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I A nonempty subspace

  1. Oct 6, 2016 #1
    I have a simple question. Say we have some subspace that is nonempty and closed under scalar multiplication and vector addition. How could we deduce that ##0 \vec{u} = \vec{0}##?
     
  2. jcsd
  3. Oct 6, 2016 #2

    Krylov

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    It holds that ##0\vec{u} = \vec{0}## for every ##\vec{u} \in V##, where ##V## is any vector space. This follows directly from the defining axioms and does not require the introduction of a subspace.

    What did you try yourself to prove it?
     
  4. Oct 6, 2016 #3

    Math_QED

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    Every subspace is non empty, closed under scalar multiplication and vector addition so no need to say that.

    You should show some effort.
     
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