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Interpretation of a Basis

  1. May 6, 2012 #1
    From what I understand, a basis is essentially a subset of a vector space over a given field.

    Now what I'm not so sure of is the linearly independence part. If the basis has two linearly independent vectors, then than means they aren't collinear: rather, they wouldn't have the same slope or be generated by each other?

    let's say a vector v[itex]\epsilon[/itex]V is (v[itex]_{1}[/itex], v[itex]_{2}[/itex],...,v[itex]_{n}[/itex])
    and a vector w[itex]\epsilon[/itex]V is (w[itex]_{1}[/itex], w[itex]_{2}[/itex],...,w[itex]_{n}[/itex])
    such that
    v [itex]\neq[/itex] cw.
    for any constant c[itex]\epsilon[/itex][itex]\textbf{F}[/itex]

    These would then be non-collinear which means there are no linear operators that can turn v into w?

    Any insight would be greatly appreciated and thank you in advance.
     
  2. jcsd
  3. May 6, 2012 #2

    tiny-tim

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    Hi Krovski! :smile:
    Yes.

    Similarly, 3 vectors u v and w are linearly independent if there are no linear operators that can turn u and v into w, ie au + bv = w (and similarly for any number of vectors).
     
  4. May 6, 2012 #3
    Actually, there are no *linear combinations* of u and v that can give you w. But you can still have a linear operator that could possibly take u to v.

    For example, suppose
    { u , v , w } is a basis ( for a 3 dimensional space, and they are all linearly independent ) then permuting this basis to { v , u , w } and extending by linearity is a linear operator
     
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