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Linear indepedent functions and complex conjugation

  1. Sep 12, 2008 #1
    Suppose [itex]\{\varphi_i\}[/itex] is an infinite set of linearly independent functions. Is [itex]\{ \varphi_i^\ast \}[/itex] linearly indepedent? How about [itex]\{ \varphi_i \} \cup \{ \varphi_i^\ast\}[/itex]?
     
  2. jcsd
  3. Sep 13, 2008 #2

    morphism

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    Well, what are your thoughts on the matter?
     
  4. Sep 13, 2008 #3
    Well, if [itex]\{ \varphi_i \}[/itex] is LI then [itex]\{ \varphi_i^\ast \}[/itex] is also trivially LI, because if it wasn't then you could just take the complex conjugate violating linear indepedence of [itex]\{ \varphi_i \}[/itex].

    It's not clear if my second claim is true, although I'd like it to be, I suspect there are counterexamples waiting to be found. I'd like to be proven wrong, however.
     
  5. Sep 13, 2008 #4

    morphism

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    What if we take, say, the set [itex]\{i f, f\}[/itex], where f is some real-valued function?
     
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