# Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis

## Problem 597

Let $F:\R^2\to \R^2$ be the function that maps each vector in $\R^2$ to its reflection with respect to $x$-axis.

Determine the formula for the function $F$ and prove that $F$ is a linear transformation.

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## Solution 1.

Let $\begin{bmatrix}

x \\

y

\end{bmatrix}$ be an arbitrary vector in $\R^2$.

Its reflection with respect to $x$-axis is the vector $\begin{bmatrix}

x \\

-y

\end{bmatrix}$.

Thus the formula for the function $F$ is given by

\[F\left(\, \begin{bmatrix}

x \\

y

\end{bmatrix} \,\right)=\begin{bmatrix}

x \\

-y

\end{bmatrix}.\]

Next, we prove that the function $F$ is a linear transformation from $\R^2$ to $\R^2$.

We need to verify the following two properties: for any $\mathbf{u}, \mathbf{v}\in \R^2$ and $c\in \R$, we have

- $F(\mathbf{u}+\mathbf{v})=F(\mathbf{u})+F(\mathbf{v})$
- $F(c\mathbf{u})=cF(\mathbf{u})$.

Let $\mathbf{u}=\begin{bmatrix}

x \\

y

\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}

x’ \\

y’

\end{bmatrix}$.

Then we have

\begin{align*}

F(\mathbf{u}+\mathbf{v})&=F\left(\, \begin{bmatrix}

x+x’ \\

y+y’

\end{bmatrix} \,\right)\\[6pt]
&=\begin{bmatrix}

x+x’ \\

-(y+y’)

\end{bmatrix} &&\text{by the formula for $F$}\\[6pt]
&=\begin{bmatrix}

x \\

-y

\end{bmatrix}+\begin{bmatrix}

x’ \\

-y’

\end{bmatrix}\\[6pt]
&=F\left(\, \begin{bmatrix}

x \\

y

\end{bmatrix} \,\right)+F\left(\, \begin{bmatrix}

x’ \\

y’

\end{bmatrix}

\,\right) &&\text{by the formula for $F$}\\[6pt]
&=F(\mathbf{u})+F(\mathbf{v}).

\end{align*}

This prove property (a).

Next, we have

\begin{align*}

F(c\mathbf{u})&=F\left(\, \begin{bmatrix}

cx \\

cy

\end{bmatrix} \,\right)\\[6pt]
&=\begin{bmatrix}

cx \\

-(cy)

\end{bmatrix} &&\text{by the formula for $F$}\\[6pt]
&=c\begin{bmatrix}

x \\

-y

\end{bmatrix}\\[6pt]
&=cF(\mathbf{u})&&\text{by the formula for $F$.}

\end{align*}

Hence property (b) is verified.

Therefore, the function $F$ is a linear transformation.

## Solution 2.

We give another proof that $F$ is a linear transformation.

From the formula that we obtained in Solution 1, we see that

\begin{align*}

F\left(\, \begin{bmatrix}

x \\

y

\end{bmatrix} \,\right)=\begin{bmatrix}

1 & 0\\

0& -1

\end{bmatrix}\begin{bmatrix}

x \\

y

\end{bmatrix}.

\end{align*}

It follows that the function $F$ is the matrix multiplication by $\begin{bmatrix}

1 & 0\\

0& -1

\end{bmatrix}$.

As the matrix multiplication function is always a linear transformation, we conclude that $F$ is a linear transformation.

Observe that the matrix $\begin{bmatrix}

1 & 0\\

0& -1

\end{bmatrix}$ is the matrix representation of the linear transformation $F$.

That is,

\[\begin{bmatrix}

1 & 0\\

0& -1

\end{bmatrix}=[T(\mathbf{e}_1), T(\mathbf{e}_2],\]
where $\mathbf{e}_1=\begin{bmatrix}

1 \\

0

\end{bmatrix}$ and $\mathbf{e}_2=\begin{bmatrix}

0 \\

1

\end{bmatrix}$ are standard basis vectors of $\R^2$.

## Related Question.

The following problem is a generalization of the above problem.

**Problem**.

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}.\]

Note that if $m=0$, then this problem is the same as the current problem.

The solution is given in the post ↴

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

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## 1 Response

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