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Thread: Basis For W Perp

  1. #1
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    Exclamation Basis For W Perp

    Given the subspace
    W=span {
    [ 4 ] [ 6 ] [ 3 ]
    [ 5 ] [ 5 ] [ 4 ]
    [ 9 ] [ 1 ] [ 8 ]
    [ -2 ] [ 12 ] [ -3 ] }

    How do you find a basis for W perp?
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  2. #2
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    Re: Basis For W Perp

    The first thing I would do is look up "W perp". I suspect you would find that it is the "orthogonal complement" of W, the set of vectors perpendicular to all vectors in W. In particular they will be perpendicular to each of those spanning vectors. If (a, b, c, d) is such a vector, you must have (a, b, c, d).(4, 5, 9, -2)= 4a+ 5b+ 9c- 2d= 0, (a, b, c, d).(6, 5, 1, 12)= 6a+ 5b+ c+ 12d= 0, (a, b, c, d).(3, 4, 8, -3)= 3a+ 4b+ 8c- 3d= 0. Solve those three equations. Since there are three equations in four unknowns you will be able to solve for some of those in terms of the others (three in terms of one, two in terms of the other two, one in terms of the other three.)
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