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Linear Algebra vector spaces

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let V be the set of all complex-valued functions, f, on the real line such that
    f(-t)= f(t) with a bar over it, which denotes complex conjugation.
    Show that V, with the operations
    (f+g)(t)= f(t)+g(t)
    is a vector space over the field of real numbers.

    2. Relevant equations

    3. The attempt at a solution
    I don't know what complex conjugation means, so I have no idea where to start with this.
  2. jcsd
  3. Apr 18, 2009 #2

    Tom Mattson

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    Let [itex]x,y=\in\mathbb{R}[/itex]. Then [itex]z=x+iy\in\mathbb{C}[/itex]. The complex conjugate of [itex]z[/itex] is [itex]\overline{z}=x-iy[/itex].

    So you have a function [itex]f:\mathbb{R}\rightarrow\mathbb{C}[/itex] such that [itex]f(-t)=\overline{f(t)}[/itex]. You can rewrite [itex]f(t)[/itex] as [itex]f(t)=u(t)+iv(t)[/itex], where [itex]u,v[/itex] are real valued functions of a real variable [itex]t[/itex].

    Chew on that and see if you don't know how to start.
  4. Apr 19, 2009 #3
    I'm still not sure what to do. I have to show all the properties of a vector space, right?
    But I'm not sure how to write it out and everything.
    Like, for vector addition, do I add another fuction g(t)=x(t)+iy(t) where x,y are real valued and then add f(t)+g(t) and show it equals g(t)+f(t)?
  5. Apr 21, 2009 #4

    Tom Mattson

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    Yes, you have to show that all the properties of a vector space hold. That f(t)+g(t)=g(t)+f(t) is obvious because the functions are complex valued. The real heart of the matter (as far as vector addition goes) is showing that f(t)+g(t) is even IN the vector space. That is, you have to show that (f+g)(-t)=(f+g)(t)-bar (sorry, Latex isn't working).
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