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Inner product proof

  1. Nov 11, 2009 #1
    Let V be an inner product space, and let W be a finite-dimensional subspace of V. If x[tex]\notin[/tex] W, prove that there exists y[tex]\in[/tex] V such that y [tex]\in[/tex] W(perp), but <x,y>[tex]\neq[/tex] 0.

    I don't have a clue....
  2. jcsd
  3. Nov 11, 2009 #2


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    you could start by using something along the lines of gram-schmidt to decompose to x into a sum of vector from W & one from W perp...
  4. Nov 12, 2009 #3
    Umm..I dont really get it. Can you explain more specifically? Thank you.
  5. Nov 12, 2009 #4


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    what don't you get?

    first you need to assume x is non-zero

    x is not contained in W, and as its non-zero, this means it must have a component in W perp , (as V = W + W perp by defintion of W perp, sloppy notation here, but hopefully you get the idea)

    now consider the dot product of x with the component of x in W perp
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