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Homework Help: [Linear Algebra] Linear Independence

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Show if S = {v1,v2,v3} is independent or dependent . . .

    2. Relevant equations

    (0,0,0,0) = k1(a,b,c,d) + k2(e,f,g,h) + k3(i,j,k,l) where {a,b,c,d,e,f,g,h,i,j,k,l [itex]\in[/itex] ℝ}

    3. The attempt at a solution

    im trying to tell if i can say that this set of 3 vectors in R4 is dependent right away, just because there are less vectors than the number of dimensions. . . i looked in the book and the opposite is true -- if you have 4 vectors in R3, then the set is linearly dependent . . .

  2. jcsd
  3. Dec 7, 2011 #2
    otherwise, do i just row reduce?
  4. Dec 7, 2011 #3
    I row reduced and got:

  5. Dec 7, 2011 #4


    Staff: Mentor

    Are the vectors in the problem <a, b, c, d> etc. or are there any constants involved? The matrix you show in another post makes me think there are some numbers.

    If you have 4 vectors in R3, the set will always be linearly dependent, but when there are fewer vectors than the dimension of the space they're in, you can't say anything without doing some investigation.

    If you have 3 vectors in R4, the vectors might be linearly independent or linearly dependent - it depends on the vectors.

    For example, if the set is { <1, 0, 0, 0>, <4, 0, 0, 0>, <0, 1, 0, 0> }, this set of vectors is clearly linearly dependent - the 2nd vector is a scalar multiple of the first.

    If the set is { <1, 0, 0, 0>, <0, 1, 0, 0>, <1, 1, 0, 0> }, this set of vectors is also linearly dependent, but it's not so obvious, as no one vector is a multiple of any other. However, the 3rd vector is the sum of the other two.

    If the set is { <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 2, 0> }, it can be shown that this set of vectors is linearly independent. Your textbook probably shows some examples of testing sets of vectors for linearly independence/dependence.
  6. Dec 7, 2011 #5
    the problem has specific numbers in place of all of those letters that i used for an example.

    my problem is of the type:

    i row reduced, and got the final matrix posted (poorly) above . . .

    my main thing is trying to learn the concepts of how to solve these problems, and what everything means -- computationally i am quite comfortable of how to set up a problem and solve it, but i need to make sure that im setting things up the right way so that i can show what i am asked to show. . .
  7. Dec 7, 2011 #6


    Staff: Mentor

    The computations are the easy part. Understanding why you're doing what you're doing is harder, since you can't operate on autopilot.

    Keep in mind that your matrix represented the vector equation you had in post #1, and you were solving for the constants k1, k2, and k3. The reduced matrix you showed tells you the value of those constants.

    There's a little missing though. Your vector equation can be represented as an augmented matrix with a column of zeroes on the right - like this: [A|0]. After doing row reduction, you got a different matrix, so the augmented matrix would look like this: [B|0], with B being a matrix with a few 1's as leading entries.

    What that augmented matrix is telling you is that k1 = k2 = k3 = 0, which should tell you something about the vectors and whether they're linearly independent.
  8. Dec 7, 2011 #7
    also, if i have 2 matricies, A and B, where one of them is the zero vector. to show that they are linearly independent i need to show that no combination of them will be 0.

    so with (essentially), A + [0,0; 0,0] = A ≠ 0 is that all i need to say to show that this set of vectors, S = {A,B}, is linearly independent? or do i have to prove it in a more formal way?

  9. Dec 7, 2011 #8


    Staff: Mentor

    No set with the 0 vector/matrix in it can possibly be linearly independent.
    Think about it in relation to the definition of linear independence. The equation c1A + c2B = 0 has a solution for the constants other than the trivial solution c1 = c1 = 0. This is very easy to demonstrate.
  10. Dec 7, 2011 #9
    this makes a ton of sense, but i really dont know how to go about showing it, like im really really bad with proofs
  11. Dec 7, 2011 #10


    Staff: Mentor

    Show it by coming up with some numbers for the two constants. I can think of 4,598 solutions just off the top of my head.
  12. Dec 7, 2011 #11
    so thats all you have to do, is show a few examples with different constants?

    sorry for being a dum bass
  13. Dec 7, 2011 #12


    Staff: Mentor

    All you have to do is show two numbers, not both 0.
  14. Dec 7, 2011 #13
    yeah, i think i got it, i set c1 = 0 and c2 as like 5, and showed that in that case the statement was true, and since it isnt the trivial solution, then the sets are dependent.

    thank you very much for your responses and time!
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