MHB The product γ is a rotation or a translation

[mathmari]

Hey! :giggle:

For $p\in \mathbb{R}^2$ let $\delta_{p,\alpha}=\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}$.

Let $p,q\in\mathbb{R}^2$ and $\alpha,\beta\in \mathbb{R}$.

(a) Show that $\gamma=\delta_{p,\alpha}\circ\delta_{q,\beta}$ is a rotation of a translation (or both). Give the center of the rotation or the translation vector of $\gamma$ in respect to $p,q,\alpha, \beta$.

(b) Show analytically that the product $\gamma$ of two line reflections is a rotation or a translation. Give the geometric interpretation of the rotation angle/translation vector of $\beta$.
For (a) I have done the following :
\begin{align*}\left(\delta_{p,\alpha}\circ\delta_{q,\beta}\right )(x)&=\left (\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}\circ\tau_q\circ \delta_{\beta}\circ\tau_q^{-1}\right )(x)\\& =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (\delta_{\beta}\left (\tau_q^{-1}(x)\right )\right )\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (\delta_{\beta}\left (x-q\right )\right )\right )\right)\right ) \\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (d_{\beta}\left (x-q\right )\right )\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (d_{\beta}\left (x-q\right )+q\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)\right )\\ & =\tau_p\left ( d_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)\right )\\ & = d_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)+p\\ & = d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p\end{align*}
Is that correct so far? Do we have to substitute the rotation matrices $d_{\alpha}$ and $d_{\beta}$ ? Or is there an other way to continue?

:unsure:

 

[I like Serena]

Hey mathmari!

It looks correct so far. (Nod)

We want to prove that it is a rotation around some point or a translation.

So let's assume that it is a rotation with some angle $\phi$ around point $r$.
Then we should be able to equate what you have so far to this unknown rotation.
Can we find $\phi$ and $r$ in that case?

 

[mathmari]

So let's assume that it is a rotation with some angle $\phi$ around point $r$.
Then we should be able to equate what you have so far to this unknown rotation.
Can we find $\phi$ and $r$ in that case?

From the equation so far we have that we subtruct from $x$ the vector $q$, then we rotate by $\alpha$ and then again by $\beta$ then we add the vector $d_{\alpha}(q-p)+p$.
So... the angle is $\alpha+\beta$, right? But the point? Is the rotation around $q$ since we subtract it at the beginning? But shouldn't we add the same vector at the end as we subtract it at the beginning? :unsure:

 

[I like Serena]

From the equation so far we have that we subtract from $x$ the vector $q$, then we rotate by $\alpha$ and then again by $\beta$ then we add the vector $d_{\alpha}(q-p)+p$.
So... the angle is $\alpha+\beta$, right? But the point? Is the rotation around $q$ since we subtract it at the beginning? But shouldn't we add the same vector at the end as we subtract it at the beginning?

Yep.
So we should set up an equation.

 

[mathmari]

Yep.
So we should set up an equation.

So does it have to hold that $d_{\alpha}(q-p)+p=q$ ? :unsure:

 

[I like Serena]

So does it have to hold that $d_{\alpha}(q-p)+p=q$ ?

Nope. (Shake)

We should set up the full equation.
The argument to $d_{\alpha+\beta}$is not supposed to be $x-q$. Instead it should be $x-r$ for the as yet unknown center of rotation $r$.

 

[mathmari]

Nope. (Shake)

We should set up the full equation.
The argument to $d_{\alpha+\beta}$is not supposed to be $x-q$. Instead it should be $x-r$ for the as yet unknown center of rotation $r$.

Ah you mean to write $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ? :unsure:

 

[I like Serena]

Ah you mean to write $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

I thought to do
\begin{align*}d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p&=d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\\ & =d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}d_{\beta}d_{\beta}^{-1}q-d_{\alpha}d_{\beta}d_{\beta}^{-1}p+p\\& =d_{\alpha}d_{\beta}(x+d_{\beta}^{-1}q -d_{\beta}^{-1}p)+
d_{\alpha}d_{\beta}q+p\\& =d_{\alpha}d_{\beta}(x-d_{\beta}^{-1}( p-q))+
d_{\alpha}d_{\beta}q+p\end{align*} but that is still not in the desired form. :unsure:

 

[I like Serena]

Suppose we solve the equation for $r$?

 

[mathmari]

A rotation is in the form $d_{\phi}(x-r)+r=d_{\phi}x+(u_n-d_{\phi})r$ so in this case it must hold \begin{align*}&d_{\phi}=d_{\alpha}d_{\beta} \\ &(I_n-d_{\phi})r=-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p \Rightarrow (I_n-d_{\alpha}d_{\beta})r=-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p \\ & \Rightarrow r=(I_n-d_{\alpha}d_{\beta})^{-1}\left (-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\right )\end{align*}
Itholdsthat $d_{\alpha}d_{\beta}=d_{\alpha+\beta}$.

Then we have a rotation around the point $(I_n-d_{\alpha}d_{\beta})^{-1}\left (-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\right )$ with angle $\alpha+\beta$. We have a transltion if $d_{\alpha}d_{\beta}=I_n$, then \begin{equation*}d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p=x-q+d_{\alpha}q-d_{\alpha}p+p=x+\left ((d_{\alpha}-I_n)q-(d_{\alpha}-I_n)p\right )=x+(d_{\alpha}-I_n)(q-p)\end{equation*}
Then we have a translation about $(d_{\alpha}-I_n)(q-p)$. Is that correct? Do we have to consider the case that we have both translation and rotation? Or do allconditions must hold simultaneously in that case? :unsure:

 

[I like Serena]

Is that correct? Do we have to consider the case that we have both translation and rotation? Or do allconditions must hold simultaneously in that case? :unsure:

Put otherwise, we've found that if $I_n-d_\alpha d_\beta$ is invertible, that we have a rotation around a point.
And if $d_\alpha d_\beta=I_n$, that we have a translation.
If there is another possibility, it must be when $I_n-d_\alpha d_\beta$ is not invertible. Note that it isn't in the case of a translation.
Are there such cases?