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Linear independence

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.

    2. Relevant equations


    3. The attempt at a solution
    We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
    Does there exist a non-trivial combination such that
    ##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
    ##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
    ##k_1+k_3 =0\Rightarrow k_1 = -k_3##
    ##k_1+k_2 =0##
    ##k_2+k_3 =0##

    Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.
     
    Last edited: May 15, 2015
  2. jcsd
  3. May 15, 2015 #2

    Mark44

    Staff: Mentor

    That works for me. (IOW, I agree that the three new vectors are linearly independent.)

    Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.

    The matrix looks like this, from your system:
    $$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
    After a few row operations, the final matrix is I3.
     
  4. May 15, 2015 #3

    Ray Vickson

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    Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
     
  5. May 15, 2015 #4
    Oh. Cramer's rule.
    ##k_n = \frac{D_{k_n}}{D}## and since the determinant of the system is non zero, the corresponding determinants for every ##k_n## would be 0 (a full column of 0-s means det = 0) and therefore ##k_1 = k_2 = k_3 = 0##
     
  6. May 15, 2015 #5

    Ray Vickson

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    No, I was not referring to Cramer's rule (which is rarely actually used when solving equations). I was referring to the theorem that if det(A) ≠ 0 then the n ×n system Ak = 0 has k = 0 as its only solution.
     
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