- #1

mathmari

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We have the below maps:

- $f_1:\mathbb{R}^2\rightarrow \mathbb{R}^2, \ \ \begin{pmatrix}x \\ y\end{pmatrix}\mapsto \begin{pmatrix}-x \\ -y\end{pmatrix}$

- $f_2:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x \\ -y\\ z\end{pmatrix}$

- $f_3:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x+2 \\ y-3 \\ z+1\end{pmatrix}$

- $f_4:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}$

I want to give the geometric interpretation of these maps. I have done the following:

- We have that $\begin{pmatrix}-x \\ -y\end{pmatrix}=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$ which means that this is a rotation of $180^{\circ}$.
- Since $x$ and $z$ are unchanged and $y$ changes the sign, it is a reflection at the $xz$-plane.
- Since we add to $\begin{pmatrix}x \\ y \\ z\end{pmatrix}$ the vector $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$ it is a translation by the vector $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$.
- We have that $\begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}=\begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x \\ y\\ z\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y\\ z\end{pmatrix}$. The first matrix describes the projection to the $xy$-plane and the second one the shear mapping parallel to the $xy$-plane by the value $-1$ in the $x$ direction and by the value $-1$ in the $y$ direction.

I want to check also if $f_4$ is injective and surjective.

For that let $a,b,c,x,y,z\in \mathbb{R}$ with $f\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}x \\ y \\ z\end{pmatrix} \Rightarrow \begin{pmatrix}a-c \\ b-c \\ 0\end{pmatrix}=\begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}$. From that we get $a-c=x-z$ and $b-c=y-z$. Subtracting these two relations we get $a-b=x-y$. This equality is also satisfied with $x=-b$ and $y=-a$ and this means that $f_4$ is not injective.

The vector $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$ has no preimage under $f_4$ and so $f_4$ is not surjective.

Are both justifivations and results correct? (Wondering)

I want to check also if the reflection to the $z$-axis is injective and surjective.

For that does it hold that each point of the plane is the image of exactly one point and so this transformationis bijective? (Wondering)