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Math Help - Inner product space proof question

  1. #1
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    Inner product space proof question

    Prove that {u,v}=0 for all v belongs to V iff u=0.

    {u,v}=  \Sigma u*v, where u* is conjugate of u

    If u=0 then {u,v} is obviously 0.

    now im not sure how to prove it the other way

    If {u,v}=0 then u=0

     \Sigma u*v=0...

    Thanks in advance
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  2. #2
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    Quote Originally Posted by firebio View Post
    Prove that {u,v}=0 for all v belongs to V iff u=0.

    {u,v}=  \Sigma u*v, where u* is conjugate of u

    If u=0 then {u,v} is obviously 0.

    now im not sure how to prove it the other way

    If {u,v}=0 then u=0

     \Sigma u*v=0...

    Thanks in advance

    Apparently you're using {u,v} to denote inner product... Anyway, if it is true that <u,v>=0\,\,\forall v\in V then this is true for v=u as well, so apply now positiveness of inner product to get that it must be u=0 .

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by firebio View Post
    Prove that {u,v}=0 for all v belongs to V iff u=0.

    {u,v}=  \Sigma u*v, where u* is conjugate of u

    If u=0 then {u,v} is obviously 0.

    now im not sure how to prove it the other way

    If {u,v}=0 then u=0

     \Sigma u*v=0...

    Thanks in advance
    What is the conjugate in this case?
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