
vector spaces
Let v1 = span{(1 0 2)} and v2 = span{(0 1 1)}.
Observe that they are both vector spaces. A new set of vectors, S, is constructed by taking any vecotr from v1 and any vector from v2 and these 2 vectors together.
(a) explain why S will also be a vector space.
help please.

Check that S satisfies all the axioms of a vector space. (You know what the definition of a vector space is, don’t you?)
Hint: Any vector in S will be of the form $\displaystyle (a,b,2a+b)$.

yes.. i know.. woould it be the definiton...... u + v is in V.
V is closed in addition?? therefore S is also a vector space?

I’m afraid you’ll have to prove much more than that (though you’re on the right track).
What are all of the axioms that define a vector space? Go through all of them, and check that the set S satisfies all of them.

no sorry...we havent really covered all of them in our course just yet..so i wouldnt know how to check

You should at least have covered the definition of a vector space – otherwise there would be no point in asking the question you asked! (Rofl)
Very well, here is an elementary definition of a vector space. Or more precisely, the definition of a real vector space.
A real vector space consists of a nonempty set V of elements called vectors which can be added together and multiplied by real numbers (called scalars) such that the following properties hold: For any vectors u and v and any scalar (real number) α, u+v and αu are in V.
 For any vectors u, v and w, we have u+v = v+u and (u+v)+w = u+(v+w).
 There is a vector called 0 such that if u is any vector, u+0 = u.
 Given any vector u, there is a vector v such that u+v = 0. (This vector v is usually denoted −u.)
 For any vectors u and v and any scalars α and β, we have α(u+v) = αu+αv, (α+β)u = αu+βu and (αβ)u = α(βu).
 1u = u for any vector u.
Now your job is to prove that the set $\displaystyle S=\{(a,b,2a+b):a,b\in\mathbb{R}\}$ is a vector space. I’ll do the first one for you. Let $\displaystyle \mathbf{u}=(a_1,b_1,2a_1+b_1)$ and $\displaystyle \mathbf{v}=(a_2,b_2,2a_2+b_2)$ be vectors in S and $\displaystyle \alpha$ be real. Then
$\displaystyle \mathbf{u}+\mathbf{v}=(a_1,b_1,2a_1+b_1)+(a_2,b_2, 2a_2+b_2)=(a_1+a_2,b_1+b_2,2(a_1+a_2)+(b_1+b_2))$
is in S and
$\displaystyle \alpha\mathbf{u}=\alpha(a_1,b_1,2a_1+b_1)=(\alpha a_1,\alpha b_1,2\alpha a_1+\alpha b_1)$
is also in S.
Now you do the rest.

hmmm.. nope i think that is too indepth for my course. i understand what you are trying to do though there. you are using pretty much any values...(a, b..etc) and using them as vectors to prove that the definition holds, and hence proving that S is also a vector space??
am i correcT?