Sketch of the Reflection Transformation of a Parallelogram

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SUMMARY

The discussion focuses on the linear transformation that reflects points through the $x_2$ axis, represented by the matrix $T = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$. It illustrates the transformation with specific vectors, $\textbf{u} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and $\textbf{v} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$, demonstrating that $T\textbf{u} = \begin{bmatrix} -3 \\ 4 \end{bmatrix}$ and $T\textbf{v} = \begin{bmatrix} -4 \\ 3 \end{bmatrix}$. The discussion concludes with the confirmation of the properties of linear transformations, including the transformation of the zero vector.

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$\textbf{Problem:}$
Let $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation that reflects each point through the $x_2$ axis. Make two sketches that illustrate properties of linear transformation.

$\textbf{Solution:}$
Let $T(\textbf{x}) = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -x_1 \\ x_2 \end{bmatrix}$

Let
$\textbf{u} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}, \textbf{v} = \begin{bmatrix} 4 \\ 3 \end{bmatrix}$
And $\textbf{u + v} = \begin{bmatrix} 7 \\ 7 \end{bmatrix}$

Then
$T\textbf{u} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} -3 \\ 4 \end{bmatrix}$
$T\textbf{v} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 4 \\ 3 \end{bmatrix} = \begin{bmatrix} -4 \\ 3 \end{bmatrix}$
$T\textbf{u + v} =\begin{bmatrix} -7 \\ 7 \end{bmatrix}$
$T\textbf{(0)} = \textbf{0}$

View attachment 3893
 

Attachments

  • Axis Reflection.png
    Axis Reflection.png
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  • Reflection About X_2 Axis.png
    Reflection About X_2 Axis.png
    6.2 KB · Views: 116
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Seems good to me. Which software did you use to make the drawing?
 
Evgeny.Makarov said:
Seems good to me. Which software did you use to make the drawing?

Geogebra
 

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