# Interpretation of maps

#### mathmari

##### Well-known member
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Hey!!

Let $\phi_1:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\cdot v$, $\phi_2:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\cdot v$ which describes the shadow of the sun on a plane.
Give the direction of the sun by giving a vector. How do we find this vector?

To give the equation of the plane do we have to find the image of the map?

How can we explain why at a projection $\psi$ it holds that $\psi\circ\psi=\psi$ and $\text{im} \psi=\text{fix} \psi$?

#### Klaas van Aarsen

##### MHB Seeker
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Hey!!

Let $\phi_1:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\cdot v$, $\phi_2:\mathbb{R}^3\rightarrow \mathbb{R}^3$, $v\mapsto \frac{1}{3}\begin{pmatrix}2 & - 1& - 1\\ - 1& 2 & -1\\ - 1&-1 & 2\end{pmatrix}\cdot v$ which describes the shadow of the sun on a plane.
Give the direction of the sun by giving a vector. How do we find this vector?
Hey mathmari !!

Suppose we have a vector in the direction of the sun.
Then its shadow will have length 0, won't it?
Put otherwise, its image must be the zero vector.

To give the equation of the plane do we have to find the image of the map?
Suppose we have a vector in the plane.
Then its shadow will coincide with this vector, won't it?
Put otherwise, its image must be the same as the vector.
It also means that the vector is a fixpoint of the transformation.

How can we explain why at a projection $\psi$ it holds that $\psi\circ\psi=\psi$ and $\text{im} \psi=\text{fix} \psi$?
What is $\psi$?
Did you mean $\phi_2$?

#### mathmari

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MHB Site Helper
Suppose we have a vector in the direction of the sun.
Then its shadow will have length 0, won't it?
Put otherwise, its image must be the zero vector.

Suppose we have a vector in the plane.
Then its shadow will coincide with this vector, won't it?
Put otherwise, its image must be the same as the vector.
It also means that the vector is a fixpoint of the transformation.
Could you explain to me these two statements? I haven't really understood why these hold.

What is $\psi$?
Did you mean $\phi_2$?
Oh yes, you are right

#### Klaas van Aarsen

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Could you explain to me these two statements? I haven't really understood why these hold.
Consider a sundial.
One that has a long thin rod that casts a shadow on the plane of the sundial.

Now suppose we change the direction of that rod, so that it points directly towards the sun.
Which shadow will it cast?

Alternatively, suppose we lower the rod, so that it comes closer and closer to the plane of the sundial, until it finally reaches the plane of the sundial.
What happens to its shadow?

#### mathmari

##### Well-known member
MHB Site Helper
Now suppose we change the direction of that rod, so that it points directly towards the sun.
Which shadow will it cast?
If that rod has the direction to the sun, then there is no shadow. Therefore, the vector has to in the kernel of the map, right?

Alternatively, suppose we lower the rod, so that it comes closer and closer to the plane of the sundial, until it finally reaches the plane of the sundial.
What happens to its shadow?
Then if we turn in that direction so that the shadow gets maximum, then the shadow is equal to the length of rod. That means thst this vector has to be in the fix set of the map, right?

#### Klaas van Aarsen

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If that rod has the direction to the sun, then there is no shadow. Therefore, the vector has to in the kernel of the map, right?

Then if we turn in that direction so that the shadow gets maximum, then the shadow is equal to the length of rod. That means that this vector has to be in the fix set of the map, right?
Yep.

#### mathmari

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MHB Site Helper
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?

From the above we get that the direction of the sun is a multiple of $\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$, or not?

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?

#### Klaas van Aarsen

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Yep.

Did you notice that the basis of the image spans the same plane as the basis of the fixed points?

#### mathmari

##### Well-known member
MHB Site Helper
Did you notice that the basis of the image spans the same plane as the basis of the fixed points?
Ahh yes! Does this mean that all the vectors of the image are fixed points?

#### Klaas van Aarsen

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Ahh yes! Does this mean that all the vectors of the image are fixed points?
Yep.
It also means that they are eigenvectors that correspond to eigenvalue 1.
And the vector to the sun is an eigenvector that corresponds to eigenvalue 0.

#### mathmari

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Yep.
It also means that they are eigenvectors that correspond to eigenvalue 1.
And the vector to the sun is an eigenvector that corresponds to eigenvalue 0.
Ah ok! Do we need these two statements? Are these equivalent with what we said before? I got stuck right now.

#### Klaas van Aarsen

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Ah ok! Do we need these two statements? Are these equivalent with what we said before? I got stuck right now.
We don't need these two statements.
They are merely equivalent with what we said before.

#### mathmari

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We don't need these two statements.
They are merely equivalent with what we said before.
Ok!

That the vectors of the image are fixed points is just in this case or does this hold every time if we consider a projection?

#### Klaas van Aarsen

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Ok!

That the vectors of the image are fixed points is just in this case or does this hold every time if we consider a projection?
It holds every time if we consider a projection.
It is diiferent for a rotation.

#### mathmari

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It holds every time if we consider a projection.
Does this hold because the projection doesn't change the length of the vector, it just "brings" it to an other position?

#### Klaas van Aarsen

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Does this hold because the projection doesn't change the length of the vector, it just "brings" it to an other position?
A projection does change the length of a vector in general.
The result of a projection is a vector in the plane of projection, which is generally shorter than the original vector.
However, the result of the projection of a vector that is already in the plane, is the same vector.

Did you perhaps mean a rotation?

#### mathmari

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A projection does change the length of a vector in general.
The result of a projection is a vector in the plane of projection, which is generally shorter than the original vector.
However, the result of the projection of a vector that is already in the plane, is the same vector.

Did you perhaps mean a rotation?
No, I mean projection.. I just want to understand why the image equal the set of fixed points and why it holds that $\phi_2\circ\phi_2=\phi_2$, but I got stuck.

#### Klaas van Aarsen

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No, I mean projection.. I just want to understand why the image equal the set of fixed points and why it holds that $\phi_2\circ\phi_2=\phi_2$, but I got stuck.
It seems there is some confusion what exactly a projection is.
What do you think it is?

For reference, wiki gives as definition:
In an abstract setting we can generally say that a projection is a mapping of a set (or of a mathematical structure) which is idempotent, which means that a projection is equal to its composition with itself.
It says in words that $\phi_2$ is called a projection if $\phi_2\circ\phi_2=\phi_2$.

#### mathmari

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For reference, wiki gives as definition:
In an abstract setting we can generally say that a projection is a mapping of a set (or of a mathematical structure) which is idempotent, which means that a projection is equal to its composition with itself.
It says in words that $\phi_2$ is called a projection if $\phi_2\circ\phi_2=\phi_2$.
I am trying to understand this because I am confused now. Can we see that graphically?

#### Klaas van Aarsen

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I am trying to understand this because I am confused now. Can we see that graphically?
How about this picture.

It shows an orthogonal projection in 2D onto a line.
(edit: just noticed the 3rd axis, making it 3D. Well forget about that axis for now, and consider it 2D. )
The vector with the person on top is 'dropped' onto the line.
As you can see, the resulting projection vector is shorter.
If we project the result again it doesn"t change.
The image of all possible vectors form the line.
And all vectors on the line are fixpoints. That is, repeated application yields the same vector. Put otherwise, the transformation is 'idempotent'.