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  1. MHB Master
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    #1
    Hey!!

    Let $\phi_1:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\cdot v$, $\phi_2:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\cdot v$ which describes the shadow of the sun on a plane.
    Give the direction of the sun by giving a vector. How do we find this vector?

    To give the equation of the plane do we have to find the image of the map?

    How can we explain why at a projection $\psi$ it holds that $\psi\circ\psi=\psi$ and $\text{im} \psi=\text{fix} \psi$?

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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    Let $\phi_1:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\cdot v$, $\phi_2:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\cdot v$ which describes the shadow of the sun on a plane.
    Give the direction of the sun by giving a vector. How do we find this vector?
    Hey mathmari!!

    Suppose we have a vector in the direction of the sun.
    Then its shadow will have length 0, won't it?
    Put otherwise, its image must be the zero vector.

    Quote Originally Posted by mathmari View Post
    To give the equation of the plane do we have to find the image of the map?
    Suppose we have a vector in the plane.
    Then its shadow will coincide with this vector, won't it?
    Put otherwise, its image must be the same as the vector.
    It also means that the vector is a fixpoint of the transformation.

    Quote Originally Posted by mathmari View Post
    How can we explain why at a projection $\psi$ it holds that $\psi\circ\psi=\psi$ and $\text{im} \psi=\text{fix} \psi$?
    What is $\psi$?
    Did you mean $\phi_2$?

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    #3 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Suppose we have a vector in the direction of the sun.
    Then its shadow will have length 0, won't it?
    Put otherwise, its image must be the zero vector.

    Suppose we have a vector in the plane.
    Then its shadow will coincide with this vector, won't it?
    Put otherwise, its image must be the same as the vector.
    It also means that the vector is a fixpoint of the transformation.
    Could you explain to me these two statements? I haven't really understood why these hold.

    Quote Originally Posted by Klaas van Aarsen View Post
    What is $\psi$?
    Did you mean $\phi_2$?
    Oh yes, you are right

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    #4
    Quote Originally Posted by mathmari View Post
    Could you explain to me these two statements? I haven't really understood why these hold.
    Consider a sundial.
    One that has a long thin rod that casts a shadow on the plane of the sundial.


    Now suppose we change the direction of that rod, so that it points directly towards the sun.
    Which shadow will it cast?

    Alternatively, suppose we lower the rod, so that it comes closer and closer to the plane of the sundial, until it finally reaches the plane of the sundial.
    What happens to its shadow?

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    #5 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Now suppose we change the direction of that rod, so that it points directly towards the sun.
    Which shadow will it cast?
    If that rod has the direction to the sun, then there is no shadow. Therefore, the vector has to in the kernel of the map, right?


    Quote Originally Posted by Klaas van Aarsen View Post
    Alternatively, suppose we lower the rod, so that it comes closer and closer to the plane of the sundial, until it finally reaches the plane of the sundial.
    What happens to its shadow?
    Then if we turn in that direction so that the shadow gets maximum, then the shadow is equal to the length of rod. That means thst this vector has to be in the fix set of the map, right?

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    #6
    Quote Originally Posted by mathmari View Post
    If that rod has the direction to the sun, then there is no shadow. Therefore, the vector has to in the kernel of the map, right?

    Then if we turn in that direction so that the shadow gets maximum, then the shadow is equal to the length of rod. That means that this vector has to be in the fix set of the map, right?
    Yep.

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    #7 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Yep.
    The echelon form of the matrix of the map is \begin{align*}\frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\ \underset{R_3:R_3+\frac{1}{2}\cdot R_1}{\overset{R_2:R_2+\frac{1}{2}\cdot R_1}{\longrightarrow}} \ \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ 0 & \frac{3}{2} & -\frac{3}{2}\\ 0&-\frac{3}{2} & \frac{3}{2}\end{pmatrix} \ \overset{R_3:R_3+R_2}{\longrightarrow} \ \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ 0 & \frac{3}{2} & -\frac{3}{2}\\ 0&0 & 0\end{pmatrix}\ \underset{R_2:2\cdot R_2}{\overset{R_1:\left (\frac{3}{2}\right )\cdot R_1}{\longrightarrow}} \ \begin{pmatrix}1 & - \frac{1}{2}& - \frac{1}{2}\\ 0 & 1 & -1\\ 0&0 & 0\end{pmatrix}\end{align*}

    The kernel of $\psi$ is $\text{ker}(\psi)=\left \{v\in \mathbb{R}^3\mid \psi (v)=0\in \mathbb{R}^3\right \} $.

    We have that \begin{equation*}\psi (v)=0\Rightarrow \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\begin{pmatrix}v_1 \\ v_2 \\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\\ 0\end{pmatrix}\end{equation*}

    Using the echelon form we get the equations:
    \begin{align*}v_1-\frac{1}{2}v_2-\frac{1}{2}v_3=&0 \\ v_2-v_3=&0 \end{align*}

    The solution vector is therefore $\begin{pmatrix}v_1 \\ v_2 \\ v_3\end{pmatrix}=\begin{pmatrix}v_3 \\ v_3 \\ v_3\end{pmatrix}$

    So the kernel of the map is $\text{Kern}(\psi)=\left \{\begin{pmatrix}v_3 \\ v_3 \\ v_3\end{pmatrix}\mid v_3\in \mathbb{R}\right \}=\left \{v_3\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}\mid v_3\in \mathbb{R}\right \}$.
    A basis of the kernel is $\left \{\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}\right \}$.

    Since the kernel doesn't contain only the zero vector, the map $\psi$ is not injective.


    The image of $\psi$ is $\text{Im}(\psi)=\left \{w\in \mathbb{R}^3\mid w=\psi (v) \ \text{ für ein } v\in \mathbb{R}^3\right \}$.

    A basis of the image contains the linear independent columns of the matrix of the map, so $\left \{\begin{pmatrix}2 \\ -1\\ -1\end{pmatrix}, \ \begin{pmatrix}-1 \\ 2\\ -1\end{pmatrix}\right \}$.

    Since the basis of the image contains only two elements and a basis of $\mathbb{R}^3$ three, the map is not surjective.


    The set of fixed points of $\psi$ is $\text{Fix}(\psi)=\left \{v\in \mathbb{R}^3\mid \psi (v)=v \right \}$.

    We have the following:
    \begin{align*}\frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix} &\Rightarrow \left (\frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}-I_3\right )\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix} \\ & \Rightarrow \left (\begin{pmatrix}\frac{2}{3} & - \frac{1}{3}& - \frac{1}{3}\\ - \frac{1}{3}& \frac{2}{3} & -\frac{1}{3}\\ - \frac{1}{3}&-\frac{1}{3} & \frac{2}{3}\end{pmatrix}-\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 1 & 0\end{pmatrix}\right )\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix} \\ & \Rightarrow \begin{pmatrix}- \frac{1}{3} & - \frac{1}{3}& - \frac{1}{3}\\ - \frac{1}{3}& - \frac{1}{3} & -\frac{1}{3}\\ - \frac{1}{3}&-\frac{1}{3} & - \frac{1}{3}\end{pmatrix}\begin{pmatrix}v_1 \\ v_2\\ v_3\end{pmatrix}=\begin{pmatrix}0 \\ 0\\ 0\end{pmatrix}\end{align*}

    We calculate the echelon form of the matrix:
    \begin{equation*}\begin{pmatrix}- \frac{1}{3} & - \frac{1}{3}& - \frac{1}{3}\\ - \frac{1}{3}& - \frac{1}{3} & -\frac{1}{3}\\ - \frac{1}{3}&-\frac{1}{3} & - \frac{1}{3}\end{pmatrix} \longrightarrow \begin{pmatrix}- \frac{1}{3} & - \frac{1}{3}& - \frac{1}{3}\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} \longrightarrow \begin{pmatrix} 1 & 1 & 1\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\end{equation*}
    And so we get the equation $v_1+v_2+v_3=0$ and so $v_1=-v_2-v_3$.

    The solution vector is $\begin{pmatrix}v_1 \\ v_2 \\ v_3\end{pmatrix}=\begin{pmatrix}-v_2-v_3 \\ v_2 \\ v_3\end{pmatrix}$.

    Therefore the set of fixed points is:
    \begin{align*}\text{Fix}(\psi)&=\left \{\begin{pmatrix}-v_2-v_3 \\ v_2 \\ v_3\end{pmatrix}\mid v_2, v_3\in \mathbb{R} \right \}=\left \{\begin{pmatrix}-v_2 \\ v_2 \\ 0\end{pmatrix}+\begin{pmatrix}-v_3 \\ 0 \\ v_3\end{pmatrix}\mid v_2, v_3\in \mathbb{R} \right \} \\ &=\left \{v_2\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}+v_3\begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix}\mid v_2, v_3\in \mathbb{R} \right \}\end{align*}
    A basis of the set of fixed points is $\left \{\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix},\ \begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix}\right \}$.


    Is so far everything correct?


    From the above we get that the direction of the sun is a multiple of $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$, or not?

    The equation of the plane is a linear combination of the basis elements of the set of fixed points, i.e. $a\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}+b\begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix}$, or not?

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    #8
    Yep.

    Did you notice that the basis of the image spans the same plane as the basis of the fixed points?

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    #9 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    Did you notice that the basis of the image spans the same plane as the basis of the fixed points?
    Ahh yes! Does this mean that all the vectors of the image are fixed points?

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    #10
    Quote Originally Posted by mathmari View Post
    Ahh yes! Does this mean that all the vectors of the image are fixed points?
    Yep.
    It also means that they are eigenvectors that correspond to eigenvalue 1.
    And the vector to the sun is an eigenvector that corresponds to eigenvalue 0.

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