MHB Analyzing the Composition of Scalings with Different Centers and Scaling Factors

[mathmari]

Hey!

Let $S_i$, $i=1,2$ the scaling with scaling factor $r_i$ and center $Z_i$.

Let $r_1r_2\neq 1$. I want to show that $S_2\circ S_1$ is a scaling and to calculate the center and the scaling factor.

When $Z_1\neq Z_2$ I want to show that $Z\in Z_1Z_2$. I don't really have an idea. Could you give me a hint how we could show that?

 

[I like Serena]

Hey!

Let $S_i$, $i=1,2$ the scaling with scaling factor $r_i$ and center $Z_i$.

Let $r_1r_2\neq 1$. I want to show that $S_2\circ S_1$ is a scaling and to calculate the center and the scaling factor.

When $Z_1\neq Z_2$ I want to show that $Z\in Z_1Z_2$. I don't really have an idea. Could you give me a hint how we could show that?

Hey mathmari! (Smile)

A scaling would be of the form:
$$S_i(\mathbf x) = \mathbf Z_i + r_i(\mathbf x - \mathbf Z_i)$$
wouldn't it?

Can we make the composition $S_2\circ S_1$? (Wondering)

 

[mathmari]

Hey mathmari! (Smile)

A scaling would be of the form:
$$S_i(\mathbf x) = \mathbf Z_i + r_i(\mathbf x - \mathbf Z_i)$$
wouldn't it?

Can we make the composition $S_2\circ S_1$? (Wondering)

We have the that \begin{align*}S_2\circ S_1&= S_2(S_1) \\ & = \mathbf Z_2 + r_2(S_1 - \mathbf Z_2) \\ & =\mathbf Z_2 + r_2(\mathbf Z_1 + r_1(\mathbf x - \mathbf Z_1) - \mathbf Z_2) \\ & =\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1(\mathbf x - \mathbf Z_1) - r_2\mathbf Z_2 \\ & = (1-r_2)\mathbf Z_2+(r_2+r_2r_1)(\mathbf x - \mathbf Z_1)\end{align*} right?

 

[I like Serena]

We have the that \begin{align*}S_2\circ S_1&= S_2(S_1) \\ & = \mathbf Z_2 + r_2(S_1 - \mathbf Z_2) \\ & =\mathbf Z_2 + r_2(\mathbf Z_1 + r_1(\mathbf x - \mathbf Z_1) - \mathbf Z_2) \\ & =\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1(\mathbf x - \mathbf Z_1) - r_2\mathbf Z_2 \\ & = (1-r_2)\mathbf Z_2+(r_2+r_2r_1)(\mathbf x - \mathbf Z_1)\end{align*} right

Yep. (Nod)
Can we write it in the form $\mathbf Z+r(\mathbf x - \mathbf Z)$?

 

[mathmari]

Yep. (Nod)
Can we write it in the form $\mathbf Z+r(\mathbf x - \mathbf Z)$?

We have that $\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2$. Since we have just one element that is multiplied with $r_1$, except of $\mathbf x$, I mean $r_2r_1\mathbf Z_1$, I don't see how we could write it in the above form. (Wondering)

 

[I like Serena]

We have that $\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2$. Since we have just one element that is multiplied with $r_1$, except of $\mathbf x$, I mean $r_2r_1\mathbf Z_1$, I don't see how we could write it in the above form. (Wondering)

A property of a scaling is that it leaves exactly 1 point unchanged, which is $\mathbf Z$.
Suppose we try to find it.
Can we solve:
$$(S_2 \circ S_1)(\mathbf x) = \mathbf x$$
(Wondering)

 

[mathmari]

A property of a scaling is that it leaves exactly 1 point unchanged, which is $\mathbf Z$.
Suppose we try to find it.
Can we solve:
$$(S_2 \circ S_1)(\mathbf x) = \mathbf x$$
(Wondering)

So, you mean that
\begin{align*}&\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2-\mathbf x=0 \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=0 \end{align*}
For what do we have to solve? (Wondering)

 

[I like Serena]

So, you mean that
\begin{align*}&\mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2-\mathbf x=0 \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=0 \end{align*}
For what do we have to solve? (Wondering)

Yes.
We have to solve for $\mathbf x$.
And then we have to verify that if we substitute it for $\mathbf Z$ that we can match $\mathbf Z + r(\mathbf x - \mathbf Z)$ after all, showing that it is indeed a scaling with this $\mathbf Z$ as its center and with $r=r_1r_2$ as its scale.
And while doing so, we should discover why it was given that $r_1r_2 \ne 1$. (Thinking)

 

[mathmari]

Yes.
We have to solve for $\mathbf x$.

We have that \begin{align*}&\Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=0 \\ & \Rightarrow (r_2r_1-1)\mathbf x = -\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2 \\ & \Rightarrow \mathbf x = \frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right )\end{align*} right? (Wondering)

 

[I like Serena]

We have that \begin{align*}&\Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=0 \\ & \Rightarrow (r_2r_1-1)\mathbf x = -\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2 \\ & \Rightarrow \mathbf x = \frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right )\end{align*} right? (Wondering)

Yep. (Nod)

 

[mathmari]

And then we have to verify that if we substitute it for $\mathbf Z$ that we can match $\mathbf Z + r(\mathbf x - \mathbf Z)$ after all, showing that it is indeed a scaling with this $\mathbf Z$ as its center and with $r=r_1r_2$ as its scale.

What exactly do we have to do? I haven't really understood that. (Wondering)

 

[I like Serena]

What exactly do we have to do? I haven't really understood that. (Wondering)

We have to prove that $S_2\circ S_1$ is a scaling don't we?
And we have to find its center $\mathbf Z$ and scaling factor $r$.
So we need to find $r$ and $\mathbf Z$ such that $(S_2\circ S_1)(\mathbf Z) = \mathbf Z$ and $(S_2\circ S_1)(\mathbf Z + \mathbf a) = \mathbf Z + r\mathbf a$. (Thinking)

 

[mathmari]

So we need to find $r$ and $\mathbf Z$ such that $(S_2\circ S_1)(\mathbf Z) = \mathbf Z$ and $(S_2\circ S_1)(\mathbf Z + \mathbf a) = \mathbf Z + r\mathbf a$. (Thinking)

So, do we have to solve the followng system?
\begin{equation*}(S_2\circ S_1)(\mathbf Z) = \mathbf Z \Rightarrow \mathbf Z=\frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*}

\begin{equation*}(S_2\circ S_1)(\mathbf Z + r\mathbf a) = \mathbf Z + r\mathbf a \Rightarrow \mathbf Z + r\mathbf a=\frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*}
(Wondering)

 

[I like Serena]

So, do we have to solve the followng system?
\begin{equation*}(S_2\circ S_1)(\mathbf Z) = \mathbf Z \Rightarrow \mathbf Z=\frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*}

Yes.

\begin{equation*}(S_2\circ S_1)(\mathbf Z + r\mathbf a) = \mathbf Z + r\mathbf a \Rightarrow \mathbf Z + r\mathbf a=\frac{1}{r_2r_1-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*}

That should be $(S_2\circ S_1)(\mathbf Z + {\color{red}\cancel r}\mathbf a)=\mathbf Z + r\mathbf a$, which has to hold for any $\mathbf a$.
That is, if we pick a vector with respect to the center, it should be scaled by a factor $r$. (Thinking)

 

[mathmari]

Yes.

So, the center is \begin{equation*} \mathbf Z=\frac{1}{r_1r_2-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_1r_2\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*} right? (Wondering)

That should be $(S_2\circ S_1)(\mathbf Z + {\color{red}\cancel r}\mathbf a)=\mathbf Z + r\mathbf a$, which has to hold for any $\mathbf a$.
That is, if we pick a vector with respect to the center, it should be scaled by a factor $r$. (Thinking)

Ahh, ok! So, we have the following:
\begin{align*}&(S_2\circ S_1)(\mathbf Z + \mathbf a) = \mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1 (\mathbf Z + \mathbf a) - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf Z + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf Z + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2= r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 -\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2 + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2= r\mathbf a \\ & \Rightarrow r_2r_1 \mathbf a = r\mathbf a \\ & \Rightarrow r=r_1r_2\end{align*} right? (Wondering)

 

[I like Serena]

So, the center is \begin{equation*} \mathbf Z=\frac{1}{r-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r\mathbf Z_1 + r_2\mathbf Z_2\right ) \end{equation*} right?

Yes.
... except that we don't know what $r$ is yet at this point, so we should leave it as $r_1r_2$. (Nerd)

Ahh, ok! So, we have the following:
\begin{align*}&(S_2\circ S_1)(\mathbf Z + \mathbf a) = \mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1 (\mathbf Z + \mathbf a) - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf Z + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf Z + r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + (r_2r_1-1)\mathbf Z + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2= r\mathbf a \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 -\mathbf Z_2 - r_2\mathbf Z_1+ r_2r_1\mathbf Z_1 + r_2\mathbf Z_2 + r_2r_1 \mathbf a - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2= r\mathbf a \\ & \Rightarrow r_2r_1 \mathbf a = r\mathbf a \\ & \Rightarrow r=r_1r_2\end{align*} right?

Yep. (Nod)

 

[mathmari]

I see! (Nerd) We have that
\begin{equation*}\mathbf Z=\frac{1}{r_1r_2-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_1r_2\mathbf Z_1 + r_2\mathbf Z_2\right ) \Rightarrow \mathbf Z=\frac{1}{r_1r_2-1}\left ((r_1r_2- r_2)\mathbf Z_1 + (r_2-1)\mathbf Z_2\right ) \end{equation*}
If $Z_1\neq Z_2$ do we have that $Z\in Z_1Z_2$ because $Z$ is a linear combination of $Z_1$ and $Z_2$ ? So, we have the constraint $r_1r_2\neq 1$ because we have in the denominatot $r_1r_2-1$, right? (Wondering)

If we would have $r_1r_2=1$, we would have the following:
\begin{align*}&(S_2 \circ S_1)(\mathbf x) = \mathbf x\\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + \mathbf x - \mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow (1-r_2)\mathbf Z_2 -(1- r_2)\mathbf Z_1 =0 \\ & \Rightarrow (1-r_2)(\mathbf Z_2 -\mathbf Z_1) =0\end{align*}
What would we get from here?
That the center would be the same in that case or $r_2=1$ ? What would that mean?

 

[I like Serena]

We have that
\begin{equation*}\mathbf Z=\frac{1}{r_1r_2-1}\left (-\mathbf Z_2 - r_2\mathbf Z_1+ r_1r_2\mathbf Z_1 + r_2\mathbf Z_2\right ) \Rightarrow \mathbf Z=\frac{1}{r_1r_2-1}\left ((r_1r_2- r_2)\mathbf Z_1 + (r_2-1)\mathbf Z_2\right ) \end{equation*}
If $Z_1\neq Z_2$ do we have that $Z\in Z_1Z_2$ because $Z$ is a linear combination of $Z_1$ and $Z_2$ ?

I'm afraid that's not good enough. A linear combination can be any point in all of $\mathbb R^2$. (Worried)
We want to know if $Z$ is a point on the line segment $Z_1Z_2$.
A possible approach is to consider that the line segment $Z_1Z_2$ is given by $(1-t)Z_1 + tZ_2$ for $t\in [0,1]$.

So, we have the constraint $r_1r_2\neq 1$ because we have in the denominatot $r_1r_2-1$, right? (Wondering)

If we would have $r_1r_2=1$, we would have the following:
\begin{align*}&(S_2 \circ S_1)(\mathbf x) = \mathbf x\\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + r_2r_1\mathbf x - r_2r_1\mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow \mathbf Z_2 + r_2\mathbf Z_1 + \mathbf x - \mathbf Z_1 - r_2\mathbf Z_2=\mathbf x \\ & \Rightarrow (1-r_2)\mathbf Z_2 -(1- r_2)\mathbf Z_1 =0 \\ & \Rightarrow (1-r_2)(\mathbf Z_2 -\mathbf Z_1) =0\end{align*}
What would we get from here?
That the center would be the same in that case or $r_2=1$ ? What would that mean?

Indeed.
It would mean that our effective scaling factor would (or should) be 1, which is not an actual scaling.
More specifically, from your equation we can see that either $Z_1=Z_2$, or $r_2=1\Rightarrow r_1=1$.
That is, either we scale up and down around the same center, or we scale both times by a factor of 1.
In both cases we have an identity transformation.

What would happen if neither of those conditions would be satisfied? (Wondering)

 

[mathmari]

I'm afraid that's not good enough. A linear combination can be any point in all of $\mathbb R^2$. (Worried)
We want to know if $Z$ is a point on the line segment $Z_1Z_2$.
A possible approach is to consider that the line segment $Z_1Z_2$ is given by $(1-t)Z_1 + tZ_2$ for $t\in [0,1]$.

So, do we have to show that $\frac{r_1r_2- r_2}{r_1r_2-1}$ and $\frac{r_2-1}{r_1r_2-1}$ are in the interval $[0,1]$ ? (Wondering)

It would mean that our effective scaling factor would (or should) be 1, which is not an actual scaling.
More specifically, from your equation we can see that either $Z_1=Z_2$, or $r_2=1\Rightarrow r_1=1$.
That is, either we scale up and down around the same center, or we scale both times by a factor of 1.
In both cases we have an identity transformation.

What exactly does it mean that we scale up and down around the same center? Does it mean that once $r$ is positive and once negative? (Wondering)

What would happen if neither of those conditions would be satisfied? (Wondering)

Can it be that we don't take cases for $r_1r_2$ ? (Wondering)

 

[I like Serena]

So, do we have to show that $\frac{r_1r_2- r_2}{r_1r_2-1}$ and $\frac{r_2-1}{r_1r_2-1}$ are in the interval $[0,1]$ ? (Wondering)

Yes. And additionally that their sum is 1.

What exactly does it mean that we scale up and down around the same center? Does it mean that once $r$ is positive and once negative? (Wondering)

Scaling up means a factor $r>1$ (distance to the center becomes larger), while scaling down means a factor $0<r<1$ (distance becomes smaller).

Can it be that we don't take cases for $r_1r_2$ ? (Wondering)

What would happen if we first scale up around the origin with a factor of 2.
And then scale down around (2,0) with a factor of 1/2?
That is, when the centers are different and $r_2\ne 1$. (Wondering)

 

[mathmari]

Yes. And additionally that their sum is 1.

To show the inequalities do we have to take cases if $r_1r_2<1$ or $r_1r_2>1$ ? (Wondering)

Scaling up means a factor $r>1$ (distance to the center becomes larger), while scaling down means a factor $0<r<1 (distance becomes smaller)$.

And is this the only possibility that $Z_1=Z_2$ ? (Wondering)

 

[I like Serena]

To show the inequalities do we have to take cases if $r_1r_2<1$ or $r_1r_2>1$ ? (Wondering)

Now that I look a bit closer, can it be that $Z_1Z_2$ represents the line rather than the line segment?
In that case we only have to verify that the sum is 1.

And is this the only possibility that $Z_1=Z_2$ ? (Wondering)

What do you mean? (Wondering)

If $Z_1=Z_2$ and $r_1r_2=1$, then either $r_1$ scales up, and $r_2$ scales down by its inverse.
Or $r_1$ scales down, and $r_2$ scales up by its inverse.
Or both $r_1$ and $r_2$ are 1, meaning they don't scale at all, and their centers are irrelevant. But then they would not be scalings to begin with.

 

[mathmari]

Now that I look a bit closer, can it be that $Z_1Z_2$ represents the line rather than the line segment?
In that case we only have to verify that the sum is 1.

So, when we have that $x$ is a linear combination of two points so that the sum of the coefficients is equal to 1, do we conculde that $x$ is on the line between the two points? (Wondering)

If $Z_1=Z_2$ and $r_1r_2=1$, then either $r_1$ scales up, and $r_2$ scales down by its inverse.
Or $r_1$ scales down, and $r_2$ scales up by its inverse.
Or both $r_1$ and $r_2$ are 1, meaning they don't scale at all, and their centers are irrelevant. But then they would not be scalings to begin with.

I understand! (Nerd)

 

[I like Serena]

So, when we have that $x$ is a linear combination of two points so that the sum of the coefficients is equal to 1, do we conculde that $x$ is on the line between the two points?

Yep.
If $\mathbf a$ and $\mathbf b$ are 2 vectors on a line, then $\mathbf a + t(\mathbf b - \mathbf a) = (1-t)\mathbf a + t\mathbf b$ for $t\in\mathbb R$ is a vector representation of that line. (Nerd)

 

[mathmari]

Yep.
If $\mathbf a$ and $\mathbf b$ are 2 vectors on a line, then $\mathbf a + t(\mathbf b - \mathbf a) = (1-t)\mathbf a + t\mathbf b$ for $t\in\mathbb R$ is a vector representation of that line. (Nerd)

Ahh I see! Thanks a lot! (flower)