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Linear dependence and inner product space

  1. Aug 2, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    The following is from the book Linear Algebra 3rd Edn by Stephen Friedberg, et al:
    Here aj are scalars of field F and vj are vectors of inner product space V.

    2. Relevant equations
    Theorem 6.3:
    3. The attempt at a solution
    Now I don't understand why theorem 6.3 implies aj = 0 = [tex]\langle 0,v_j \rangle / ||v_j||^2 [/tex] for all j. I can see how this is zero if the numerator [tex]\langle 0,v_j \rangle[/tex] = 0, but the inner product isn't even defined yet and I don't see anywhere in the axioms that the inner product of the zero vector and an orthogonal vector would always be zero. So how does theorem 6.3 apply here?
     
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  3. Aug 2, 2008 #2

    CompuChip

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    I find that hard to believe, since I don't see how you can talk about an inner product space without having defined an inner product :)
    So I guess you should go back a bit and find that definition and you will see that it immediately follows from the properties of an inner product that <v, 0> = 0 for all v. For example:
    • Let x be any vector, then <v, 0> = <v, 0*x> (see definition of vector space) = 0 * <v, x> (by linearity of the inner product) = 0 (by multiplicative properties of the reals :smile:)
    • <v, 0> = <v, v - v> (by definition of the vector space) = <v, v> + <v, -v> (by additivity of the inner product) = <v, v> - <v, v> = 0. Of course, technically, v - v = v + (-v) with the existence of -v asserted by the definition of vector space, etc.
     
  4. Aug 2, 2008 #3

    Defennder

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    Well I meant the specific inner product function, not the general notion of an inner product on a vector space.
    This is just what I need. Thanks!
     
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