## Could a saint please guide me work through my research on AFFINE SETS AND MAPPINGS

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:

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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

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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
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

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

## Could a saint please guide me work through my research on AFFINE SETS AND MAPPINGS

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?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by micromass 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

Blog Entries: 8
Recognitions:
Gold Member
Staff Emeritus
 Quote by tamintl 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.

 Quote by micromass 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?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by micromass 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.

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

Thanks micro
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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).

 Quote by micromass 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?
 bump?
 Recognitions: Science Advisor is it permissible to set λ = 1/2?

 Quote by Deveno 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)

Is that sufficient?

Thanks

Recognitions:
 Quote by tamintl 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.....

 Quote by Deveno 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