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Linear Algebra HELP

  1. Nov 17, 2004 #1
    Ok need help with a question that could be in tomorrow's exam:
    In Euclidean space R^4 equipped with Euclidean Inner Product, let W be subspace which has basis {(1,0,-1,0), (0,-1,0,1), (2,1,-3,0)}.
    Determine the vector in W which is closest to v=(1,0,0,-1)
     
  2. jcsd
  3. Nov 17, 2004 #2

    arildno

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    That vector must be the orthogonal projection of v into W.
    Proof:
    Consider the squared distance vector between an arbitrary element in W and v:
    [tex](v-\alpha{W}_{1}-\beta{W}_{2}-\gamma{W}_{3})^{2}[/tex]
    This is minimized by differentiating with respect to [tex]\alpha,\beta,\gamma[/tex] and setting the three results equal to zero.
    For example, we gain by differentiating with respect to [tex]\alpha[/tex]
    [tex]-2(v-\alpha{W}_{1}-\beta{W}_{2}-\gamma{W}_{3})\cdot{W}_{1}=0[/tex]
    That is, the distance vector is orthogonal to [tex]W_{1}[/tex]

    By solving the system of equations, you gain the minimizing values of [tex]\alpha,\beta,\gamma[/tex] and hence, the vector in W with shortest distance to v.
    [tex]W_{1},W_{2},W_{3}[/tex] is of course the basis of W.
     
    Last edited: Nov 17, 2004
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