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Math Help - Linear combination and dot product

  1. #1
    Senior Member Shanks's Avatar
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    Linear combination and dot product

    let N be the intersection of the sets \{x:<v_i,x>=0\},i=1,2...m.
    Prove or disprove that a vector v can be expressed as a linear combination of vectors v_i,i=1,...m iff for any x in N, <v,x>=0.
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  2. #2
    Junior Member
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    Quote Originally Posted by Shanks View Post
    let N be the intersection of the sets \{x:<v_i,x>=0\},i=1,2...m.
    Prove or disprove that a vector v can be expressed as a linear combination of vectors v_i,i=1,...m iff for any x in N, <v,x>=0.
    The first direction is easy. If v is a linear combination of v_i's then it's clear that <v,x>=0 for any x in N.

    The other direction is the tricky one. Assume that <v,x>=0 for any x in N. First, define a subspace U as follows:

    U:=span\{v_1, ..., v_m\}

    It's clear that N=U^{\perp } (I'll leave you to check that if you're not convinced). So in particular, N has a basis, say \{x_1, ..., x_n\}. We can also assume, without loss of generality, that this basis is orthonormal.

    Now, we use the fact that U+N gives the whole vector space (I'm assuming this is a fact you know). Then, v must be a linear combination of v_i's and x_j's. So write:

    v=a_1v_1+...+a_mv_m+b_1x_1+...+b_nx_n

    Take any one of the x_j.

    <v,x_j>=b_j

    Since this must be zero, we get that b_j=0 and since our choice of x_j was arbitrary, all the b's must be zero. So we're left with:

    v=a_1v_1+...+a_mv_m

    And so v is indeed a linear combination of the v_i's
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