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Inner product space question

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let V be a complex inner product space and let S be a subset of V. Suppose that v in V is a vector for which
    <s,v > + <v,s> [itex]\leq[/itex] <s,s>
    Prove that v is in the orthogonal set S[itex]\bot[/itex]
    2. Relevant equations

    We have the three inner product relations:
    1) conjugate symmetry
    <x,y> = [itex]\overline{<y,x>}[/itex]

    2) linearity
    <x+y,z> = <x,z>+<y,z>

    3) Def of a norm
    ||x|| = √<x,x>

    There may be more that apply such as triangle inequality or the Cauchy–Schwarz inequality
    but im not sure

    3. The attempt at a solution

    I know that if v is in the set S[itex]\bot[/itex]
    then s is orthogonal to v so <s,v> = <v,s> = 0
    Therefor im guessing through all these equations and the given inequality it can be shown that <s,v>+<v,s> needs to be both ≥ 0 and ≤ 0 therefor it will be zero

    Am I thinking of the way forward correctly?
    Any help towards a solution would be great!

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 14, 2012 #2

    HallsofIvy

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    For all s in S? In that case, your conclusion is not true.
    If v is orthogonal to S, this is saying that <s, s>= 0 for all s in S.

     
  4. Dec 14, 2012 #3
    I did for get to state that this was for all s in S.

    Could you explain a little more why <s,s>=0

    would you be referring to the fact that we may be able to pick a case where v=s and therefor
    2<s,s> ≤ <s,s>
    and this would only happen when <s,s> = 0 because <x,x> ≥ 0?
     
  5. Dec 14, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If v is in the orthogonal complement to S, then <v, s>= <s, v>= 0 so <s, s>= <v, s>+ <s, v>= 0+ 0= 0.

    (No, I was not reffering to the case where v= s. If s is in S and v is in the orthogonal complement to S, then v= s only if v= s= 0.)
     
  6. Dec 14, 2012 #5
    We are given that <s, s>≥ <v, s>+ <s, v> not that they are equal so really saying that
    <v, s>+ <s, v> =0 ( something we know will be true seeing as we are proving v is in the orthogonal complement of S) only shows that <s,s> can be 0 or anything greater than zero, not defiantly 0
     
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