Understand Affine Subsets & Mappings: Research Project for Undergrads

tamintl
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This research project is to help me (I'm an undergraduate) get my head around this topic. It is concerned with affine subsets of a vector space and the mappings between them. As an
application, the construction of certain fractal sets in the plane is considered. It would be considered pretty basic to a seasoned maths student.

I am wanting to learn this so I will be sticking around. I will not just leave. I want to commit to this. Thanks

There are two parts: A and B

If someone is willing to help, I will post each topic AFTER I have fully understood the previous topic. This way it will run in a logical order.

PART A:

----------------------------------------------------------------------------------
Throughout Part A, V will be a real vector space and, for a non-empty subset S of V and
a ε V , the set {x+a: x ε S} will be denoted by S + a

----------------------------------------------------------------------------------

TOPIC 1: Definition of Affine Subset:

An affine subset of V is a non-empty subset M of V with the property that λx+(1-λ)y ε M whenever x,y ε M and λ ε ℝ

To illustrate this concept, show that:

M = { x = (x1,...x4) ε ℝ4 : 2x1-x2+x3 = 1 and x1+4x3-2x4 = 3}

is an affine subset of ℝ4.

I'm not so sure where to start. Opinions welcome

Regards
Tam
 
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Take x and y in M. You must show that \lambda x+ (1-\lambda) y\in M. Call this number z for convenience.

To show that z is in M, you need to show that

2z_1-z_2+z_3=1~\text{and}~z_1+4z_2-2z_4=3

You know that z_i=\lambda x_i + (1-\lambda) y_i so substitute that in.
 


micromass said:
Take x and y in M. You must show that \lambda x+ (1-\lambda) y\in M. Call this number z for convenience.

To show that z is in M, you need to show that

2z_1-z_2+z_3=1~\text{and}~z_1+4z_2-2z_4=3

You know that z_i=\lambda x_i + (1-\lambda) y_i so substitute that in.

Okay great.

So subbing in z we get:

LHS:

2(λx1+(1-λ)y1) - (λx2+(1-λ)y2) + (λx3+(1-λ)y3)

Now taking:

λ(2x1-x2+x3) We know that the bold part = 1

(1-λ)(2y1-y2+y3) We again know that the bold part = 1

so we have λ + (1-λ) = 1 = RHS

AND NOW DO THE SAME WITH THE SECOND PART X1+4X3-2X4 = 3... IVE DONE THAT IN MY OWN TIME.
-------------------------------------------------------------------------------------------------------

So I think I've grasped that. I will look at the next topic and report back when I've had a go. Thanks Micro
 


Now Topic A2

Let M be an affine subset of V.

QUESTION: Prove that M+a is affine for every a ε V and that, if 0 ε M, then M is a subspace

So my attempt:

Proof: x,y is in M+a

take: x = m1+a and y = m2+a for some m1,m2M

Therefore, λ(m1+a) + (1-λ)(m2+a)

Now rearranging gives:

(i) λm1 + (1-λ)m2 which must be in M by definition.

(ii) λa + (1-λ)a
=a(λ+1-λ)
=a

Hence, λm1 + (1-λ)m2 + a is in M+a. So M+a is affine.


I'm unsure of what to do with the zero part of the question?
 


So assume that 0 is in M. You must prove that it is a subspace. So you must check the axioms of being a subspace.
 


micromass said:
So assume that 0 is in M. You must prove that it is a subspace. So you must check the axioms of being a subspace.

OKay, using the definition: Let M be a subspace of vecotr space V. Then M is a subspace of V IFF

i) 0 ε M
ii) x+y ε M for all x,y ε M
iii) λx ε M for all x ε M

(i) holds since we are assuming 0 ε M
(ii) holds since we showed this in the last part of the question
(iii) holds since in the last part of the question λx ε M

Is this enough? I'm unsure of (iii)

Regards
Tam
 


tamintl said:
OKay, using the definition: Let M be a subspace of vecotr space V. Then M is a subspace of V IFF

i) 0 ε M
ii) x+y ε M for all x,y ε M
iii) λx ε M for all x ε M

(i) holds since we are assuming 0 ε M
(ii) holds since we showed this in the last part of the question
(iii) holds since in the last part of the question λx ε M

Is this enough? I'm unsure of (iii)

Regards
Tam

Could you explain (ii) and (iii)?? You have to use the assumption that 0 is in M for all of these questions.
Begin by showing (iii). Apply the definition of M affine on x and 0.
 


micromass said:
Could you explain (ii) and (iii)?? You have to use the assumption that 0 is in M for all of these questions.
Begin by showing (iii). Apply the definition of M affine on x and 0.

Okay, on x,

λx + (1-λ)x will be in M by definition

on 0,

λ(0) + (1-λ)(0) = 0 which is in M since we are assuming 0 ε M

Have I understood you?
 


No. You have to show that for any x and for any λ, that λx is in M.

You know that M is affine, so you know that for any x and for any y, we have that λx+(1-λ)y is in M.
Now choose a special value of y.
 
  • #10
micromass said:
No. You have to show that for any x and for any λ, that λx is in M.

You know that M is affine, so you know that for any x and for any y, we have that λx+(1-λ)y is in M.
Now choose a special value of y.

Oh okay. If we take y=0 (using the condition 0€M)

Then we get (lambda)x + (1-lambda)(0) which is just (lambda)x

So we know for any x and lambda that it will be in M. So that is iii done.

What about ii

Ps: I'm on my phone so sorry for weak notation.

Thanks micro
 
  • #11


For (ii), you need to prove that if x and y are in M, then x+y is in M.

You know that for each r and s in M that

\lambda r+(1-\lambda)s\in M

Now choose the right r and s such that we can conclude that x+y is in M. Use (iii).
 
  • #12
micromass said:
For (ii), you need to prove that if x and y are in M, then x+y is in M.

You know that for each r and s in M that

\lambda r+(1-\lambda)s\in M

Now choose the right r and s such that we can conclude that x+y is in M. Use (iii).

Take r=x and s=0 so since we know (lambda)x is in M, x+0 is in M.

Or could we use the M+a proof?
 
  • #13


bump?
 
  • #14


is it permissible to set λ = 1/2?
 
  • #15


Deveno said:
is it permissible to set λ = 1/2?

Yes, I think...

Edit: taking λ = 1/2

f(x+y) = f(1/2(2x)) + f(1/2(2y))

= 1/2 [ f(2x) + f(2y) ]

taking 2 out gives:

= f(x) + f(y)

Hence closed under addition

Is that sufficient?

Thanks
 
Last edited:
  • #16


tamintl said:
Yes, I think...

Edit: taking λ = 1/2

f(x+y) = f(1/2(2x)) + f(1/2(2y))

= 1/2 [ f(2x) + f(2y) ]

taking 2 out gives:

= f(x) + f(y)

Hence closed under addition

Is that sufficient?

Thanks

where does "f" come from?

my reasoning goes like this: 1/2 and 1/2 sum to 1, so (1/2)x + (1/2)y is an affine combination, that is: (x+y)/2 is in M.

now, use part (iii) to conclude that...
 
  • #17


Deveno said:
where does "f" come from?

my reasoning goes like this: 1/2 and 1/2 sum to 1, so (1/2)x + (1/2)y is an affine combination, that is: (x+y)/2 is in M.

now, use part (iii) to conclude that...

Yeah forget about the 'f's.. yeah that makes sense.

Deveno, if I sent you the question sheet it may be easier for both you and I to understand. Of course, only if you are happy to help. Would that be okay? The reason I ask is that it is hard for me to get my points across since I don't know latex.

Regards
 
  • #18


Hi

I am doing a similar assignment and have been finding it difficult to find relevant material to the questions. However I have found the guidance on this thread very useful so far and was hoping you could send me any further information on this assignment as I think it would be a great help.

Thanks
 
  • #19


I think this is the same assignment as the one I'm doing - I'm basically in the same position as matt90, and have spent hours doing research on this with no luck. I would also really appreciate any additional help you have to offer.
 
  • #20


Thread closed temporarily for Moderation...
 

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