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Linear Dependence of Equation Vectors

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the members of the given set of vectors are linearly independent for -[tex]\infty[/tex] < t < [tex]\infty[/tex]. If they are linearly dependent, find the linear relation among them.

    x(1)(t) = (e-t, 2e-t), x(2)(t) = (e-t, e-t), x(3)(t) = (3e-t, 0)

    (the vectors are written as row vectors)

    2. Relevant equations

    The section in my book about linear dependence of equation vectors is immediately followed by a discussion of eigenvalues. Wronskians are not covered until the next chapter.

    3. The attempt at a solution

    I set up a matrix of equations, each vector in the problem statement a column of the matrix, augmented it with the 0 vector, and row-reduced, resulting in

    [e-t, 0, -3e-t | 0; 0, e-t, 6e-t | 0]

    The book doesn't give any examples, and I'm having a hard time with where to go from here, or if this is the right approach in the first place.
     
  2. jcsd
  3. Oct 11, 2009 #2

    Mark44

    Staff: Mentor

    What does your reduced matrix tell you about the linear independence/dependence of your three vectors?

    Your answer should address the question in your problem.

    Has your book discussed linear independence and dependence for ordinary vectors? The idea here is about the same.
     
  4. Oct 11, 2009 #3

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    The e-t is just a red herring. You can factor it out and your problem amounts to asking whether (1,2), (1,1), and (3,0) are independent. You probably already know that you can't have 3 independent vectors in R2. So they are going to be dependent. If you try the condition directly for independence you check:

    a(1,2) + b(1,1) + c(3,0) = 0 or

    a + b + 3c = 0
    2a + b + 0c = 0

    The row reduction you have done is the same as writing these as

    a + 0b -3c = 0
    0a +b + 6c = 0

    Clearly you can take the c on the other side and let it be most anything except 0. Say c = 1, giving a = 3 and b = -6.

    So your row reduction really does the work, you just need to interpret it right.
     
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