Welcome to our community

Be a part of something great, join today!

Geometric interpretation of maps

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! :eek:

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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)

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)

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)

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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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)


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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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)

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.

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)