Rotations, Reflections and Translations

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In summary: Thinking)Yes, I think it would be better to show the graph of the transformations. (Clap)We can start by drawing a graph of the transformation $\tau_{-v}$ followed by the transformation $d_a$. (Writing)Then we can show the graph of $d_{a,v}$.Finally, we can show the graph of the transformation $\tau_v\circ d_a\circ \tau_{-v}$. (Nod)From the graphs, we can explain why the relationships hold. (Thinking)Also, we can identify the matrix $A$ and the vector $w$ from the graphs. (Nod)In summary, we discussed the linear mapping $\sigma_a$ which is a reflection on
  • #1
mathmari
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Hey! :eek:

Let $ \tau_v: \mathbb{R}^2 \rightarrow \mathbb{R}^2, \ \ x \mapsto x + v $ be the shift by the vector $ v \in \mathbb{R}^2 $.

Let $ \sigma_a: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $ be the reflection on the straight line through the origin, where $ a $ describes the angle between the straight line and the positive $ x $ axis.

One can show that the mapping $\sigma_a$ is linear.

  1. Determine $\sigma_a\begin{pmatrix} 1\\ 0\end{pmatrix}$ and $\sigma_a\begin{pmatrix} 0\\ 1\end{pmatrix}$ by making a sketch. Determine the matrix $s_a: = M (\sigma_a)$.
  2. Show that
    • $\tau_v\circ \tau_w=\tau_{v+w}$
    • $\delta_a\circ \tau_v=\tau_{\delta_a(v)}\circ \delta_a$
    • $\sigma_a\circ \sigma_a=\text{id}_{\mathbb{R}^2}$
  3. Let's consider rotation around an arbitrary point. Let $d_{a,v}:\mathbb{R}^2\rightarrow \mathbb{R}^2, \ \ x\mapsto Ax+w$ the rotation around the angle $a$ and the pont $v$. Show at a graph that $d_{a,v}=\tau_v\circ d_a\circ \tau_{-v}$. Determine then $A\in M_2(\mathbb{R})$ and $w\in \mathbb{R}^2$ in respect to $ a$ and $ v$ .
I have done the following:

  1. For $\sigma_a\begin{pmatrix} 1\\ 0\end{pmatrix}$ we do the following:

    We draw an unit circle that goes through the point $(1,0)$.

    $a$ is the angle between the x-axis and the line, say $g$. From that line $g$ we draw a line such that the angle between these two lines is equal to $a$. The intersection point of that line and the unit circle is the desired image point $(x,y)$.

    View attachment 9540

    We have a rotation around the origin by $a+a=2a$.

    We calculate the coordinates of the image point using the rotation matrix \begin{equation*}\begin{pmatrix}\cos (2a) & -\sin (2a) \\ \sin (2a)& \cos (2a)\end{pmatrix}\begin{pmatrix} 1\\ 0\end{pmatrix}=\begin{pmatrix} \cos (2a)\\ \sin (2a)\end{pmatrix}\end{equation*}
    For $\sigma_a\begin{pmatrix} 0\\ 1\end{pmatrix}$ do we consider the following graph:

    View attachment 9541
  2. For the first bullet:
    $$(\tau_v\circ \tau_w)(x)=\tau_v(\tau_w(x))=\tau_v(x+w)=(x+w)+v=x+(w+v)=\tau_{w+v}(x)$$

    For the second bullet:
    \begin{align*}&(\delta_a\circ \tau_v)(x)=\delta_a(\tau_v(x))=\delta_a(x+v)=d_a\cdot (x+v)=d_a\cdot x+d_a\cdot v \\ &(\tau_{\delta_a(v)}\circ \delta_a
    )(x)=\tau_{\delta_a(v)}( \delta_a(x))=\tau_{\delta_a(v)}( d_a\cdot x)=d_a\cdot x+\delta_a(v)=d_a\cdot x+d_a\cdot v\end{align*}

    For the third bullet:
    \begin{align*}(\sigma_a\circ \sigma_a)(x)&=\sigma_a(\sigma_a(x))=\sigma_a\left (\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}x\right )=\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}x \\ & =\begin{pmatrix}\cos^2 (2a)+\sin^2(2a) & \cos(2a)\sin (2a)-\sin(2a)\cos(2a) \\ \sin (2a)\cos(2a)-\cos(2a)\sin(2a) &\sin^2(2a)+\cos^2 (2a)\end{pmatrix}x=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}x=x\end{align*}
  3. We translate by v to the origin to make the rotation by the origin. Then we make the rotation at the origin by the angle a. Then we translate by v such that we get the rotation around the point v by the angle a.

    We have that \begin{equation*}d_{a,v}(x)=\tau_v(\delta_a(\tau_{-v}(x)))=\tau_v(\delta_a(x-v))=\tau_v(d_a\cdot (x-v))=d_a\cdot (x-v)+v=d_a x+ (I-d_a)v \end{equation*}
    So we get $A=d_a$ and $w=(I-d_a)v$.

    Is everything correct and complete?
 

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  • #2
mathmari said:
Let $ \sigma_a: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $ be the reflection on the straight line through the origin, where $ a $ describes the angle between the straight line and the positive $ x $ axis.
One can show that the mapping $\sigma_a$ is linear.

1. Determine $\sigma_a\begin{pmatrix} 1\\ 0\end{pmatrix}$ and $\sigma_a\begin{pmatrix} 0\\ 1\end{pmatrix}$ by making a sketch. Determine the matrix $s_a: = M (\sigma_a)$.

I have done the following:
For $\sigma_a\begin{pmatrix} 1\\ 0\end{pmatrix}$ we do the following:
We draw an unit circle that goes through the point $(1,0)$.
$a$ is the angle between the x-axis and the line, say $g$. From that line $g$ we draw a line such that the angle between these two lines is equal to $a$. The intersection point of that line and the unit circle is the desired image point $(x,y)$.
We have a rotation around the origin by $a+a=2a$.
We calculate the coordinates of the image point using the rotation matrix \begin{equation*}\begin{pmatrix}\cos (2a) & -\sin (2a) \\ \sin (2a)& \cos (2a)\end{pmatrix}\begin{pmatrix} 1\\ 0\end{pmatrix}=\begin{pmatrix} \cos (2a)\\ \sin (2a)\end{pmatrix}\end{equation*}

Hey mathmari!

I think we are not supposed to use the matrix. (Worried)
Aren't we supposed to find the matrix as the last part of the question?

Instead we can find the image geometrically.
That is, we can draw the unit circle and read off the x- and y-coordinates of the image, which are the cosine respectively the sine of the angle of $2a$ in the picture.
View attachment 9543
Alternatively we can draw a right triangle and find those coordinates with trigonometry.

From your picture we can see that the coordinates are $(\cos(2a),\sin(2a))$.
It corresponds to the point $(\cos t,\sin t)$ in the picture of the unit circle.

Since it is the image of $(1,0)$, it follows that it must be the first column in the matrix $M(\sigma_a)$. (Thinking)

mathmari said:
For $\sigma_a\begin{pmatrix} 0\\ 1\end{pmatrix}$ do we consider the following graph:

Yes.
We can find the image of (0,1) from it, which will be the second column of the matrix $M(\sigma_a)$. (Thinking)

mathmari said:
2. For the first bullet:
$$(\tau_v\circ \tau_w)(x)=\tau_v(\tau_w(x))=\tau_v(x+w)=(x+w)+v=x+(w+v)=\tau_{w+v}(x)$$

For the second bullet:
\begin{align*}&(\delta_a\circ \tau_v)(x)=\delta_a(\tau_v(x))=\delta_a(x+v)=d_a\cdot (x+v)=d_a\cdot x+d_a\cdot v \\ &(\tau_{\delta_a(v)}\circ \delta_a
)(x)=\tau_{\delta_a(v)}( \delta_a(x))=\tau_{\delta_a(v)}( d_a\cdot x)=d_a\cdot x+\delta_a(v)=d_a\cdot x+d_a\cdot v\end{align*}

For the third bullet:
\begin{align*}(\sigma_a\circ \sigma_a)(x)&=\sigma_a(\sigma_a(x))=\sigma_a\left (\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}x\right )=\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}\begin{pmatrix}\cos (2a) & \sin (2a) \\ \sin (2a) &-\cos (2a)\end{pmatrix}x \\ & =\begin{pmatrix}\cos^2 (2a)+\sin^2(2a) & \cos(2a)\sin (2a)-\sin(2a)\cos(2a) \\ \sin (2a)\cos(2a)-\cos(2a)\sin(2a) &\sin^2(2a)+\cos^2 (2a)\end{pmatrix}x=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}x=x\end{align*}

All correct. (Nod)

Alternatively, we could show the relationships geometrically. (Thinking)

mathmari said:
3. Let's consider rotation around an arbitrary point. Let $d_{a,v}:\mathbb{R}^2\rightarrow \mathbb{R}^2, \ \ x\mapsto Ax+w$ the rotation around the angle $a$ and the pont $v$. Show at a graph that $d_{a,v}=\tau_v\circ d_a\circ \tau_{-v}$. Determine then $A\in M_2(\mathbb{R})$ and $w\in \mathbb{R}^2$ in respect to $ a$ and $ v$ . We translate by v to the origin to make the rotation by the origin. Then we make the rotation at the origin by the angle a. Then we translate by v such that we get the rotation around the point v by the angle a.

We have that \begin{equation*}d_{a,v}(x)=\tau_v(\delta_a(\tau_{-v}(x)))=\tau_v(\delta_a(x-v))=\tau_v(d_a\cdot (x-v))=d_a\cdot (x-v)+v=d_a x+ (I-d_a)v \end{equation*}
So we get $A=d_a$ and $w=(I-d_a)v$.

Is everything correct and complete?

It is correct. (Nod)

But didn't the question ask for a graph?
And to show the relationship through the graph?
And subsequently find the matrix $A$ and the vector $w$. (Wondering)
 

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  • #3
Klaas van Aarsen said:
It is correct. (Nod)

But didn't the question ask for a graph?
And to show the relationship through the graph?
And subsequently find the matrix $A$ and the vector $w$. (Wondering)

Do we draw the following graph for that?

View attachment 9545

(Wondering)
 

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  • #4
That is the intended graph yes. (Nod)
 
  • #5
Klaas van Aarsen said:
That is the intended graph yes. (Nod)

Great! Thanks! (Yes)
 

FAQ: Rotations, Reflections and Translations

1. What are rotations, reflections, and translations?

Rotations, reflections, and translations are three types of transformations used in geometry to move and manipulate figures on a coordinate plane. Rotations involve rotating a figure around a fixed point, reflections involve flipping a figure across a line, and translations involve sliding a figure in a certain direction.

2. What is the difference between a rotation and a reflection?

The main difference between a rotation and a reflection is the type of movement that occurs. In a rotation, the figure is turned around a fixed point, while in a reflection, the figure is flipped across a line. Additionally, rotations preserve the shape and size of a figure, while reflections may change the orientation and orientation of a figure.

3. How do you perform a translation on a figure?

To perform a translation on a figure, you must know the direction and distance of the translation. You then move each point of the figure the specified distance in the specified direction. Another way to think of it is sliding the figure without changing its shape or orientation.

4. What is the difference between a rigid transformation and a non-rigid transformation?

A rigid transformation is a transformation that preserves the shape and size of a figure. This includes rotations, reflections, and translations. On the other hand, a non-rigid transformation is a transformation that changes the shape and/or size of a figure. This includes transformations such as dilations and shears.

5. How are rotations, reflections, and translations used in real life?

Rotations, reflections, and translations are used in many real-life applications, including art, engineering, and architecture. For example, architects may use reflections to create symmetrical designs, while engineers may use translations to move and position objects in a precise manner. These transformations are also used in computer graphics to create animations and special effects.

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