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Prove that W is a subspace of P_4(t)

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Let W = {p(t) ∈ P4(t): p(0)=0 }. Prove that W is a subspace of P4(t)


    2. Relevant equations
    none


    3. The attempt at a solution
    I know three things have to be true in order to be a subspace:
    1. zero vector must exist as an element
    2. if u and v are elements, u+v must be an element
    3. if u is an element cu is an element

    I'm a bit confused. I've done some searching on this forum + my textbook, and I haven't seen any questions in this particular format. Any help would be greatly appreciated.
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2

    micromass

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    Let's start with this:

    This is a bit weirdly phrased. The zero vector always exists in the larger space. The issue is that it must be an element of the subspace.

    The zero vector is here the zero polynomial. Thus p(x)=0 for all x. Does that p lie in our subspace? What do we need to check for that?
     
  4. Apr 23, 2012 #3
    I edited my original post for the first condition. I'm not sure how to check for p(x)=0 to lie in our subspace. I think I may be a bit confused by P_4(x). What does the P_4 mean?
     
  5. Apr 23, 2012 #4

    micromass

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    Likely, it is the vector space of polynomial function of degree ≤4.

    So examples of things in [itex]P_4[/itex] are

    [itex]p(x)=0,~p(x)=x+1,~p(x)=4x^4+2x+1[/itex]

    An example of something not in there is

    [itex]p(x)=x^6+x^5+1[/itex] (since the degree is 6 and thus not ≤4)

    or

    [itex]p(x)=\sin(x)[/itex] (since it is not a polynomial)
     
  6. Apr 23, 2012 #5
    so P(x)=0 lies in our subspace of P_4 because it is an example element in P4
     
  7. Apr 23, 2012 #6

    micromass

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    Not so quick!!

    P(x) certainly lies in P4, no problem with that.
    But now you have a subspace of P4.

    That is, you only look at elements of P4 which satisfy p(0)=0.

    So, for example

    [tex]p(x)=x^2+x[/tex]

    lies in P4 AND it satisfies p(0)=02+0=0. So it is in our subspace.

    But

    [tex]p(x)=x^2+x+1[/tex]

    does lie in P4, but it has p(0)=1. So it is not in our subspace.
     
  8. Apr 23, 2012 #7
    Alright! I get that. However, what I don't get is the question. I see you listed a few polynomials as examples that satisfy vector 0 being an element of the subspace, but what is the question asking for? There is no polynomial or equation associated with the question.
     
  9. Apr 25, 2012 #8
    Ok, for anyone who has come across my problem via search, there are a few things that you need to know.

    Span of vectors = subspace

    p(t) = a0 + a1t1..+ a4t4

    With p(0)=0 a0=0

    W={a1t...a4t4
    span = { t, t^2, t^3, t^4 }
     
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