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Math Help - Perpendicular Vectors

  1. #1
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    Perpendicular Vectors

    Problem:

    Consider the vector

    \vec{v}=\begin{bmatrix}1\\ 2\\ 3\\ 4\end{bmatrix} in \mathbb{R}^{4}

    Find a basis of the subspace of R^4 consisting of all vectors perpendicular to \vec{v}.

    I'm not very sure about how to solve this problem. I assume that I need to find a set of vectors \vec{u_{1}},...,\vec{u_{m}} for which the case \vec{v} \cdot \vec{u_{i}} = 0 holds.

    I tried using the formula,

    \vec{v}^\perp=\vec{v}-\vec{v}^\parallel

    Since I already know \vec{v}, I would just need to figure out \vec{v}^\parallel. To do this, I used projection.

    proj_x(\vec{v}) = \sum_{i=1}^{m}(\vec{u}_{i} \cdot \vec{v})\vec{u}_{i}

    This is where I get stuck, because I'm LOOKING for the u vectors... correct?

    Am I far off the path I should be on here or am I on the right track?

    Any help is greatly appreciated. Thanks for reading.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    \begin{bmatrix}-2\\ 1\\ 0\\ 0\end{bmatrix} is one member of the basis. Can you find the rest?
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  3. #3
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    I know, I saw that in the back of the book, but I can't figure out where that's coming from. Perhaps a hint to push me in the right direction would be good.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by tangibleLime View Post
    I know, I saw that in the back of the book, but I can't figure out where that's coming from. Perhaps a hint to push me in the right direction would be good.
    Set two components of the basis vector to zero and a third component to 1. Solve for the fourth one.

    This way, you should be able to find two more members of the basis.
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  5. #5
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    here is one way the vector (-2,1,0,0) might have been found:

    we want the dot product (1,2,3,4).(x,y,z,w) to be 0. for our first vector, we can choose any x,y,z and w that will work. so we can choose z = w = 0,

    and find x,y so that x + 2y = 0. if y = 1, x will have to be -2.

    now, if you choose x = y = 0, clearly whatever vector you wind up with, will be linearly independent of (-2,1,0,0) (unless the only vector that works is (0,0,0,0)).

    that will give you 2 elements of the basis, a similar approach will yield the 3rd.
    Last edited by Deveno; May 4th 2011 at 04:48 PM.
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  6. #6
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    Thanks! I took an arbitrary column vector, dotted it with the given vector and solved for each variable which gave me the correct answer.
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