
vector space prove!
prove wether the follow is a vector space or not.
could you please show me how to prove the following question? Intuituively, i think it is a vector space, but i don't know how to prove the 10 axioms step by step. Thank you very much.
Q1) S = {[x p(x)]  x is an element of R}, Where p(x) is any polynomial, vector addition is defined as
[x p(x)] + [y p(y)] = [x+y p(x+y)]
and scalar multiplication defined as
$\displaystyle alpha$ * [x p(x)] = [$\displaystyle alpha$*x p($\displaystyle alpha$*x)]

Just to verify, vector addition is defined as: $\displaystyle xp(x) + yp(y) = (x+y)p(x+y)$?
Let's go through some of the axioms and see if you can do the rest.
1. $\displaystyle \bold{u}, \bold{v} \in S \: \Rightarrow \: \bold{u} + \bold{v} \in S$
Let's choose two arbitrary vectors, ap(a) and bp(b). Then, $\displaystyle ap(a) + bp(b) = (a+b)p(a+b) \in S$ as it is in the form xp(x) (x = a + b) in this case.
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2. Given scalar k, $\displaystyle \bold{u} \in S$, then $\displaystyle k\bold{u} \in S$ as well.
Again, pick an arbitrary vector ap(a). Then $\displaystyle k \cdot ap(a) = kap(ka) \in S$ (imagine x = ka).
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3. $\displaystyle \bold{u} + \bold{v} = \bold{v} + \bold{u}$
Pick two arbitrary vectors ap(a) and bp(b). Then:
$\displaystyle ap(a) + bp(b) = (a+b)p(a+b) = (b+a)p(b+a) = bp(b) + ap(a)$
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4. $\displaystyle (c+k)\bold{u} = c\bold{u} + k\bold{u}$
Pick arbitrary vector ap(a). Then, $\displaystyle (c+k)ap(a) = (c+k)a p\left[(c+k)a\right] \in S$ (imagine x = (c+k)a)
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5. There exists a zero vector, $\displaystyle \bold{0}$, such that $\displaystyle \bold{u} + \bold{0} = \bold{0} + \bold{u} = \bold{u}$
Again, choose arbitrary vector ap(a). Then:
$\displaystyle ap(a) + 0p(0) = (a+0)p(a+0) = ap(a)$ (just going by definition)
$\displaystyle 0p(0) + ap(a) = ...$
etc. etc.
As you can see, verifying some of these are pretty trivial. Now, it looks like the set is indeed a vector space but I may be wrong. Go through all the axioms and see if they are all satisfied.