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

Linear Transformation problems and solutions

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

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

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

  1. 11/01/2017

    […] The solution is given in the post ↴ Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis […]

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