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Math Help - [SOLVED] If S is linearly dependent, prove that one of its vectors must be the zero v

  1. #1
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    [SOLVED] If S is linearly dependent, prove that one of its vectors must be the zero v

    Let S=\{v_1,v_2,...,v_k\} be an orthogonal set of vectors in R^n. If S is linearly dependent, prove that one of the v_j must be the zero vector.
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  2. #2
    Junior Member nimon's Avatar
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    Since \{v_{1},\ldots,v_{k}\} is a linearly dependent set, there is a non-trivial solution to the equation

    \lambda_{1}v_{1}+\ldots+\lambda_{k}v_{k}=0,

    i.e. at least one \lambda_{j} is non-zero.

    Then compute

    \begin{array}{rcl} 0&=&\langle v_{j},0\rangle \\ &=&\langle v_{j}, \sum_{i=1}^{k} \lambda_{i}v_{i} \rangle \\ &=&\sum_{i=1}^{k} \lambda_{i}\langle v_{j},v_{i} \rangle \\ &=&\lambda_{i}\langle v_{j},v_{j} \rangle,\\ \end{array}

    which implies that \|v_{j}\|^{2} = 0 because \lambda_{j}\neq 0, so v_{j}=0.
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  3. #3
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    Quote Originally Posted by nimon View Post
    Since \{v_{1},\ldots,v_{k}\} is a linearly dependent set, there is a non-trivial solution to the equation

    \lambda_{1}v_{1}+\ldots+\lambda_{k}v_{k}=0,

    i.e. at least one \lambda_{j} is non-zero.

    Then compute

    \begin{array}{rcl} 0&=&\langle v_{j},0\rangle \\ &=&\langle v_{j}, \sum_{i=1}^{k} \lambda_{i}v_{i} \rangle \\ &=&\sum_{i=1}^{k} \lambda_{i}\langle v_{j},v_{i} \rangle \\ &=&\lambda_{i}\langle v_{j},v_{j} \rangle,\\ \end{array}

    which implies that \|v_{j}\|^{2} = 0 because \lambda_{j}\neq 0, so v_{j}=0.
    Our course material didn't use inner product brackets, but your proof seems good. Thanks for the help.
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  4. #4
    Junior Member nimon's Avatar
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    No problem

    It sounds like you already know this, but if you're using the dot product you can just rewrite \langle v, w \rangle as v\cdot w and \| v \| as |v| and everything still works.

    Sorry if this is obvious! Just in case you're really unfamiliar with inner-product notation.
    Last edited by nimon; April 13th 2010 at 12:47 PM. Reason: Spelling, punctuation and grammar!
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