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  1. MHB Master
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    #1
    Hey!!

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

    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?

    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 $\sigma_a$ ?

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  3. MHB Seeker
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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

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

    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?

    Quote Originally Posted by mathmari View Post
    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?

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    #3 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    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?

    Quote Originally Posted by Klaas van Aarsen View Post
    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?

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    #4
    Quote Originally Posted by mathmari View Post
    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?

    Quote Originally Posted by mathmari View Post
    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 $x$?

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    #5 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    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?

    Quote Originally Posted by Klaas van Aarsen View Post
    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?

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    #6
    Quote Originally Posted by mathmari View Post
    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?

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

    Quote Originally Posted by mathmari View Post
    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.

    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?

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    #7 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    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?

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    #8
    Quote Originally Posted by mathmari View Post
    $x$ is a directional vector along the line of reflection, or not?
    Do we have $\sigma_x(x)=x$?
    What is $\sigma_x(x)$?

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    #9 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Do we have $\sigma_x(x)=x$?
    What is $\sigma_x(x)$?
    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?

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    #10
    Quote Originally Posted by mathmari View Post
    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.

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