
MHB Master
#1
February 10th, 2020,
02:04
Hey!!
Let $1\leq n\in \mathbb{N}$. For $0_{\mathbb{R}^n}\neq x\in \mathbb{R}^n$ we define the map $$\sigma_x:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ v\mapsto v2\frac{x\cdot v}{x\cdot x}x$$
Show that:
 The map is linear.
 It holds that $\sigma_x\in \text{Sym}(\mathbb{R}^n)$ and $\sigma_x=\sigma_x^{1}$.
 For all $v,w\in \mathbb{R}$ it holds that $v\cdot w=\sigma_x(v)\cdot \sigma_x(w)$.
 Let $n=2$. Determine a vector $v\in \mathbb{R}^2$ such that $\sigma_v=\sigma_a$, where $\sigma_a$ is
the reflection on the straight line through the origin, where $a$ describes the angle between the straight line and the positive $x$axis.
I have done the following:
 $\sigma_x$ is additive:
\begin{align*}\sigma_x(v+w)&=(v+w)2\frac{x\cdot (v+w)}{x\cdot x}x=(v+w)2\frac{\sum_{i=1}^nx_i\cdot (v+w)_i}{x\cdot x}x\\ & =(v+w)2\frac{\sum_{i=1}^nx_i\cdot v_i+\sum_{i=1}^nx_i\cdot w_i}{x\cdot x}x=(v+w)2\left(\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}+\frac{\sum_{i=1}^nx_i\cdot w_i}{x\cdot x}\right )x \\ & =(v+w)2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}x2\frac{\sum_{i=1}^nx_i\cdot w_i}{x\cdot x}x=v2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}x+w2\frac{\sum_{i=1}^nx_i\cdot w_i}{x\cdot x}x \\ & = \sigma_x(v)+\sigma_x(w)\end{align*}
$\sigma_x$ is homogeneous:
\begin{align*}\sigma_x(rv)&=(rv)2\frac{x\cdot (rv)}{x\cdot x}x=rv2\frac{\sum_{i=1}^nx_i\cdot (rv)_i}{x\cdot x}x\\ & =rvr2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}x=r\left (v2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}\right ) \\ & = r\sigma_x(v)\end{align*}
Therefore the map $\sigma_x$ is linear.
 Could you give me a hint how we can show that?
 \begin{align*}\sigma_x(v)\cdot \sigma_x(w)&=\left (v2\frac{x\cdot v}{x\cdot x}x\right )\cdot \left (w2\frac{x\cdot w}{x\cdot x}x\right )=v\cdot w 2\frac{x\cdot w}{x\cdot x}v\cdot x2\frac{x\cdot v}{x\cdot x}w\cdot x+4\frac{x\cdot v}{x\cdot x}\frac{x\cdot w}{x\cdot x}x\cdot x \\ & =v\cdot w 2\frac{(x\cdot w)(v\cdot x)}{x\cdot x}2\frac{(x\cdot v)(w\cdot x)}{x\cdot x}+4\frac{(x\cdot v)(x\cdot w)}{x\cdot x}=v\cdot w 4\frac{(x\cdot w)(v\cdot x)}{x\cdot x}+4\frac{(x\cdot v)(x\cdot w)}{x\cdot x}\\ & =v\cdot w\end{align*}
 How can we find such a vector? Do we use the matrix of the map of $\sigma_a$ ?

February 10th, 2020 02:04
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MHB Seeker
#2
February 10th, 2020,
05:23
Originally Posted by
mathmari
Hey!!
Let $1\leq n\in \mathbb{N}$. For $0_{\mathbb{R}^n}\neq x\in \mathbb{R}^n$ we define the map $$\sigma_x:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ v\mapsto v2\frac{x\cdot v}{x\cdot x}x$$
Show that:
 The map is linear.
 It holds that $\sigma_x\in \text{Sym}(\mathbb{R}^n)$ and $\sigma_x=\sigma_x^{1}$.
 For all $v,w\in \mathbb{R}$ it holds that $v\cdot w=\sigma_x(v)\cdot \sigma_x(w)$.
 Let $n=2$. Determine a vector $v\in \mathbb{R}^2$ such that $\sigma_v=\sigma_a$, where $\sigma_a$ is
the reflection on the straight line through the origin, where $a$ describes the angle between the straight line and the positive $x$axis.
.
2. Could you give me a hint how we can show that?
Hey mathmari!!
The symmetry group is the group of bijections from a set to the same set.
Is it a bijection?
Originally Posted by
mathmari
3. How can we find such a vector? Do we use the matrix of the map of $\sigma_a$ ?
Suppose $\sigma_x$ is a projection. What would be a vector along its line of reflection?
And what would be a vector perpendicular to its line of reflection?

MHB Master
#3
February 10th, 2020,
11:19
Thread Author
Originally Posted by
Klaas van Aarsen
The symmetry group is the group of bijections from a set to the same set.
Is it a bijection?
Do we check that using the definitions? As we do that when we consider functions?
Originally Posted by
Klaas van Aarsen
Suppose $\sigma_x$ is a projection. What would be a vector along its line of reflection?
And what would be a vector perpendicular to its line of reflection?
The one will be the vector x and the other the vector v?

MHB Seeker
#4
February 10th, 2020,
13:14
Originally Posted by
mathmari
Do we check that using the definitions? As we do that when we consider functions?
Suppose we verify that $\sigma_x(\sigma_x)=\operatorname{id}$.
Then that proves that $\sigma_x^{1}=\sigma_x$ doesn't it?
And additionally it proves that $\sigma_x$ is invertible, which implies it is bijective, doesn't it?
Originally Posted by
mathmari
The one will be the vector x and the other the vector v?
Isn't $v$ the parameter to the function?
It's not a constant vector, is it?
Which one is the vector $x$?

MHB Master
#5
February 10th, 2020,
13:33
Thread Author
Originally Posted by
Klaas van Aarsen
Suppose we verify that $\sigma_x(\sigma_x)=\operatorname{id}$.
Then that proves that $\sigma_x^{1}=\sigma_x$ doesn't it?
And additionally it proves that $\sigma_x$ is invertible, which implies it is bijective, doesn't it?
Knowing that the map is invertible do we know that the map is bijective or injective?
Originally Posted by
Klaas van Aarsen
Isn't $v$ the parameter to the function?
It's not a constant vector, is it?
Which one is the vector $x$?
I haven't understood that part. Could you explain it further to me?

MHB Seeker
#6
February 10th, 2020,
14:55

MHB Master
#7
February 12th, 2020,
18:24
Thread Author
Originally Posted by
Klaas van Aarsen
We have the map $\sigma_x$ and it is given by a formula of the form $\sigma_x:v\mapsto \sigma_x(v)$.
In this formula $x$ is a fixed vector.
And $v$ is an arbitrary vector that is mapped to a vector in the image. In particular $v$ is
not a fixed vector.
Assuming the $\sigma_x$ is a reflection in $\mathbb R^2$, there must be a vector $n$ normal to the line of reflection such that $\sigma_x(n)=n$, mustn't it?
And there must be a directional vector $d$ along the line of reflection, such that $\sigma_x(d)=d$, which is a fixpoint.
Is $x$ a normal vector to the line of reflection, or is it a directional vector along the line of reflection?
$x$ is a directional vector along the line of reflection, or not?

MHB Seeker
#8
February 12th, 2020,
18:28
Originally Posted by
mathmari
$x$ is a directional vector along the line of reflection, or not?
Do we have $\sigma_x(x)=x$?
What is $\sigma_x(x)$?

MHB Master
#9
February 12th, 2020,
18:31
Thread Author
Originally Posted by
Klaas van Aarsen
Do we have $\sigma_x(x)=x$?
What is $\sigma_x(x)$?
We have that $\sigma_x(x)=x2\frac{x\cdot x}{x\cdot x}x=x2x=x$.
That means that $x$ is a normal vector to the line of reflection, right?

MHB Seeker
#10
February 12th, 2020,
18:32
Originally Posted by
mathmari
We have that $\sigma_x(x)=x2\frac{x\cdot x}{x\cdot x}x=x2x=x$.
That means that $x$ is a normal vector to the line of reflection, right?
Yep.