Reflection Matrix

Reflection Matrix for Horizontal Axis

Recall that reflecting a point $(x,y)$ across the $x$-axis results in the image $(x,-y)$. The $x$-coordinate remains unchanged, while the $y$-coordinate changes sign.

Start with a matrix whose entries are unknown:

For reflection across the $x$-axis, the matrix must satisfy:

By equating the coefficients, we get:

• $rx + sy = 1x + 0y \implies r=1, s=0$
• $tx + uy = 0x - 1y \implies t=0, u=-1$

The reflection matrix for the $x$-axis is:

Reflection Matrix for Vertical Axis

Reflecting a point $(x,y)$ across the $y$-axis results in $(-x,y)$.

This gives $r=-1, s=0, t=0, u=1$.

The matrix is:

Reflection Matrix for Main Diagonal Line

Reflecting a point $(x,y)$ across the line $y=x$ results in $(y,x)$.

This gives $r=0, s=1, t=1, u=0$.

The matrix is:

Reflection Matrix for Negative Diagonal Line

Reflecting a point $(x,y)$ across the line $y=-x$ results in $(-y,-x)$.

This gives $r=0, s=-1, t=-1, u=0$.

The matrix is:

Basic Reflection Matrices

Each matrix below acts on a column vector that holds the coordinates of a point. Read one row as the rule for one output coordinate, so the first row produces the new $x$ and the second row produces the new $y$. The matrices for the coordinate axes and for the origin follow.

• Reflection matrix for the $x$-axis:

  $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

• Reflection matrix for the $y$-axis:

  $$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$

• Reflection matrix for line $y=x$:

  $$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

• Reflection matrix for line $y=-x$:

  $$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$$

Finding the Image of a Point using Matrix

Find the image of $(3,-4)$ reflected across the $x$-axis.

Solution:

Using the reflection matrix for the $x$-axis:

The image is $(3,4)$.

Finding the Image of a Triangle using Matrix

Determine the image of triangle $ABC$ with vertices $A(3,1)$, $B(-2,3)$, and $C(2,-1)$ reflected across the $y$-axis!

Solution:

The reflection matrix for the $y$-axis is:

The matrix of triangle $ABC$'s vertices:

The columns represent $A$, $B$ and $C$, from left to right. Multiply each column by the reflection matrix to find both coordinates of its image.

The image is triangle $A'B'C'$ with vertices $A'(-3,1)$, $B'(2,3)$, and $C'(-2,-1)$.

Exercises

1. Find the image of point $(3,5)$ reflected across the $x$-axis using the following reflection matrix:

   $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

2. Determine the image of triangle $ABC$ with vertices $A(3,1)$, $B(-2,3)$, and $C(2,-1)$ reflected across the line $y=-x$!

Solutions

1. Multiply the position vector of $(3,5)$ by the reflection matrix for the $x$-axis:

   $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$$

   Image: $(3,-5)$.

2. The reflection matrix for $y=-x$ is:

   $$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$$

   Matrix of $ABC$ vertices:

   $$\begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}$$

   $$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 3 & -2 & 2 \\ 1 & 3 & -1 \end{pmatrix}$$

   Compute each entry by multiplying a row by a column:

   $$\begin{aligned} &m_{11}=(0)(3)+(-1)(1) \\ &m_{12}=(0)(-2)+(-1)(3) \\ &m_{13}=(0)(2)+(-1)(-1) \\ &m_{21}=(-1)(3)+(0)(1) \\ &m_{22}=(-1)(-2)+(0)(3) \\ &m_{23}=(-1)(2)+(0)(-1) \end{aligned}$$

   $$=\begin{pmatrix}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\end{pmatrix}$$

   $$= \begin{pmatrix} -1 & -3 & 1 \\ -3 & 2 & -2 \end{pmatrix}$$

   Image: $A'(-1,-3)$, $B'(-3,2)$, $C'(1,-2)$.