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2 Linear Algebra Proofs about Linear Independence

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Proof 1:
    Show that S= {v1, v2, ... vp} is a linearly independent set iff Ax = 0 has only the trivial solution, where the columns of A are composed of the vectors in S. Be sure to state the relationship of the vector x to the vectors in S

    2. The attempt at a solution
    As far as I can tell (from my book and class), the actual definition of linear independence is
    "A set of vectors is linearly independent iff there exists only the trivial solution to c1v1 + c2v2+ ... + cpvp = 0, or similarly only a trivial solution exists to the equation Ax = 0 when the columns of A are composed of the vectors in S. "

    I know many implications of this property, but not the steps in between the term and the definition. Is there a more basic definition that I need to start this proof out with? Are the two options I gave in my definition of linear independence really different enough to merit in-between steps?

    1. The problem statement, all variables and given/known data

    Proof 2:
    Prove or disprove: If {v1, v2, ... vp}, p>=2 is a linearly independent set, then vp is not an element of Span {v1, v2, ... vp-1}.


    2. The attempt at a solution

    I just took a guess at this one:

    Assume that vp}, p>=2 is linearly independent, and let vp be an element of Span {v1, v2, ... vp-1}.

    By definition of spanning, vp is a linear combination of {v1, v2, ... vp-1} .

    By definition of linear combination, there exists c1v1 + c2v2+ ... + cpvp-1 = vp.

    Algebra give us:
    c1v1 + c2v2+ ... + cpvp-1 - vp = 0.

    This gives a solution other than the trivial solution (where {c1, c2, ... cp} are all zero) to the equation c1v1 + c2v2+ ... + cpvp = 0, because cp must equal -1.

    This is a contradiction. {v1, v2, ... vp}, p>=2 can't be a linearly independent set while vp is an element of Span {v1, v2, ... vp-1}. , so if {v1, v2, ... vp}, p>=2 is a linearly independent set, then vp is not an element of Span {v1, v2, ... vp-1}, as desired. Q.E.D.


    Any help or hints on either of these would be greatly appreciated. The first one I just need a starting point, and the second one I need to know if I made any mistakes, or if that is a valid proof. Thanks a lot.
     
  2. jcsd
  3. Sep 19, 2007 #2

    EnumaElish

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    I agree with the 2nd proof. In the first problem, the statement and the predicate (the definition) seem to be identical -- I am not clear on what is being asked.
     
  4. Sep 19, 2007 #3

    radou

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    Regarding proof 1, do you know anything about the dimension of the vector space of solutions to a homogenous system? How does it relate to the rank of the system matrix A? What *is* the rank of your matrix A, since it consists of p linear independent vectors?
     
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