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Subspaces of P spaces and C[a,b] spaces

  1. Dec 5, 2006 #1
    1. The problem statement, all variables and given/known data
    Determine whether the set of polynomials of degree 3 form a subspace of P(4)

    2. Relevant equations

    [tex]P(4) = c_3 x^3 + c_2 x^2 + c_1 x + c_0[/tex]

    3. The attempt at a solution

    [tex]\alpha P(4_1) = \alpha c_3 x^3 + \alpha c_2 x^2 + \alpha c_1 x + \alpha c_0[/tex]

    This just scales the coefficients, right? It would still be a polynomial of degree 3 I think...

    For addition, wouldn't you just be adding two polynomials? So wouldn't you just obtain another degree 3 polynomial?

    Thanks, Sean.
    Last edited: Dec 5, 2006
  2. jcsd
  3. Dec 5, 2006 #2
    Note that [tex] \Re^{2} [/tex] is not a subspace of [tex] \Re^{3} [/tex]
  4. Dec 5, 2006 #3
    I don't see how that helps. I suspect I will, but, why is what you said true? Why don't the set of vectors (a,b,0) doesn't span a subspace in R3?
  5. Dec 5, 2006 #4
    Look at vector addition and scalar multiplication in [tex] \Re^{2} [/tex] and [tex] \Re^{3} [/tex]. What are the conditions to show that if [tex] W [/tex] is a nonempty subset of [tex] V [/tex] if [tex] V [/tex] is a vector space over [tex] \Re [/tex] then [tex] W [/tex] is a subspace of [tex] V [/tex]?
    Last edited: Dec 5, 2006
  6. Dec 5, 2006 #5


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    Imagine two polynomials of the third degree with leading coefficients a and -a. What happens is you add them?
  7. Dec 5, 2006 #6

    Hmm. Do you mean W is nonempty, the sum of two vectors in W must lie in W, and W times a scalar must lie in W?

    So take two vectors in R2, say (x1,x2),(y1,y2). Their sum equals (x1+y1,x2+y2), which lies in R2, and (cx1,cx2) also lies in R2.
  8. Dec 5, 2006 #7

    matt grime

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    A polynomial of degree 3 is a cubic, ax^3+bx^2+cx+d where a must be non-zero. Thus the space fails to be a vector subspace of the polys for lots of reasons. Note that the space of polynomials of degree at most 3 *is* a subspace, but this is strictly different from the space of polys of degree 3.
  9. Dec 5, 2006 #8
    thanks : )
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