# Linear Transformation problems (1 Viewer)

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#### Mathman23

Hi Guys,

I have these linear transformation problems which have caused me some trouble today.

I hope You can help me.

a) $$(x,y) \rightarrow (x+3,y+5)$$ is called a linear translation according to my Linear Algebra textbook.

I'm tasked with showing that the above can't be done as linear transformation by using regular coordinates. Secondly I'm tasked with providing a linear transformation using homogeneous coordinants, which does the translation.

Any hits idears on how I do that??

I know what the translation can be written as

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{c} x \\ y \\ 1 \end{array} \right] = \left[ \begin{array}{c} x + 3 \\ y+5 \\ 1 \end{array} \right]$$

But does that help me in any way proving the above ??

b) I'm tasked provinding a linear transformation which rotates the following rectangle including edges 90 degress clockwise around the center of the figure.

I'm provided the following coordinants for the figure.

A(3,1) B(5/2, 2), C(9/2) and D(5,2).

In hits ideers on how I do this ?

Finally C)

Show that the following linear transformation t(x) is given in regular coordinants,

$$T(x) = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] \cdot x$$

such that it can be completed by this linear transformation

$$T(x) = \left[ \begin{array}{ccc} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & 0 \\ 0 & 0 & 1 \end{array} \right] \cdot x$$

using homogeneous coordinants.

In hints or ideer on how I do that?

Sincerely and best Regards,

Fred

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#### Galileo

Homework Helper
For a) Where does the zero vector get mapped to under a lienar transformation?

b) The center of the figure is not the origin. Use ideas from the previous exercise to answer this one.

#### Mathman23

Hi

In a) Doesn't the zero-vector map itself onto itself??

/Fred

Galileo said:
For a) Where does the zero vector get mapped to under a lienar transformation?

b) The center of the figure is not the origin. Use ideas from the previous exercise to answer this one.

Homework Helper
Exactly!

...

#### Mathman23

Hi again,

But how does this aide me in answering question a) ??

/Fred

Galileo said:
Exactly!

...

#### Mathman23

Hi

I have re-thought question a.

Here is my surgestion.

The translation $$(x,y) \rightarrow (x+3,y+5)$$ can be written as $$(x,y) \rightarrow (x+3,y+5,1)$$ in homogeneous coordinants.

As mentioned in the intial post my task is to prove that the translation can't be done using the regular coordinants which implies that
$$(x,y) \rightarrow (x+3,y+5) \notin \mathbb{R}^2$$ ???

Can I do this by proving that the lines are not linear dependent in $$\mathbb{R}^2$$ ??

#### HallsofIvy

How about using what Galileo suggested? What does this map (0,0) to?

I don't know what you mean by "$$(x,y) \rightarrow (x+3,y+5) \notin \mathbb{R}^2$$". (x+3,y+ 5) certainly is in $$\mathbb{R}^2$$!

And when you say "proving that the lines are not linear dependent in $$\mathbb{R}^2$$", what lines are you talking about?

#### Mathman23

HallsofIvy said:
How about using what Galileo suggested? What does this map (0,0) to?

I don't know what you mean by "$$(x,y) \rightarrow (x+3,y+5) \notin \mathbb{R}^2$$". (x+3,y+ 5) certainly is in $$\mathbb{R}^2$$!

And when you say "proving that the lines are not linear dependent in $$\mathbb{R}^2$$", what lines are you talking about?
Hi

First of all (0,0) maps itself to (0,0) so I guess that means the translation to (3,5,1) ??

What I'm tasked with doing is proving that the mapping can't be done using regular coordinants, but with homogeneous coordinants. Thereby implying that the translation can be written as $$(x,y,1) \rightarrow (x+3, y+5,1)\in \mathbb{R}^3$$. Therefore I thought I that need to prove something about lack of linear dependents in $$R^2$$.

I hints to what I then can do ??

Best Regards

/Fred

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Homework Helper

#### HallsofIvy

Mathman23 said:
Hi

First of all (0,0) maps itself to (0,0) so I guess that means the translation to (3,5,1) ??
We must be talking about completely different things! (0,0) doesn't map "itself" into anything! I thought we were talking about the transformation $$(x,y)\rightarrow (x+3,y+5)$$ which obviously maps (0,0) into (0+3,0+5)= (3, 5). Since you just said that a linear transformation must map (0,0) into (0,0) this is clearly not a linear transformation.

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#### Mathman23

I finally got it.

The translation $$(x,y) \rightarrow (x+3, y+5)$$ can be written in matrix form:

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{cc} x + 3 \\ y+5 \end{array} \right]$$

If $$\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right]$$ then obviously the mapping can't take place using regular coordinants.

However doing the translastion using homogeneous coordinants (x,y,1):

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{c} x \\ y \\ 1 \end{array} \right] = \left[ \begin{array}{cc} x + 3 \\ y+5 \\ 1 \end{array} \right]$$

Then the mapping can be done since:

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{cc} 3 \\ 5 \\ 1 \end{array} \right]$$

I guess that this answers my question a) in my initial post ???

/Fred

#### Galileo

Homework Helper
You can't give that function a matrix respresentation since a nonzero translation is NOT a linear transformation.

I don't get your matrix multiplication either. You cannot multiply a 2x3 matrix with a 2x1 matrix. Are we talking about the same things here?

#### Mathman23

I will write again what I'm supposed to do according to my assigment.

a) The translation $$(x,y) \rightarrow (x+3, y+5)$$ can't be done as a linear mapping using regular coordinants, but only with homogeneous coordinants.

Anyhow if I write the translation as:

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \end{array} \right]\left[ \begin{array}{c} x \\ y \\ 0 \end{array} \right] = \left[ \begin{array}{cc} x + 3 \\ y+5 \end{array} \right]$$

next if I set (x,y) = (0,0) then as You wrote in Your post then the linear transformation doesn't take place if I use regular coordinants.

Doesn't that prove the point put forward in my assignment ??

/Fred

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#### HallsofIvy

Mathman23 said:
$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \end{array} \right]\left[ \begin{array}{c} x \\ y \\ 0 \end{array} \right] = \left[ \begin{array}{cc} x + 3 \\ y+5 \end{array} \right]$$
Except that that matrix multiplication is incorrect!

$$\left[ \begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 5 \end{array} \right]\left[ \begin{array}{c} x \\ y \\ 0 \end{array} \right] = \left[ \begin{array}{cc} x\\ y\\0 \end{array} \right]$$

The whole point is that, since (0,0)-> (3, 5), not (0, 0), as Galileo pointed out in his first response, this not a linear transformation and can not be written as a matrix multiplication!

One important point about "homogeneous coordinates" is that translation (that's what this transformation is) is a linear transformation in homogeneous coordinates.

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