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Proof about inner product spaces

  1. Jun 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose V is a real inner-product space and (v1, . . . , vm) is a
    linearly independent list of vectors in V. Prove that there exist
    exactly 2^m orthonormal lists (e1, . . . , em) of vectors in V such
    that
    span(v1, . . . , vj) = span(e1, . . . , ej)
    for all j ∈ {1, . . . , m}.


    2. Relevant equations



    3. The attempt at a solution

    Alright, just to be sure I have the right idea (this is not the proof attempt)
    I'll consider an orthogonal basis {u1} with j=1 so the non orthogonal basis for V
    is {v1}.
    I guess that if j=1 then let u1=v1 so e1=u1/llu1ll so two possible orthonormal lists are {e1} and{-e1} then after adding an additional element to the basis {v1} (call it
    v2) j becomes 2, the non-orthogonal becomes {v1,v2}, the orthogonal basis is {u1,u2}
    and u2=v2-proj_u2(v2) and e2=u2/llu2ll (with the other possible orthonormal basis
    being -e2) but since e1 and -e1 cannot share
    the same list, they must be put in separate (but similar) lists which double
    the amount of lists from when j was 1 so now we have {e1, e2} and {e1, -e2}
    and {-e1, e2} finally {-e1, -e2} four possible orthonormal lists. If m=j, thats it,
    but if not, eventually when it does, we will keep multiplying 2^j by 2 as each vj is added to the basis {v1..vj-1}.
    This isn't the proof, just want to see if I'm seeing this right.
     
  2. jcsd
  3. Jun 27, 2009 #2

    Hurkyl

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    Re: proof

    Are you sure you stated the problem right? Consider, for example, that any two-dimensional real inner-product space has infinitely many orthonormal bases.

    (edit: aha, nevermind)
     
    Last edited: Jun 27, 2009
  4. Jun 27, 2009 #3

    Office_Shredder

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    Re: proof

    Yeah, that's the basic idea. Given [itex]e_1,..., e_{j-1},[/itex] there are two choices for [itex]e_j[/itex] and they are additive opposites of each other
     
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