Finding a Vector Perpendicular to the Line of Reflection

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Properties
In summary, a map $\sigma_x$ is defined as a linear function that maps a vector $v$ to $v-2\frac{x\cdot v}{x\cdot x}x$. It is shown to be linear, symmetric, and preserves the dot product of two vectors. In the case where $n=2$, a vector $v$ can be determined to satisfy the condition $\sigma_v=\sigma_a$, where $\sigma_a$ is a reflection on a straight line through the origin with a given angle $a$. Further discussion includes verifying the bijectivity of $\sigma_x$ and determining the vectors $n$ and $d$ that are normal and directional, respectively, to the line of reflection.
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

Let $1\leq n\in \mathbb{N}$. For $0_{\mathbb{R}^n}\neq x\in \mathbb{R}^n$ we define the map $$\sigma_x:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ v\mapsto v-2\frac{x\cdot v}{x\cdot x}x$$

Show that:
  1. The map is linear.
  2. It holds that $\sigma_x\in \text{Sym}(\mathbb{R}^n)$ and $\sigma_x=\sigma_x^{-1}$.
  3. For all $v,w\in \mathbb{R}$ it holds that $v\cdot w=\sigma_x(v)\cdot \sigma_x(w)$.
  4. 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:

  1. $\sigma_x$ 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}x-2\frac{\sum_{i=1}^nx_i\cdot w_i}{x\cdot x}x=v-2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}x+w-2\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=rv-2\frac{\sum_{i=1}^nx_i\cdot (rv)_i}{x\cdot x}x\\ & =rv-r2\frac{\sum_{i=1}^nx_i\cdot v_i}{x\cdot x}x=r\left (v-2\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.
  2. Could you give me a hint how we can show that? (Wondering)
  3. \begin{align*}\sigma_x(v)\cdot \sigma_x(w)&=\left (v-2\frac{x\cdot v}{x\cdot x}x\right )\cdot \left (w-2\frac{x\cdot w}{x\cdot x}x\right )=v\cdot w -2\frac{x\cdot w}{x\cdot x}v\cdot x-2\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*}
  4. How can we find such a vector? Do we use the matrix of the map of $\sigma_a$ ? (Wondering)
 
Physics news on Phys.org
  • #2
mathmari said:
Hey! :eek:

Let $1\leq n\in \mathbb{N}$. For $0_{\mathbb{R}^n}\neq x\in \mathbb{R}^n$ we define the map $$\sigma_x:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ v\mapsto v-2\frac{x\cdot v}{x\cdot x}x$$

Show that:
  1. The map is linear.
  2. It holds that $\sigma_x\in \text{Sym}(\mathbb{R}^n)$ and $\sigma_x=\sigma_x^{-1}$.
  3. For all $v,w\in \mathbb{R}$ it holds that $v\cdot w=\sigma_x(v)\cdot \sigma_x(w)$.
  4. 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? (Wondering)

mathmari said:
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? (Wondering)
 
  • #3
Klaas van Aarsen said:
The symmetry group is the group of bijections from a set to the same set.
Is it a bijection? (Wondering)

Do we check that using the definitions? As we do that when we consider functions? (Wondering)

Klaas van Aarsen said:
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? (Wondering)

The one will be the vector x and the other the vector v? (Wondering)
 
  • #4
mathmari said:
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? (Wondering)
And additionally it proves that $\sigma_x$ is invertible, which implies it is bijective, doesn't it? (Wondering)

mathmari said:
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? (Worried)

Which one is the vector $x$? (Wondering)
 
  • #5
Klaas van Aarsen said:
Suppose we verify that $\sigma_x(\sigma_x)=\operatorname{id}$.
Then that proves that $\sigma_x^{-1}=\sigma_x$ doesn't it? (Wondering)
And additionally it proves that $\sigma_x$ is invertible, which implies it is bijective, doesn't it? (Wondering)

Knowing that the map is invertible do we know that the map is bijective or injective? (Wondering)

Klaas van Aarsen said:
Isn't $v$ the parameter to the function?
It's not a constant vector, is it? (Worried)

Which one is the vector $x$? (Wondering)

I haven't understood that part. Could you explain it further to me? (Wondering)
 
  • #6
mathmari said:
Knowing that the map is invertible do we know that the map is bijective or injective?

Do we have any propositions or some such saying so? (Wondering)

Or can we prove it?
That is, can we prove that $\sigma_x$ is both injective and surjective given that it is invertible? (Wondering)

mathmari said:
I haven't understood that part. Could you explain it further to me?

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. (Nerd)

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. (Thinking)

Is $x$ a normal vector to the line of reflection, or is it a directional vector along the line of reflection? (Wondering)
 
  • #7
Klaas van Aarsen said:
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. (Nerd)

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. (Thinking)

Is $x$ a normal vector to the line of reflection, or is it a directional vector along the line of reflection? (Wondering)

$x$ is a directional vector along the line of reflection, or not? (Wondering)
 
  • #8
mathmari said:
$x$ is a directional vector along the line of reflection, or not?

Do we have $\sigma_x(x)=x$?
What is $\sigma_x(x)$? (Wondering)
 
  • #9
Klaas van Aarsen said:
Do we have $\sigma_x(x)=x$?
What is $\sigma_x(x)$? (Wondering)

We have that $\sigma_x(x)=x-2\frac{x\cdot x}{x\cdot x}x=x-2x=-x$.

That means that $x$ is a normal vector to the line of reflection, right? (Blush)
 
  • #10
mathmari said:
We have that $\sigma_x(x)=x-2\frac{x\cdot x}{x\cdot x}x=x-2x=-x$.

That means that $x$ is a normal vector to the line of reflection, right?

Yep. (Nod)
 
  • #11
Klaas van Aarsen said:
Yep. (Nod)

Ok! Can we get a specific vector? Maybe using the matrix of map $\sigma_a$ ? (Wondering)
 
  • #12
mathmari said:
Ok! Can we get a specific vector? Maybe using the matrix of map $\sigma_a$ ? (Wondering)

The question only asks for a fixed vector.
We know now that we need a vector perpendicular to the line of reflection.
The line has an angle a with respect to the positive x-axis.
So $(\cos a,\,\sin a)$ is a directional vector along the line.
Can we find a vector perpendicular to it? (Wondering)
 

FAQ: Finding a Vector Perpendicular to the Line of Reflection

1. What is σ?

σ, also known as sigma, is the symbol used to represent the standard deviation of a set of data. It is a measure of how spread out the data points are from the mean.

2. How is σ calculated?

To calculate σ, you first find the mean of the data set. Then, for each data point, you subtract the mean and square the result. Next, you find the sum of all these squared differences and divide it by the total number of data points. Finally, take the square root of this value to get the standard deviation.

3. What does σ tell us about a data set?

σ tells us how much the data points deviate from the mean. A smaller σ indicates that the data points are closer to the mean, while a larger σ indicates that the data points are more spread out.

4. How is σ used in statistical analysis?

σ is used to measure the variability or dispersion of a data set. It is an important tool in statistical analysis as it helps us understand the distribution of the data and make comparisons between different sets of data.

5. Can σ be negative?

No, σ cannot be negative. It is always a positive value as it is the square root of the sum of squared differences, which are always positive.

Similar threads

Replies
52
Views
3K
Replies
4
Views
1K
Replies
2
Views
2K
Replies
4
Views
471
Replies
2
Views
968
Replies
23
Views
1K
Replies
5
Views
1K
Replies
4
Views
1K
Back
Top