# Showing it is a vector space or not

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.

Related Calculus and Beyond Homework Help News on Phys.org
lanedance
Homework Helper
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
Homework Helper
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...

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?

Last edited:
lanedance
Homework Helper
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

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
Homework Helper
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

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

lanedance
Homework Helper
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.