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Homework Help: Inner Product Spaces

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\beta[/tex] be a basis for a finite dimensional inner-product space.

    b) Prove that is < x, z > = < y, z> for all z [tex]\in[/tex] [tex]\beta[/tex], then x = y

    2. Relevant equations

    3. The attempt at a solution
    start with the Cauchy-Schwarz:
    |< x, z >| [tex]\leq[/tex] ||x|| ||z||

    then because <x,z> = <y,z>

    |< y, z >| [tex]\leq[/tex] ||x|| ||z||

    so y = x, is this correct?
  2. jcsd
  3. Jun 5, 2010 #2


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  4. Jun 5, 2010 #3
    Any guidance?
  5. Jun 5, 2010 #4
    What is < x, z> - < y, z>? When is < x, x> = 0?
  6. Jun 5, 2010 #5
    <x,z> - <y,z> = 0 and <x,x> = 0 iff x =0

    <x,z> + <-y,z> = <x-y, z> = 0

    and |<x-y, z>| [tex]\leq[/tex] ||x-y|| [tex]\cdot[/tex] ||z||

    if z [tex]\neq[/tex] 0 the ||x-y|| must = 0. and ||x-y|| = 0 iff x-y = 0 so x = y
  7. Jun 5, 2010 #6
    C.S. gives you an upper bound, via rearrangement, a lower bound for ||x-y|| of 0. That doesn't mean ||x-y|| = 0. Use the fact that <x-y,z> = 0 for every basis vector.
  8. Jun 5, 2010 #7
    if z is a basis vector, then x-y is orthogonal to any and every z [tex]\in[/tex] [tex]\beta[/tex] but im not sure where to go from there.
  9. Jun 5, 2010 #8
    Hint: x-y is also orthogonal to every linear combination of z's.
  10. Jun 6, 2010 #9
    Im completely stuck. I know that x-y [tex]\in[/tex] ortho complement of [tex]\beta[/tex]

    But in this section, im not supposed to know that yet, so I dont wanna go down that route.
  11. Jun 6, 2010 #10


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    how about assuming you have a suitable orthonormal basis for the space, then consider the product of x and y with each of ths basis vectors...

    or if you know <x-y,z> = 0 for all z, how about the case z = x-y?
  12. Jun 6, 2010 #11
    in the case z = x-y, then <x-y, x-y> = 0 and thats only if x-y = 0 meaning x = y.

    But heres my question, a few posts ago I thought about this, but then I got to thinking why x-y would necessarily be in [tex]\beta[/tex]. I know x-y must = some linear combination of z's [tex]\in[/tex] [tex]\beta[/tex] since it is a basis. but why is x-y necessarily in [tex]\beta[/tex]?

    let x' = x-y and x-y = c1z1 +...+ cnzn then

    <x', c1z1 +...+ cnzn> = <x', c1z1> +...+ <x', cnzn>

    with each term inner product = 0

    but c1z1 +...+ cnzn isn't necessarily in [tex]\beta[/tex], rather in the span([tex]\beta[/tex])
    Last edited: Jun 6, 2010
  13. Jun 6, 2010 #12


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    Thats a fair point and I think the above is only true if x & y are in B.

    As it says innner product space as opposed to a "subspace" its probably fair to assume that x & y are in B. Once x & y are in B, clearly x-y is in B.

    As a counter example say B is the subspace of R^3 spanned by [0,0,1]. Then let x = [1,0,1] and y = [0,1,1]

    clearly <x,z> = <y,z> = 0 for all z in B, but x != y

    note that the projection of x & y onto B will still be equal though...
    Last edited: Jun 6, 2010
  14. Jun 6, 2010 #13
    Got it... I guess I read the problem wrong. I think you are write. The wya the problem is written, I think its meaning to imply x and y are in the inner product space. Geez, I had thought of this solution long ago, but I never figuered x and y were implied as being in B
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