Klaas van Aarsen said:
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? (Wondering) From the above we get that the direction of the sun is a multiple of $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$, or not? (Wondering)
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? (Wondering)
