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Linear independence again

  1. Nov 1, 2006 #1

    radou

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    Let V be a vector space over a field F, [tex]v_{1}, \cdots, v_{n} \in V[/tex] and [tex]\alpha_{1}, \cdots, \alpha_{n} \in F[/tex]. Further on, let the set [tex]\left\{v_{1}, \cdots, v_{n}\right\}[/tex] be linearly independent, and b be a vector defined with [tex]b=\sum_{i=1}^n \alpha_{i}v_{i}[/tex]. One has to find necessary and sufficient conditions on the scalars [tex]\alpha_{1}, \cdots, \alpha_{n}[/tex] such that the set S=[tex]\left\{b, v_{2}, \cdots, v_{n}\right\}[/tex] is linearly independent, too.

    Well, I just used a simple proposition which states that a set is dependent if there exists at least one vector from that set which can be shown as a linear combination of the rest of the vectors from the same set. So, obviously, for [tex]\alpha_{1} = 0[/tex], the set S is dependent, which makes [tex]\alpha_{1} \neq 0[/tex] a necessary condition for S to be independent. Further on, a sufficient condition would be [tex]\alpha_{1} = \cdots = \alpha_{n} = 0[/tex], which leaves us with the set S\{b}. This set must be linearly independent, since it is a subset of {v1, ..., vn}, which we know is linearly independent.

    I may be boring, but I'm just checking if my reasoning is allright.. :smile:

    Edit. I just realized, for [tex]\alpha_{1} = \cdots = \alpha_{n} = 0[/tex] we have b = 0, which makes the set dependent! So [tex]\alpha_{1} \neq 0[/tex] is a necessary condition, but what's the sufficient condition? Is it that at least one of the scalars [tex]\alpha_{2}, \cdots, \alpha_{n}[/tex]must not be equal zero?
     
    Last edited: Nov 1, 2006
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  3. Nov 1, 2006 #2

    Office_Shredder

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    The sufficient condition certainly includes this [tex]\alpha_{1} \neq 0[/tex]

    Basically, if a condition is necessary and sufficient, it means that the statement is true iff the condition is true. So search for a set of conditions that are true iff the set b, v 2-n is linearly independent.

    Note that if all the alphas except [tex] \alpha_1[/tex] are zero, then b is just a multiple of v1, and the new set is certainly linearly independent
     
  4. Nov 1, 2006 #3

    radou

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    Yes, I realized that.. So, it seems [tex]\alpha_{1} \neq 0[/tex] is both necessary and sufficient condition, since the rest of the scalars can be any elements of F, all zero, all non zero, or combined.
     
    Last edited: Nov 1, 2006
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