# The Matrix for the Linear Transformation of the Reflection Across a Line in the Plane

## Problem 498

Let $T:\R^2 \to \R^2$ be a linear transformation of the $2$-dimensional vector space $\R^2$ (the $x$-$y$-plane) to itself of the reflection across a line $y=mx$ for some $m\in \R$.

Then find the matrix representation of the linear transformation $T$ with respect to the standard basis $B=\{\mathbf{e}_1, \mathbf{e}_2\}$ of $\R^2$, where

\[\mathbf{e}_1=\begin{bmatrix}

1 \\

0

\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}

0 \\

1

\end{bmatrix}.\]

Add to solve later

Sponsored Links

Contents

## Solution.

Let $A$ be the matrix representation of $T$ with respect to the standard basis $B$.

Observe that each vector on the line $y=mx$ does not move under the linear transformation $T$.

Since the vector $\begin{bmatrix}

1 \\

m

\end{bmatrix}$ is on the line $y=mx$, it follows that

\[A\begin{bmatrix}

1 \\

m

\end{bmatrix}=\begin{bmatrix}

1 \\

m

\end{bmatrix}. \tag{*}\]

Note that if $m\neq 0$ then the line $y=\frac{-1}{m}x$ is perpendicular to the line $y=mx$ at the origin.

If $m=0$, then the line $x=0$ is perpendicular to the line $y=0$ at the origin.

In either case the vector $\begin{bmatrix}

-m \\

1

\end{bmatrix}$ is on the perpendicular line.

Thus, by the reflection across the line $y=mx$, this vector is mapped to $\begin{bmatrix}

m \\

-1

\end{bmatrix}$.

That is, we have

\[A\begin{bmatrix}

-m \\

1

\end{bmatrix}=\begin{bmatrix}

m \\

-1

\end{bmatrix}. \tag{**}\]

It follows from (*) and (**) that

\begin{align*}

A\begin{bmatrix}

1 & -m\\

m& 1

\end{bmatrix}&=\begin{bmatrix}

A\begin{bmatrix}

1 \\

m

\end{bmatrix}& A \begin{bmatrix}

-m \\

1

\end{bmatrix}

\end{bmatrix}

=\begin{bmatrix}

1 & m\\

m& -1

\end{bmatrix}.

\end{align*}

The determinant of the matrix $\begin{bmatrix}

1 & -m\\

m& 1

\end{bmatrix}$ is $1+m^2\neq 0$, hence it is invertible.

(Note that since column vectors are nonzero orthogonal vectors, we knew it is invertible.)

The inverse matrix is

\[\begin{bmatrix}

1 & -m\\

m& 1

\end{bmatrix}^{-1}=\frac{1}{1+m^2}\begin{bmatrix}

1 & m\\

-m& 1

\end{bmatrix}.\]

Therefore, we have

\begin{align*}

A&=\begin{bmatrix}

1 & m\\

m& -1

\end{bmatrix}

\begin{bmatrix}

1 & -m\\

m& 1

\end{bmatrix}^{-1}\\[6pt]
&=\begin{bmatrix}

1 & m\\

m& -1

\end{bmatrix}

\cdot \frac{1}{1+m^2}\begin{bmatrix}

1 & m\\

-m& 1

\end{bmatrix}\\[6pt]
&=\frac{1}{1+m^2}\begin{bmatrix}

1-m^2 & 2m\\

2m& m^2-1

\end{bmatrix}.

\end{align*}

\[A=\frac{1}{1+m^2}\begin{bmatrix}

1-m^2 & 2m\\

2m& m^2-1

\end{bmatrix}.\]

## Comment.

The matrix representation $A$ of a linear transformation $T:\R^2 \to \R^2$ with respect to the standard basis $B=\{\mathbf{e}_1, \mathbf{e}_2\}$ of $\R^2$ is given by

\[A=\begin{bmatrix}

T(\mathbf{e}_1) & T(\mathbf{e}_2)\end{bmatrix}

.\]
So we can technically find the matrix $A$ by finding $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$, which can be found by elementary plane geometry.

However, it is easier to determine the outputs of $T$ for vectors on the line $y=mx$ and vectors on the perpendicular line.

That why we didn’t compute the vectors $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ directly in the above solution.

Add to solve later

Sponsored Links