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Thread: Linearly Independent Vectors

  1. #1
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    Linearly Independent Vectors

    Let a system of vectors $\displaystyle \bold{v}_{1},\bold{v}_{2}, \ldots, \bold{v}_{r} $ be linearly independent but not generating. Show that it is possible to find a vector $\displaystyle \bold{v}_{r+1} $ such that the system $\displaystyle \bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1} $ is linearly independent.

    So let $\displaystyle \bold{v}_{r+1} $ be any vector that cannot be represented as a linear combination $\displaystyle \sum_{k=1}^{r} \alpha_{k} \bold{v}_{k} $. So $\displaystyle \bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1} $ does not form a basis for some vector space $\displaystyle V $.

    To show linear independence, we have to show that $\displaystyle \sum_{i=1}^{k+1} \alpha_{k} \bold{v}_{k} = \bold{0} $ with $\displaystyle \alpha_{k} = 0 $.
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  2. #2
    Super Member PaulRS's Avatar
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    $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\alpha _k \cdot \bold{v_k} } = \bold{0}
    $

    By absurd, assume that there is such a linear combination with $\displaystyle
    \sum {\left| {\alpha _k } \right|} > 0
    $ (1) (that is, not all the scalars are 0). We have that $\displaystyle
    \alpha _{n + 1} = 0
    $ is impossible by the hypothesis ( the others are LI and we have (1) ), so $\displaystyle
    \alpha _{n + 1} \ne 0
    $, and therefore $\displaystyle
    \bold{v_{n + 1}} = \sum\limits_{k = 1}^n {\left( -{\tfrac{{\alpha _k }}
    {{\alpha _{n + 1} }}} \right) \cdot \bold{v_k} }
    $ that is $\displaystyle
    \left[ {\bold{v_1} ,...,\bold{v_n} } \right]
    $ spans $\displaystyle \bold{v_{n+1}}$ ABSURD!
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