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# elements in base does not depend on the basis

  1. Jul 11, 2013 #1
    Essentially, I have to show that where {[itex]e_1,...,e_n[/itex]} forms the basis of [itex]L[/itex], no family of vectors {[itex]e'_1 ,..., e'_m[/itex]} with [itex]m[/itex]>[itex]n[/itex] can serve as the basis of [itex]L[/itex]. The book shows this by saying there exists a [itex]0[/itex] vector such that [itex]0 = \sum_{i=1}^{m}x_ie'_i[/itex], where not all [itex]x_i[/itex] vanish. I wanted to show it by distinguishing the sets to which [itex]e[/itex] and [itex]e'[/itex] belong and showing that if one belongs to a set in n-dimension and the other belongs to a different set in g-dimension where g>n, the n-dimension cannot encompass the g-dimension. I'm not sure if this is doable or too longwinded. Also, I do not understand their reasoning or where the [itex]0[/itex] vector comes from. Help?
     
    Last edited: Jul 11, 2013
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  3. Jul 12, 2013 #2

    mfb

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    If n vectors form a basis of L, then the dimension of L is n, and not g. Where does that g come from, and what is it supposed to be?
    It is possible to assume that there is a different basis with m>n vectors, and show that the first n vectors cannot form a basis then (proof by contradiction).

    What do you mean with "where the 0 vector comes from"? The 0 vector is part of L.
     
  4. Jul 12, 2013 #3

    Stephen Tashi

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    You should should state exactly what book's proof says.

    That's doesn't describe a specific method, so I'm not sure either.


    You'll have to reveal the book's proof if you want someone to evaluated it.
     
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