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.

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

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.

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

The post Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis first appeared on Problems in Mathematics.

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 which is 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
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}.\]

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*}

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.

Related Question.

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

The post The Matrix for the Linear Transformation of the Reflection Across a Line in the Plane first appeared on Problems in Mathematics.