Geometric interpretation of maps

• MHB
• mathmari
In summary, the conversation discusses four different maps: $f_1, f_2, f_3,$ and $f_4$ that each have a different geometric interpretation. $f_1$ is a rotation of $180^{\circ}$, $f_2$ is a reflection at the $xz$-plane, $f_3$ is a translation by the vector $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$, and $f_4$ is a projection to the $xy$-plane followed by a shear mapping parallel to the $xy$-plane with a factor of $-1$ in the $x$ and $y$ directions. It
mathmari
Gold Member
MHB
Hey!

We have the below maps:

1. $f_1:\mathbb{R}^2\rightarrow \mathbb{R}^2, \ \ \begin{pmatrix}x \\ y\end{pmatrix}\mapsto \begin{pmatrix}-x \\ -y\end{pmatrix}$
2. $f_2:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x \\ -y\\ z\end{pmatrix}$
3. $f_3:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x+2 \\ y-3 \\ z+1\end{pmatrix}$
4. $f_4:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y\\ z\end{pmatrix}\mapsto \begin{pmatrix}x-z \\ y-z \\ 0\end{pmatrix}$

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

1. 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 $180^{\circ}$.
2. Since $x$ and $z$ are unchanged and $y$ changes the sign, it is a reflection at the $xz$-plane.
3. 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 $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$.
4. 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.
Is everything correct? (Wondering)
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)

mathmari said:
1. 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 $180^{\circ}$.

Hey mathmari!

Yep.
And it is also a point reflection in the origin. (Nerd)

mathmari said:
2. Since $x$ and $z$ are unchanged and $y$ changes the sign, it is a reflection at the $xz$-plane.

3. 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 $\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}$.

4. 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.

All correct.
Let's make it a factor $-1$ rather than a value. Otherwise it seems as if it is a translation. (Nerd)

mathmari said:
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?

Yep. (Nod)

mathmari said:
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?
Each point "of the plane" is not good enough. (Worried)
We need that each point in the co-domain has a preimage of exactly one element.
Then it is both injective and surjective. (Thinking)

I'm afraid that the fact that "each point in the domain has an image of exactly one point" is also not good enough.
It means there could still be points in the co-domain that do not have an original. (Worried)
It does mean it is injective, but we still need to verify that it is surjective.
Put otherwise, it is both injective and surjective iff it is 1-1 and onto. (Thinking)

Klaas van Aarsen said:
Yep.
And it is also a point reflection in the origin. (Nerd)
All correct.
Let's make it a factor $-1$ rather than a value. Otherwise it seems as if it is a translation. (Nerd)

Is there a tool where we can see these translations graphically? Do we have to draw the images of the unit vectors for that? (Wondering)
Klaas van Aarsen said:
Each point "of the plane" is not good enough. (Worried)
We need that each point in the co-domain has a preimage of exactly one element.
Then it is both injective and surjective. (Thinking)

I'm afraid that the fact that "each point in the domain has an image of exactly one point" is also not good enough.
It means there could still be points in the co-domain that do not have an original. (Worried)
It does mean it is injective, but we still need to verify that it is surjective.
Put otherwise, it is both injective and surjective iff it is 1-1 and onto. (Thinking)
The fact that each point in the domain has an image of exactly one point, means thst it is injective? (Wondering) Can we do do that also as below? (Wondering)

For that we consider the map $f:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y \\ z\end{pmatrix}\mapsto \begin{pmatrix}-x \\ -y \\ z\end{pmatrix}$.

For injectivity: Let $a,b,c,x,y,z\in \mathbb{R}$ with $f\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}x \\ y \\ z\end{pmatrix}\Rightarrow \begin{pmatrix}-a \\ -b \\ c\end{pmatrix}=\begin{pmatrix}-x \\ -y \\ z\end{pmatrix}\Rightarrow \begin{cases}-a=-x \\ -b=-y \\ c=z\end{cases}\Rightarrow \begin{cases}a=x \\ b=y \\ c=z\end{cases} \Rightarrow \begin{pmatrix}a \\ b \\ c\end{pmatrix}=\begin{pmatrix}x \\ y \\ z\end{pmatrix}$.

That means that $f$ is injective. For surjectivity: Let $\begin{pmatrix}a \\ b \\ c\end{pmatrix}\in \mathbb{R}^3$. We have that $\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}-a \\ -b \\ c\end{pmatrix}$. So for each $w\in \mathbb{R}^3$ there is a $v\in \mathbb{R}^3$ such that $f(v)=w$.

That means that $f$ is surjective. Is everything correct? (Wondering)

mathmari said:
Is there a tool where we can see these translations graphically? Do we have to draw the images of the unit vectors for that?

I don't really know of a specific tool to show the effect of a linear transformation.
Drawing the unit vectors and their images should work yes. (Thinking)

mathmari said:
The fact that each point in the domain has an image of exactly one point, means thst it is injective?

Let me rephrase that a bit. (Blush)

A function is injective if each point in the domain has a unique image of one point. That is, each such image does not have another original. We also say that the function is 1-1. (Thinking)

Equivalently, a function is injective if each point in the co-domain has at most 1 original.

mathmari said:
Can we do do that also as below?

For that we consider the map $f:\mathbb{R}^3\rightarrow \mathbb{R}^3, \ \ \begin{pmatrix}x \\ y \\ z\end{pmatrix}\mapsto \begin{pmatrix}-x \\ -y \\ z\end{pmatrix}$.

For injectivity: Let $a,b,c,x,y,z\in \mathbb{R}$ with $f\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}x \\ y \\ z\end{pmatrix}\Rightarrow \begin{pmatrix}-a \\ -b \\ c\end{pmatrix}=\begin{pmatrix}-x \\ -y \\ z\end{pmatrix}\Rightarrow \begin{cases}-a=-x \\ -b=-y \\ c=z\end{cases}\Rightarrow \begin{cases}a=x \\ b=y \\ c=z\end{cases} \Rightarrow \begin{pmatrix}a \\ b \\ c\end{pmatrix}=\begin{pmatrix}x \\ y \\ z\end{pmatrix}$.

That means that $f$ is injective.

For surjectivity: Let $\begin{pmatrix}a \\ b \\ c\end{pmatrix}\in \mathbb{R}^3$. We have that $\begin{pmatrix}a \\ b \\ c\end{pmatrix}=f\begin{pmatrix}-a \\ -b \\ c\end{pmatrix}$. So for each $w\in \mathbb{R}^3$ there is a $v\in \mathbb{R}^3$ such that $f(v)=w$.

That means that $f$ is surjective.

Is everything correct?

Yep. (Nod)

1. What is a geometric interpretation of maps?

A geometric interpretation of maps refers to the process of representing the physical features of the Earth's surface using geometric shapes and figures. This includes the use of points, lines, and polygons to represent landmasses, bodies of water, and other features on a map.

2. How is a geometric interpretation of maps different from other types of map projections?

Unlike other map projections that use mathematical equations to flatten the Earth's curved surface onto a 2-dimensional map, a geometric interpretation of maps relies on the use of geometric shapes and figures to represent features on a map. This approach allows for a more visually appealing representation of the Earth's surface, but may result in some distortion of size and shape.

3. What are some advantages of using a geometric interpretation of maps?

One advantage of using a geometric interpretation of maps is that it allows for a more intuitive understanding of the Earth's surface. By using familiar shapes and figures, it is easier to interpret and analyze the information presented on a map. Additionally, this approach can also be used to create aesthetically pleasing maps that are visually appealing.

4. Are there any limitations to using a geometric interpretation of maps?

Yes, there are some limitations to using a geometric interpretation of maps. One major limitation is the potential for distortion of size and shape in the representation of features on a map. This can make it difficult to accurately measure distances or compare the sizes of different areas on the map. Additionally, this approach may not be suitable for representing large-scale or highly detailed maps.

5. How is a geometric interpretation of maps used in scientific research?

A geometric interpretation of maps is commonly used in scientific research to visually represent and analyze data related to the Earth's surface. This can include mapping geological features, land use patterns, and environmental changes. By using geometric shapes and figures, scientists can better understand and communicate complex information about the Earth's surface.

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