Linear Transformation problems

In summary: Hi Guys, I have these linear transformation problems which have caused me some trouble today.I hope You can help me.In summary, the conversation discusses three tasks related to linear transformations: showing that a linear translation cannot be done using regular coordinates, providing a linear transformation using homogeneous coordinates, and completing a given linear transformation using homogeneous coordinates. The participants also discuss the mapping of the zero vector and the center of a figure under a linear transformation. They also clarify that a nonzero translation is not a linear transformation and that a 2x3 matrix cannot be multiplied with a 2x1 matrix.
  • #1
Mathman23
254
0
Hi Guys,

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

I hope You can help me.


a) [tex](x,y) \rightarrow (x+3,y+5)[/tex] 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

[tex]\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][/tex]

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,

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

such that it can be completed by this linear transformation


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

using homogeneous coordinants.

In hints or ideer on how I do that?

Sincerely and best Regards,

Fred
 
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  • #2
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.
 
  • #3
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.
 
  • #4
Exactly!

...
 
  • #5
Hi again,

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

/Fred

Galileo said:
Exactly!

...
 
  • #6
Hi

I have re-thought question a.

Here is my surgestion.

The translation [tex](x,y) \rightarrow (x+3,y+5)[/tex] can be written as [tex](x,y) \rightarrow (x+3,y+5,1)[/tex] 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
[tex](x,y) \rightarrow (x+3,y+5) \notin \mathbb{R}^2[/tex] ?

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

/Mads
 
  • #7
How about using what Galileo suggested? What does this map (0,0) to?

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

And when you say "proving that the lines are not linear dependent in [tex]\mathbb{R}^2[/tex]", what lines are you talking about?
 
  • #8
HallsofIvy said:
How about using what Galileo suggested? What does this map (0,0) to?

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

And when you say "proving that the lines are not linear dependent in [tex]\mathbb{R}^2[/tex]", 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 [tex](x,y,1) \rightarrow (x+3, y+5,1)\in \mathbb{R}^3[/tex]. Therefore I thought I that need to prove something about lack of linear dependents in [tex]R^2[/tex].

I hints to what I then can do ??

Best Regards

/Fred
 
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  • #9
Mathman23 said:
[tex](x,y) \rightarrow (x+3,y+5)[/tex]
Mathman23 said:
First of all (0,0) maps itself to (0,0)

Does it?..
 
  • #10
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 [tex](x,y)\rightarrow (x+3,y+5)[/tex] 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.
 
Last edited by a moderator:
  • #11
I finally got it.

The translation [tex](x,y) \rightarrow (x+3, y+5)[/tex] can be written in matrix form:

[tex]\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][/tex]

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

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

[tex]\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][/tex]

Then the mapping can be done since:

[tex]\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][/tex]

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

/Fred
 
  • #12
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?
 
  • #13
I will write again what I'm supposed to do according to my assigment.

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

Anyhow if I write the translation as:

[tex]\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][/tex]

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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  • #14
Mathman23 said:
[tex]\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][/tex]

Except that that matrix multiplication is incorrect!

[tex]\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][/tex]


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.
 
Last edited by a moderator:

1. What is a linear transformation?

A linear transformation is a mathematical operation that maps one vector space to another while preserving the basic structure of the original space. In other words, it is a function that takes in a vector and outputs another vector, with the property that the sum of two vectors is mapped to the sum of their transformed versions and a scalar multiple of a vector is mapped to the scalar multiple of its transformed version.

2. How is a linear transformation represented mathematically?

A linear transformation is represented by a matrix, which is a rectangular array of numbers. The matrix is used to transform a vector by multiplying it with the matrix, resulting in a new transformed vector. The transformation is determined by the values in the matrix, which can be modified to achieve different transformations.

3. What are some real-life applications of linear transformations?

Linear transformations have various applications in fields such as physics, engineering, economics, and computer graphics. In physics, they are used to describe the movement of objects in space and time. In engineering, they are used to model and analyze systems such as electrical circuits and control systems. In economics, they are used to model the relationships between different variables. In computer graphics, they are used to manipulate images and create 3D animations.

4. How is a linear transformation different from other types of transformations?

A linear transformation preserves the properties of vector addition and scalar multiplication, while other types of transformations may not. For example, a linear transformation of a line segment will result in another line segment, while a nonlinear transformation may result in a curve. Linear transformations also have the property of being reversible, meaning the original vector can be obtained from the transformed vector by multiplying it with the inverse of the transformation matrix.

5. How can linear transformations be used to solve real-world problems?

Linear transformations can be used to solve real-world problems by transforming the problem into a system of linear equations, which can then be solved using various methods such as Gaussian elimination or matrix inversion. This can be applied to a wide range of problems, from optimizing production processes in manufacturing to predicting stock market trends in finance. Linear transformations provide a powerful tool for analyzing and solving complex problems in various fields.

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