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For which value of k are the vectors linearly dependent?

  1. Oct 22, 2014 #1

    kosovo dave

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    Gold Member

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

    For which value of k are the vectors linearly dependent?
    v1=[1,0,1] v2=[1,1,2] v3=[1,2,k] <--Originally written as column vectors

    2. Relevant equations


    3. The attempt at a solution
    None of the vectors look like multiples of each other to me...so time to row reduce.

    Since each vector was written as a column vector, each row should correspond to a different variable.
    I set up the augmented matrix:
    1 1 1 0
    1 2 k 0
    0 1 2 0
    (If I made an error at this point let me know. But since the vectors were originally represented as columns, I figured row 1 from each vector would form column 1 in the augmented matrix.)

    row reduce:
    1 0 -1 0
    0 1 2 0
    0 0 k-3 0

    k=3 would give a row of all zeros at the bottom meaning that we would have a free variable right? Therefore the vectors should be linearly dependent for k=3? I know linear dependence is more formally defined by having a linear equation like c1v1+ c2v2 +c3v3 = 0, where the weights are non-zero, but it's also my understanding that the presence of a free variable indicates linear dependence.
     
  2. jcsd
  3. Oct 22, 2014 #2

    Mark44

    Staff: Mentor

    No, your matrix is wrong. v1 should be the first column, v2 the second column, and v3 the third colum. You don't need a fourth column, as it would be all zeros and won't change, so it doesn't hurt to omit it.

    The matrix represents the equation Ac = 0, where the columns of A are the vectors [v1 v2 v3] and c represents a column vector of the three constants.

    Edit: Actually, your matrix will work - it just has rows two and three switched. I didn't realize at first that you did that.
    Yes.

    Assuming that k = 3, the bottom row is all zeros, as you said. The rest of the reduced matrix says that
    c1 = c3
    c2 = -2c3
    c3 is arbitrary

    Since c3 is arbitrary, let's arbitrarily set it to 1, so c1 = 1, c2 = -2.
    Notice that 1*v1 - 2*v2 + 1*v3 = 0. Since the constants in this equation are not all zero, then we conclude that the vectors v1, v2, and v3 are linearly dependent.

    Another way to see this is to note that v1 = 2v2 - v3, so v1 is a linear combination of the other two. You can solve for any of the vectors in terms of the other two.
     
    Last edited: Oct 22, 2014
  4. Oct 22, 2014 #3

    kosovo dave

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    Gold Member

    I just caught that. I will try again and report my answer.
     
  5. Oct 22, 2014 #4

    kosovo dave

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    Gold Member

    Okay so I set the matrix up correctly this time (hopefully):
    1 1 1 0
    0 1 2 0
    1 2 k 0

    which still reduces to

    1 0 -1 0
    0 1 2 0
    0 0 k-3 0
     
  6. Oct 22, 2014 #5

    Mark44

    Staff: Mentor

    Check my post again - I made an edit and also added quite a bit.
     
  7. Oct 22, 2014 #6

    kosovo dave

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    Thanks for the quick and thorough response @Mark44 ! In general though, it's probably just easier to "jam" the given column vectors v1, v2, v3 together into a matrix, right?

    Also, did you immediately notice that v1 = 2v2 - v3? Or was it plugging values in for c1, c2, c3 that allowed you to see that?
     
  8. Oct 22, 2014 #7

    Mark44

    Staff: Mentor

    Yes, that makes more sense.
    No, I didn't notice that right away.
     
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