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Showing it is a vector space or not

  • Thread starter caljuice
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  • #1
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I'm still confused as to what a vector space is. Is it just some set of numbers that meet the conditions to be closed?

So my main problem:

V is the set of all polynomials of the form at2+bt+c, where a,b,c are real numbers wtih b=a + 1

(a1t2+b1t+c1) + (a2t2+b2t+c2) = (a1+a2)t2+(b1+b2)t+(c1+c2)

also r*(a1t2+b1t+c1)=ra1t2+rb1t+rc1

Is it closed?

So I guess I go through every condition and see if it holds off? I know standard polynomials satisify all the conditions. But the book says it is not a vector space becauase b1+b2= (a1+1)+(a2+1)= a1 + a2 + 2
so our polynomial is not in V and not closed. I don't get what b= a+1 changes. It still seems like every operation/condition is still satisfied. What am I missing? Any help is appreciated.
 

Answers and Replies

  • #2
lanedance
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tex is useful & pretty easy to use, see below (click for code)
[tex] (a_1 t^2+b_1 t+c_1) + (a_2 t^2+b_2 t+c_2) = (a_1+a_2)t^2+(b_1+b_2)t+(c_1+c_2) [/tex]

now how about the condition a = b+1, does the sum satisfy that condition?
 
  • #3
lanedance
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as for the general vector space question, there's a few more things you much show, such as:
- Associativity and commutativity of the addition operation
- distributivity of scalar multiplication
These are usually trival, but not always, also you need
- Existence of inverse and identity
- And closure as you mention, as well as a few others

for a full break down, see
http://en.wikipedia.org/wiki/Vector_space

the basic idea is you can prove a lot of general properties about a system that follows these axioms

the question you've been given is most likely to make you realise its not only n-tuples and vectors in R^n that can be vector spaces...
 
  • #4
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Thanks for the help. I'll use tex next time, definitely beats hitting sup and sub every few seconds.

So it is mainly vector that satisfy conditions. I don't know why but it feels very unintuitive, maybe it's the word "space," making it seem like it should be more of a concept instead of conditions.

For the problem are you saying:

[tex]
b_1+b_2 = (a_1 + a_2 +2)
[/tex]
and if you plug in a=b+1

[tex]
b_1+b_2 = b_1 + b_2 + 4
[/tex]

not equal so it is not closed? Which condition is that breaking? u+v = v+u?

But how did you get b=a+1? Was it from b=a+1, but would it not be a= b-1 instead?
 
Last edited:
  • #5
lanedance
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V is the set of all polynomials of the form at2+bt+c, where a,b,c are real numbers wtih b=a + 1
I could be missing something, but this is a condition to be part of your set
 
  • #6
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lol I don't think you're missing anything, I'm just confused on my goal. From above are you saying since it has to equal that [tex]at^2+bt+c [/tex] but the polynomial we get is in the form [tex] at^2+bt+c+2[/tex], that extra +2 violates the form?
 
  • #7
lanedance
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now how about the condition a = b+1, does the sum satisfy that condition?
apologies I miss quoted, this should be b = a + 1

as as you have shown the set is not closed under addition or multiplication eg. adding 2 polynomials in the set will results in a polynomial which is not in the set
 
  • #8
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No problem. Okay I think I'm getting a bit better now, I felt a small click. I think the part that confused me most was that it seemed like you could always reverse it back to the old form.

Damn satisfying. Thanks again for answering everything.
 
  • #9
lanedance
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no worries, always good to call it when something doesn't seem right, best way to learn
 
  • #10
lanedance
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So it is mainly vector that satisfy conditions. I don't know why but it feels very unintuitive, maybe it's the word "space," making it seem like it should be more of a concept instead of conditions.
on this, i generally think of it as (terms may not be exact but hopefully it helps):
- when you have a general collection of elements, it is called a set
- a set can then be endowed with an addition operation and scalar multiplication
- these generate other elements, which may or may not be in the original set
- when the set with addition & scalar mult satisifes all the vector space axioms, it is know as a vector space and you can know call its elements, vectors

- If you take a subset of the vector space the elements will automatically satisfy several of the axioms, however it won't necessarily have an identity, inverses or be closed under addition & scalar mult.

- However if the subset does satisfy have an identity, inverses and is closed under addition & scalar mult. it is know as a subspace and is also a vector space, related to its parent and of equal or lower dimension.
 

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