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

by tamintl
Tags: affine, guide, mappings, research, saint, sets, work
 P: 74 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
 Emeritus Sci Advisor Thanks PF Gold P: 15,876 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.
P: 74
 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

P: 74

## 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?
 Emeritus Sci Advisor Thanks PF Gold P: 15,876 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.
P: 74
 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
Emeritus
Thanks
PF Gold
P: 15,876
 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.
P: 74
 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?
 Emeritus Sci Advisor Thanks PF Gold P: 15,876 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.
P: 74
 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
 Emeritus Sci Advisor Thanks PF Gold P: 15,876 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).
P: 74
 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?
 P: 74 bump?
 Sci Advisor P: 906 is it permissible to set λ = 1/2?
P: 74
 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
P: 906
 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.....
P: 74
 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
 P: 1 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

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