1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: LINEAR_ALGEBRA: What is a subspace of V = F(R, R), the vector space of all real funcs
  1. Vector Space R^2 (Replies: 74)

Loading...