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Homework Help: Linear Combinations of Dependent Vectors

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data

    If (u,v,w) is a family of linearly dependent vectors in vector space V and vector x is in the span of (u,v,w), then x=αu+βv+γw has infinitely-many choices for α,β, and γ.

    2. Relevant equations

    If (u,v,w) is linearly dependent, then there exists an α, β, and γ, not all equal to zero, such that αu+βv+γw=0.

    3. The attempt at a solution

    My first attempt, which didn't go anywhere, was to assume that there were only a finite number of choices and see if that led to a contradiction.

    For my second attempt, I started with the fact that (u,v,w) is linearly dependent. Then I multiplied both sides by an arbitrary scalar n. Then I thought I could add x to both sides and manipulate the equation somehow, but that didn't lead anywhere either.

    If I was dealing with the vector space R^n, then x=αu+βv+γw would have infinitely-many solutions because you would end up with at least one free variable since at least one of the vectors is just a linear combination of the others. Can you apply that reasoning to all vector spaces?
    Last edited: Feb 14, 2009
  2. jcsd
  3. Feb 14, 2009 #2


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    As you say, since (u,v,w) is linearly independent so there exist [itex]\alpha[/itex], [itex]\beta[/itex], [itex]\gamma[/itex], not all 0, such that [itex]\alpha u+ \beta v+ \gamma w[/itex]= 0.

    Further, since (u,v,w) spans V there exist, for any x in V, scalars a, b, c such that au+ bv+ cw= x.

    Now, what can you say about [itex]au+ bv+ cw+ R(\alpha u+ \beta v+ \gamma w)[/itex] for any real number R?
  4. Feb 14, 2009 #3
    It equals x?

    (Rα+a)u +(Rβ+b)v +(Rγ+c)w = x

    And since Rα or Rβ or Rγ is not zero, the equation has infinitely-many solutions?
  5. Feb 14, 2009 #4


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    Yes, every different choice for R gives a different linear combination but they are all equal to x!

    And it is doing exactly what YOU suggested:
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