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Thread: vector space prove!

  1. #1
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    Question 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)]
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  2. #2
    o_O
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    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.

    __________________________________________________ ________

    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).

    __________________________________________________ ________

    3. $\displaystyle \bold{u} + \bold{v} = \bold{v} + \bold{u}$

    Pick two arbitrary vectors ap(a) and bp(b). Then:
    $\displaystyle ap(a) + bp(b) = (a+b)p(a+b) = (b+a)p(b+a) = bp(b) + ap(a)$

    __________________________________________________ ________

    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)

    __________________________________________________ ________

    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.
    Last edited by o_O; Aug 18th 2008 at 11:04 PM.
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