Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $\displaystyle{\begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}-\frac{1}{2}z \\ -x+y-\frac{1}{2}z\\ z\end{pmatrix}}$ be a linear map and let $\displaystyle{B=\left \{\begin{pmatrix}2 \\ 2\\ 0\end{pmatrix}, \begin{pmatrix}-1 \\ 3\\ 2\end{pmatrix}, \begin{pmatrix}0 \\ 1\\ 2\end{pmatrix}\right \}}$ be a basis of $\mathbb{R}^3$.
I have shown that \begin{equation*}\text{Kern}(f) =\left \{\begin{pmatrix}y \\ y\\ 0\end{pmatrix}\mid y\in \mathbb{R} \right \}=\left \{y\begin{pmatrix}1 \\ 1\\ 0\end{pmatrix}\mid y\in \mathbb{R} \right \} \ \text{ and } \\ \text{Fix}(f) =\left \{z\begin{pmatrix}-\frac{1}{2} \\ 0\\ 1\end{pmatrix}+y\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}\mid y,z\in \mathbb{R} \right \}\end{equation*}
Then I calculated the transformation matrices \begin{equation*}BfB=\begin{pmatrix}0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \ \text{ and } \ B(f\circ f)B=\begin{pmatrix}0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}
Let $k_B(v)=\begin{pmatrix}2 \\ -3 \\ -1\end{pmatrix}$ be the coordinate vector of $v$ in respect of $B$.
This means \begin{equation*}v=2\cdot \begin{pmatrix}2 \\ 2\\ 0\end{pmatrix}-3\cdot \begin{pmatrix}-1 \\ 3\\ 2\end{pmatrix}-1\cdot \begin{pmatrix}0 \\ 1\\ 2\end{pmatrix}\end{equation*}
Then we get: \begin{equation*}f(v) = \frac{1}{2}\cdot \begin{pmatrix}2 \\ 2\\ 0\end{pmatrix}-3\cdot \begin{pmatrix}-1 \\ 3\\ 2\end{pmatrix}-1\cdot \begin{pmatrix}0 \\ 1\\ 2\end{pmatrix}\end{equation*} So we get $k_B(f(v))=\begin{pmatrix}\frac{1}{2} \\ -3 \\ -1\end{pmatrix} \text{ and } \ f(v) = \begin{pmatrix}1 \\ 1\\ 0\end{pmatrix}+ \begin{pmatrix}3 \\ -9\\ -6\end{pmatrix}+ \begin{pmatrix}0 \\ -1\\ -2\end{pmatrix}=\begin{pmatrix}4 \\ -9\\ -8\end{pmatrix}$.
We have the transformation matrix \begin{equation*}CfC=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\end{equation*} and such a basis $C$ can be \begin{equation*}\left \{\begin{pmatrix}0 \\ 1\\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 0\\ -2\end{pmatrix}, \begin{pmatrix}0 \\ 0\\ 1\end{pmatrix} \right \}\end{equation*}
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Then the question is:
Interpret the map $f$ geometrically, using the previous results.
Could you give me a hint for that?