noob1990 Since you are reflecting about a horizontal line, the $x$ coordinate won't change, as the $x-$axis is parallel to the line of reflection. Now, $2-y$ is the difference between the $y$ coordinate and the line $y=2$. This comes from observing the graph(e.g. if $y=0.4$, then it is $1.6=2-0.4=2-y$ from the line $y=2$. The derivation I gave in the edit will work for any line. Your specific example can be greatly simplified. – Daryl Mar 28 '13 at 22:06