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Math Help - Basis for a perpendicular vector

  1. #1
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    Basis for a perpendicular vector

    Problem:

    Find a basis for W^{\perp}, where

    W = span\left \{ \begin{bmatrix}1\\ 2\\ 3\\4\end{bmatrix},\begin{bmatrix}5\\ 6\\ 7\\8\end{bmatrix} \right \}

    Attempt:

    First I tried taking an arbitrary 4x1 matrix, and dotting it with a vector from the span of W. Then I realized that it must be orthogonal to all of the vectors in the span of W, so I trashed that idea.

    I know that W^{\perp} is the kernel of the orthogonal projection onto W. So therefore, W^{\perp} = proj_{W}(\vec{x}) = (\vec{u}_{1} \cdot \vec{x})\vec{u}_{m}+...+(\vec{u}_{m} \cdot \vec{x})\vec{u}_{m}

    This is where I get lost, and I'm not sure how to continue or even if I'm on the right track. Any help is greatly appreciated.

    Here is my work:

    Last edited by tangibleLime; May 4th 2011 at 04:54 PM.
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  2. #2
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    suppose v = (v1,v2,v3,v4) in R^4 is perpendicular to (1,2,3,4) and (5,6,7,8).

    then v.(a(1,2,3,4) + b(5,6,7,8)) = a(v.(1,2,3,4)) + b(v.(5,6,7,8)) = a0 + b0 = 0 + 0 = 0.

    so it suffices to pick v perpendicular to each element of the basis {(1,2,3,4), (5,6,7,8)} of W.

    you should be able to create a system of 2 linear equations in 4 variables, which will reduce to a system in two parameters.

    chose those two parameters such that you get 2 linearly independent vectors in R^4, and you're done.
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  3. #3
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    Sorry, I'm having a little trouble understanding the formatting.

    Is v a column vector with values v1->v4? And where are 'a' and 'b' coming from in the second line?
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  4. #4
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    yes, v is a column vector with coordinates v1,v2,v3,v4 in the standard basis. a and b are just arbitrary real numbers. any element of W, which is spanned by those two vectors, is a linear combination of the two, some scalar multiple of the first, plus some scalar multiple of the second.

    the point is, if you get a vector perpendicular to both, it will be perpendicular to any linear combination of the two.

    so v.(1,2,3,4) = 0 is the linear equation:

    v1 + 2v2 + 3v3 + 4v4 = 0

    and v.(5,6,7,8) = 0 is the linear equation:

    5v1 + 6v2 + 7v3 + 8v4 = 0.

    you need to solve these 2 equations simultaneously.

    well, as you might expect, the dimension of W┴, is going to be 2, so you should wind up with a solution set that looks like:

    (v1(s,t), v2(s,t), v3(s,t), v4(s,t)), where each vj(s,t) is some linear expression in s and t.

    or, to put it another way, you can eliminate one variable, and express a second variable in terms of the remaining 2.

    back-substitution will then give the "eliminated" variable in terms of those same two "undetermined" variables.

    converting to unit vectors isn't necessary, you're looking for orthognal vectors, not orthonormal ones.
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