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Math Help - perpendicular subspaces

  1. #1
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    perpendicular subspaces

    Im lost on this problem:
    Let B be a basis for a subspace W of an inner product space V, and let z exist in V. Prove that z exists in W^perp if and only if <z,v>=0 for all v in B.
    All I have is that <z,x>=0 for all x in W.
    cant seem to figure out how to prove either direction of the statement though.
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  2. #2
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    Quote Originally Posted by dannyboycurtis View Post
    Im lost on this problem:
    Let B be a basis for a subspace W of an inner product space V, and let z exist in V. Prove that z exists in W^perp if and only if <z,v>=0 for all v in B.
    All I have is that <z,x>=0 for all x in W.
    cant seem to figure out how to prove either direction of the statement though.

    Write an element of W as a lin. combination of elements in B, and use linearity of the inner product.

    Tonio
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  3. #3
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    so I have that
    x exists in W, where x=a_1v_1+...+a_nv_n for a's existing in the field V is defined on and v's vectors in the basis B.
    then <z,x>=0 implies
    <z,a_1v_1+...+a_nv_n>
    =<z,a_1v_1>+...+<z,a_nv_n>
    I cant see where to go after this, if I take out the scalar values, Im not sure where to proceed...
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  4. #4
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    Quote Originally Posted by dannyboycurtis View Post
    so I have that
    x exists in W, where x=a_1v_1+...+a_nv_n for a's existing in the field V is defined on and v's vectors in the basis B.
    then <z,x>=0 implies
    <z,a_1v_1+...+a_nv_n>
    =<z,a_1v_1>+...+<z,a_nv_n>
    I cant see where to go after this, if I take out the scalar values, Im not sure where to proceed...

    Yes, take the scalars out and in every summand you get <z,v_i> = 0, because this is precisely what you're assuming!

    Tonio
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