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

Linear Transformation problems and solutions

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

 
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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*}

In summary, the matrix representation $A$ of the linear transformation $T$ across the line $y=mx$ with respect to the standard basis is
\[A=\frac{1}{1+m^2}\begin{bmatrix}
1-m^2 & 2m\\
2m& m^2-1
\end{bmatrix}.\]

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.


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