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 $\text{im} \psi=\text{fix} \psi$?   Reply With Quote

2.

3. Originally Posted by mathmari 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. Originally Posted by mathmari 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. Originally Posted by mathmari 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 $\phi_2$?   Reply With Quote Originally Posted by Klaas van Aarsen 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.  Originally Posted by Klaas van Aarsen What is $\psi$?
Did you mean $\phi_2$? Oh yes, you are right  Reply With Quote

5. Originally Posted by mathmari 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?   Reply With Quote Originally Posted by Klaas van Aarsen 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?  Originally Posted by Klaas van Aarsen 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?   Reply With Quote

7. Originally Posted by mathmari 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.   Reply With Quote Originally Posted by Klaas van Aarsen 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?   Reply With Quote

9. Yep. Did you notice that the basis of the image spans the same plane as the basis of the fixed points?   Reply With Quote Originally Posted by Klaas van Aarsen 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?   Reply With Quote

11. Originally Posted by mathmari 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.   Reply With Quote

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