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Homework Help: LINEAR_ALGEBRA: What is a subspace of V = F(R, R), the vector space of all real funcs

  1. Sep 25, 2006 #1
    V = F(R, R), the vector space of all real valued functions f(x) of a real variable x. Which are subspaces of V?

    (A) {f | f(0) = 0}

    (B) {f | f(0) = 1}

    (C) {f | f(0) = f(1)}

    (D) [itex]C^0(R)[/itex] = {f | f is continous}

    (E) [itex]C^1(R)[/itex] = {f | f is differentiable and f' is continous}

    (F) P = {f | f is a polynomial}

    (G) [itex]P_d\,\,\,\,=\,\,\,\,{f\,\in\,P\,|\,deg(f)\,\le\,d}[/itex]

    (H) [itex]{f\,\in\,C^1(R)\,|\,f'\,=\,f}[/itex]

    I have no idea what the last five of these instances mean. Case (C) is not a subspace because it only satifies the first rule that the set {0} be in the space before it can be considered a subspace, right?

    Please help, I don't understand the terminology of the last five examples!
     
  2. jcsd
  3. Sep 25, 2006 #2

    AKG

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    You don't know what "polynomial", "differentiable", and "continuous" mean? If f is a polynomial, you don't know what deg(f), the degree of f, is?
     
  4. Sep 25, 2006 #3
    Yes, I know those terms, but not the [itex]C^0,\,C^1[/itex] terms or the P or [itex]P_d[/itex] terms.
     
  5. Sep 25, 2006 #4

    quasar987

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    Well that's why they are defined for you

    For instance, [itex]C^1(R)[/itex] denoted the set of all continuously differentiable (meaning f' is continuous) functions.
     
  6. Sep 25, 2006 #5
    (G) would be polynomials that look like

    [itex]
    f(x) = \sum_{i=0...d} a_i x^i
    [/itex]
     
  7. Sep 26, 2006 #6
    Both (A) and (B) are subspaces of V, but (C) is not because each element must be unique, right? f(0) = f(1) means that it is not unique and not 1-1. But, how would I prove that (A) and (B) are subspaces of V?
     
  8. Sep 26, 2006 #7
    If f(0) = f(1), it's still a function (think of a horizontal line), and there's no requirement for it to be 1-1.

    For (B), what is the 0 element of the set?
     
  9. Sep 26, 2006 #8
    There will be no 0 element because it always equals 1. So (B) is NOT a subset. (C) could be a horizontal line. But how would I prove these items? what would an arbitrary f(x) function look like?

    [tex]f(x)\,=\,a_1\,x^{n\,+\,2}\,+a_2\,x^{n\,+\,1}\,+\,...\,+a_n\,x^n[/tex]
     
  10. Sep 26, 2006 #9
    It's just like the matrix case. You have functions in the set [itex]f[/itex] and [itex]g[/itex] and a real number [itex]a[/itex]. Are [itex]af[/itex], [itex]f + g[/itex],and the zero function in the set?
     
  11. Sep 26, 2006 #10
    0 function would be f = 0 or [itex]f = 0 x^2 + 0 x + 0[/itex]?

    How would I write it out (prove) for case (A)?

    {f | f(0) = 0}

    1.) 0 [itex]\in[/itex] V.

    2.) a f(0) = 0 [itex]\in[/itex] V.

    3.) f(0) + g(0) = 0 + 0 = 0 [itex]\in[/itex] V.
     
    Last edited: Sep 26, 2006
  12. Sep 26, 2006 #11
    Those are all pretty clear, right? You just need a sentence after each of those computations saying "so <function> is in the set <whatever>." You may need a sentence explaining that the sum of a C^0 functions is also C^0. The same for C^1 functions. And similarly for a real number times a function (should be able to look this up in your calculus text.) For the polynomials you show directly by combining terms that the new function is a polynomial of whatever type is required.
     
    Last edited: Sep 26, 2006
  13. Sep 26, 2006 #12
    What about case (B)?

    It says that a function with zero plugged-in as the variable is equal to one. But this does not mean that the function cannot equal zero at some other plugged-in constant, right? So rule 1 checks out.

    However, rule 3 (additive) is not satisfied becuase f(0) + g(0) = (1) + (1) = 2. Which is not 1, obviously.

    Does that sound right?
     
  14. Sep 26, 2006 #13

    radou

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    Yes, it does. Further on, for scalar multiplication, i.e. a real scalar [tex]a \neq 1[/tex], you'd have [tex]a\cdot f(0) = a[/tex].
     
  15. Sep 26, 2006 #14
    So, is every case except (B) and (H) a subspace of V?
     
  16. Sep 26, 2006 #15

    HallsofIvy

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    No, it doesn't. In function spaces with f+ g defined by (f+g)(x)= f(x)+ g(x) (the usual definition), the 0 vector is the function f(x)= 0 for all x. Is that function in this set?

    Yes, that is true.
     
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