1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

  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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook