1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra: Linear Independence

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Let u,v,w be three linearly independent vectors in ℝ7. Determine a value of k,

    k= , so that the set S={u-3v,v-5w,w-ku} is linearly dependent.

    2. Relevant equations

    3. The attempt at a solution
    I don't really know why knowing that we're in ℝ7 will help. I know a couple of ways to solve this, but I don't think that they really apply. If we go:

    If it was square we could set the determinant equal to zero and solve for k. Otherwise, we have the matrix,


    If the matrix was little friendlier we could do some G-J elimination. Er... actually I'm not sure if that would help.

    Any hints?
  2. jcsd
  3. Mar 23, 2010 #2
    The first equation you listed is the right way to go. Just solve for A,B,C, and k using:


    You don't need any matrices, since u, v, and w are linearly independent, you can just set all the coefficients in front of the individual u, v, and w to 0 and solve for the A,B,C, and k in the coefficients.

    For example, (-3A+B) = 0, for the v term in that above equation.
  4. Mar 23, 2010 #3
    Wow. Okay so I did it and got 1/15 for k and got the right answer, but am still a bit clueless as to why that works. Are we saying that u,v, and w equal 1 and saying that each term equals zero? From my understanding of linear (in)dependence a set of vectors is linearly independent if the only way that their sum can equal zero is if the linear combination where A, B, and C equal zero. er.. I don't know if I said that right. What I meant was that vectors x,y, z are linearly independent provided that

    Ax+By+Cz=0 only if A,B,and C = 0.

    How can we get to the method used to solve this problem knowing only that?
  5. Mar 23, 2010 #4
    You know:

    Ru+Sv+Tw = 0

    is impossible since u,v,w are linearly independent unless R=S=T=0. That is the only way a linear combination of linearly independent vectors can sum to 0.

    But you also wrote out:


    which can be reorganized as:

    (A-Ck)u + (-3A+B)v + (-5B+C)w = 0

    which looks like Ru+Sv+Tw = 0

    Therefore each of the coefficients must be 0, since u,v,w are linearly independent.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook