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Inner Product once again :S

  1. Nov 2, 2005 #1
    A question reads:

    Let V be a vector in an inner product space V

    show that ||v|| >= ||proj(u) v|| holds for all finitie dimensional subspaces of U.

    Hint: Pythagorean Theorm.

    Okay... where on earth do i begin?

    I thought perhaps I should expand the RIGHT side of the equation, but that dosn't seem to be getting me anywhere really.

    The LEFT side seems pretty useless too, so im stuck trying to show that the RIGHT side is >= ...

    anybody have any ideas?

    Thanks
     
  2. jcsd
  3. Nov 2, 2005 #2

    NateTG

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    Do you mean
    [itex]\vec{V}[/itex] is a vector in an inner product space [itex]U[/itex]?
    otherwise that doesn't make sense?
    Maybe you can find some easy cases, and work from there?
     
  4. Nov 2, 2005 #3
    If you draw these two vectors out, it should be very obvious. The magnitude of the projection is just the component:

    [tex]\left|\text{proj}_{u}v\right|=\frac{\left(u,v\right)}{u}[/tex]

    ...which is really just saying v*cosθ. What do you know about the cosine function? Well, for one, it is less than or equal to 1 for all values of θ. Does this help?
     
  5. Nov 2, 2005 #4

    Gokul43201

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    apmcavoy : There are no "two vectors". A vector v in U is being projected onto some subspace (W, say) of U. However, with some additional construction, your method would work.

    playboy : Call the subspace W, and write the projection down in terms of the basis vectors [itex]w_i[/itex] of W. similarly write down the expansion of the norm in terms of the basis vectors [itex]u_i[/itex] of U.

    What do you know about these two sets of basis vectors ?
     
    Last edited: Nov 2, 2005
  6. Nov 2, 2005 #5
    okay thanks for the help everyone!

    Gokul43201: ill try your metod and get give back my results!
     
  7. Nov 2, 2005 #6
    Now i am even more lost than ever :S

    For the RHS....i get...

    <v, w1> + <v, w2> + ....
    ||w1||.....||w2||

    For the left hand side...

    <u,u>^0.5

    Am i on the right track?
     
  8. Nov 2, 2005 #7

    Gokul43201

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    First, we need to confirm that we're all solving the same problem. Where did U come from. Did you mean to write (in the OP) : "...for all finite dimensional subspaces U of the space V" ?
    The length (or norm) is not equal to the sum of its components.

    [tex]proj_U(v) = \sum _{i=1}^k \langle v,w_i \rangle w_i [/tex]

    [tex]\implies ||proj_U(v)|| = \sqrt{\sum _{i=1}^k \langle v,w_i \rangle ^2} [/tex]

    Expand ||v|| similarly.
     
  9. Nov 2, 2005 #8
    Noooo.... this is why its not making sense.

    Isn't the projection....

    <v wi>
    ______ wi
    <wi wi>


    and not


    <v wi> wi



    And on the LHS... (v is just a vecotr)
    so ||v|| = (v1^2 + v2^2 + ..... + vn^2)^0n5
     
  10. Nov 2, 2005 #9
    Yes, right. I'm sorry if that confused anyone who read this. I guess I was just assuming some vector u in U.
     
  11. Nov 3, 2005 #10

    Gokul43201

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    If [itex]w_i[/itex] is a basis vector, what's [itex]\langle w_i,w_i \rangle [/itex] ?
    What are v1, v2, etc ?

    PS : Why are you not addressing the problem that no one really knows what the actual question is? The question as stated in the OP is incorrect and needs to be fixed.
     
    Last edited: Nov 3, 2005
  12. Nov 3, 2005 #11
    The Question Reads:

    Let v be a vector in an inner product space V

    show that ||v|| >= ||proj u (v)|| holds for all finitie dimensional subspaces of U.

    Hint: Pythagorean Theorm.
    _____________________________________________
    how i approached it:

    Let (e1,.....,en) be an orthonognal basis:

    ||proj u (V)||^2 = <proj u (V), proj u (V)>
    ||proj u (V)|| = ( <v e1>^2/<e1 e1> + .... + <v en>^2/<en en>)^0.5

    Note that i did not show my full computation from the beginning, that would
    just be too long to type out.

    and ||v|| = (v1^2 + .... + Vn^2)^0.5 since v is a Vector

    Im lost after this point :( Please help
     
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