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Lenear algebra theoretical question

  1. Feb 27, 2009 #1
    there is a space V=Q^4 over Q (i thing Q means rational)


    there is a vector series v with length 6 on V
    does v linearly independant?
    if it is explain is not ,give a counter example ??



    Q^4 means we have 4 coordinates in each vector

    but v has vectors with 6 coordinates

    so i cant put them together

    ??
     
  2. jcsd
  3. Feb 27, 2009 #2

    HallsofIvy

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    What exactly do you mean by "a vector series v with length 6"? Do you mean 6 vectors in V or one or more vectors with 6 components? vectors with other than 4 components are not even in Q^4.

    You appear to mean 6 vectors in V. Do you mean to ask if a set of 6 vectors can be linearly independent? You know, I hope, that Q^4 has dimension 4 as a vector space over Q. Look at av1+ bv2+ cv3+ dv4+ ev4+ fv5+ gv6= 0. Rewriting v1, v2, v3, v4, v5, and v6 in terms of a basis for V and combine the same basis vectors. That will give you 4 equations for the 6 numbers a, b, c, d, e, f, g.



    If that is true then they are not even vectors in V. I feel sure there are supposed to be 6 vectors, not vectors with 6 coordinates.

     
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