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Math Help - Set proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Set proof

    Q: Prove that if S_{1} is a nonempty subset of the finite set S_{2}, and S_{1} is linearly dependent, then so is S_{2}

    A: By definition S_{1}\subset{S_{2}} \iff \forall{\vec{v}}\in{U} (\vec{v}\in{S_{1}} \Rightarrow \vec{v}\in{S_{2}}). Since we know every vector \vec{v_{i}}\in{S_{1}} can be written as a linear combination of at least one vector \vec{u}\in{S_{1}} and S_{1}\subset{S_{2}} we are guaranteed \vec{u}\in{S_2}. So, we have shown exists a linear dependent vector in S_{2}.

    \therefore S_{2} must also be linearly dependent.

    Have I shown enough work to prove my claim?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Linear dependence

    Hello Danneedshelp
    Quote Originally Posted by Danneedshelp View Post
    Q: Prove that if S_{1} is a nonempty subset of the finite set S_{2}, and S_{1} is linearly dependent, then so is S_{2}

    A: By definition S_{1}\subset{S_{2}} \iff \forall{\vec{v}}\in{U} (\vec{v}\in{S_{1}} \Rightarrow \vec{v}\in{S_{2}}). Since we know every vector \vec{v_{i}}\in{S_{1}} can be written as a linear combination of at least one vector \vec{u}\in{S_{1}} and S_{1}\subset{S_{2}} we are guaranteed \vec{u}\in{S_2}. So, we have shown exists a linear dependent vector in S_{2}.

    \therefore S_{2} must also be linearly dependent.

    Have I shown enough work to prove my claim?
    This is OK, except that it's not every vector \vec{v_{i}}\in{S_{1}}, but \exists\,\vec{v_{i}}\in{S_{1}}.

    Grandad
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