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How to prove stuff about linear algebra?

  1. Sep 24, 2005 #1
    How to prove stuff about linear algebra???

    Question:

    Suppose [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent in [itex]V[/itex] and [itex]w\in V[/itex].
    Prove that if [itex](v_1 +w, v_2 +w, ..., v_n +w)[/itex] is linearly dependent, then [itex]w\in span(v_1, ...,v_n)[/itex].

    To prove this I tried...

    If [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent then [itex]a_1 v_1 + ...+a_n v_n =0[/itex] for all [itex](a_1 , ..., a_n )=0[/itex].
    then,
    [itex]a_1 (v_1 +w)+a_2 (v_2 +w)+...+a_n (v_n +w)=0[/itex]
    is not linearly independent, but can be rewritten as,
    [itex]a_1 v_1 + ...+a_n v_n +(\sum a_i )w=0[/itex]
    so,
    [itex]a_1 v_1 + ...+a_n v_n = -(\sum a_i )w[/itex].
    Since [itex]w[/itex] is a linear combination of vectors in [itex]V[/itex], [itex]w\in span(V)[/itex].

    Did I do this right?
    Is there a better way of doing this?
    Any input is much appreciated!
     
  2. jcsd
  3. Sep 24, 2005 #2

    LeonhardEuler

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    Gold Member

    Your proof is pretty much correct, but in this sentence:
    I think you mean to say:
    If [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent then [itex]a_1 v_1 + ...+a_n v_n =0[/itex] only when each [itex]a_i=0[/itex]
     
  4. Sep 24, 2005 #3
    Yes, that would make a bit more sense. Sometimes I understand what I mean to do, but don't know how to say it. :rolleyes:

    Thanks a bunch!
     
  5. Sep 24, 2005 #4

    Hurkyl

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    This is wrong. If the collection of vectors is independent, and if [itex]a_1 v_1 + ...+a_n v_n =0[/itex] then [itex]a_1 = a_2 = \cdots = 0[/itex].
     
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