Showing it is a vector space or not

[caljuice]

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.

 

[lanedance]

tex is useful & pretty easy to use, see below (click for code)
$$(a_1 t^2+b_1 t+c_1) + (a_2 t^2+b_2 t+c_2) = (a_1+a_2)t^2+(b_1+b_2)t+(c_1+c_2)$$

now how about the condition a = b+1, does the sum satisfy that condition?

 

[lanedance]

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 realize its not only n-tuples and vectors in R^n that can be vector spaces...

 

[caljuice]

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:

$$b_1+b_2 = (a_1 + a_2 +2)$$
and if you plug in a=b+1

$$b_1+b_2 = b_1 + b_2 + 4$$

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?

 

[lanedance]

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

 

[caljuice]

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 $$at^2+bt+c$$ but the polynomial we get is in the form $$at^2+bt+c+2$$, that extra +2 violates the form?

 

[lanedance]

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

 

[caljuice]

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.

 

[lanedance]

no worries, always good to call it when something doesn't seem right, best way to learn

 

[lanedance]

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.