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Math Help - Linearly Independent Vectors

  1. #1
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    Linearly Independent Vectors

    Let a system of vectors  \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  \bold{v}_{r+1} such that the system  \bold{v}_{1}, \bold{v}_{2}, \ldots, \bold{v}_{r}, \bold{v}_{r+1} is linearly independent.

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

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

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