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Linear Independence/Dependence of Vectors

  1. Sep 29, 2008 #1
    Hi all. I'm having a really tough time figuring out how to solve this problem:

    Suppose that {v1, v2, v3} are linearly independent vectors in R7.
    a1 = v1 + 2v2

    a2 = 3v2 – v3

    a3 = v1 – v2 + v3,

    determine directly from the definitions whether the vectors {a1, a2, a3} are linearly independent or linearly dependent.

    Can anyone help me? I know what linear dependence and linear independence are, and I know how to check for either using Gauss-Jordan elimination. But I'm not sure where to start on this problem.
  2. jcsd
  3. Sep 29, 2008 #2


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    In this case, you are given a1, a2 and a3 in terms of v1...v3. So that means to say you can represent a_i, where i is 1,2 or 3, in terms of a coordinate matrix of a_i with respect to the ordered set (or basis B) of {v1, v2, v3}. So to start yourself off, write out the respective coordinate vectors of a_i.

    Then invoke the definition of linear independence:

    a1,a2,a3 are linearly independent iff

    k1a1 + k2a2 + k3a3 = 0 (the zero coordinate vector with respect to basis B)

    only has the trivial solution for all ki.

    So from the above, you can write out a square matrix from which you can then apply those techniques you know to determine if only the trivial solution exists.
  4. Sep 30, 2008 #3


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    As Defennder said, the definition of "independent" says that these vectors are independent if and only if [itex]k_1a_1+ k_2a_2+ k_3a_3= 0[/itex] implies [itex]a_1= a_2= a_3= 0[/itex]. Since you are given that [itex]a_1= v_1+ 2v_2[/itex], [itex]a_2= 3v_2- v_3[/itex] and [itex]a_3= v_1- v_2+ v_3[/itex] , that equation becomes
    [itex]k_1(v_1+ 2v_2)+ k_2(3v_2- v_3)+ k_3(v_1- v_2+ v_3)= 0[/itex]
    Multiplying that out and combining "like" terms (combining same vn), will give coefficients in terms of k1, k2, and k3 mutltiplying v1, v2, and v3 equal to 0. Since you are given that v1, v2, and v3 are independent, their coefficients MUST be 0. That gives you 3 equations for k1, k2, and k3. Solve those equations. If you get that they are all 0, the vectors are independent. That is basically just what Defennder said but you did say "directly from the definitions" and I consider this more "fundamental" than reducing matrices as Defennder suggested.
  5. Mar 22, 2009 #4
    Let A = [2 0 -1 1 -3
    1 1-3 0 -2
    1 0 -1 -1 3]
    Find linearly independant vectors {u1,u3,...,um} such that row(A) = span{u1,u2,...,um}.
    Note: You must justify that your vectors are linearly independant.

    I have know idea how to even begin. Could someone really help me out here? Thanks heaps.
  6. Mar 22, 2009 #5


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    Well, the first thing you would have to do is tell us what u1, u2, ..., um are! Here you give a matrix A, but the problem says nothing about A or any matrix. Are we to assume that u1, u2, ..., um are the rows of that matrix? The columns?
  7. Mar 22, 2009 #6
    I am trying to find {u1,u2,...,um} so that col(A) = span{u1,u2,...,um}
  8. Apr 23, 2011 #7
    I was learning about the same topic and found this problem.
    As per HallsofIvy,

    k1(1v1 + 2v2 + 0v3) + k2(0v1 + 3v2 - 1v3) + k3(1v1 - 1v2 + 1v3) = 0


    v1(1k1 + 0k2 + 1k3) + v2(2k1 + 3k2 - 1k3) + v3(0k1 - 1k2 + 1k3) = 0

    now, coefficients of v must be 0

    1k1 + 0k2 + 1k3 = 0
    2k1 + 3k2 - 1k3 = 0
    0k1 - 1k2 + 1k3 = 0

    which leaves me with

    1k1 + 1k3 = 0
    3k2 - 3k3 = 0

    as you see, these are just 2 equations with 3 different variables. Can you say just from this that they are dependent?
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