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Linearly dependent vectors

  1. Dec 9, 2004 #1
    I have a couple of questions on this subject that I need help with.

    1. Let v1, v2 and v3 be three linearly dependent vectors. Prove or disprove that the following vectors are also linearly dependent:
    v1+v2, v1+v3, v2+v3

    2. Let S = {v1, v2, v3, v4, v5} be a set of five vectors in a vector space V over a field F. Prove or disprove that if every subset T of S so that T!=S is linearly independent, S is also linearly idependent.
     
  2. jcsd
  3. Dec 9, 2004 #2

    Hurkyl

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    1. Start with the definitions.

    2. There probably isn't anything special about 5. Try a simpler problem first.
     
  4. Dec 9, 2004 #3
    I really don't know what to do with 1. If I did I wouldn't come here and ask about it.

    This is a good counter example for the 2nd question, right?
    (1, 0, 0, 0)
    (0, 1, 0, 0)
    (0, 0, 1, 0)
    (0, 0, 0, 1)
    (1, 1, 1, 1)

    Also there's another problem I need help with.
    [tex]A = \left(\begin{array}{ccc}1&1&1\\1&1&1\\a&b&c\end{array}\right)[/tex]
    Under which conditions for a, b and c are A and At row equivalent?
    I tried bringing both matrices to their canonical form (echelon form), but I get lost with all the different cases.
    [tex]A^t = \left(\begin{array}{ccc}1&1&a\\1&1&a\\1&1&c\end{array}\right)[/tex]
     
    Last edited: Dec 9, 2004
  5. Dec 9, 2004 #4

    Hurkyl

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    1. And I'm telling you what to do... start with the definitions. :tongue:

    2. Your counterexample looks good.

    3. What do you mean by row equivalent? That you can convert between A and A' by row operations?
     
  6. Dec 9, 2004 #5
    Yup. Our hunch is that a=b=1, and c is free to be whatever it wishes to be, except for a dog or a racoon maybe.
     
    Last edited: Dec 9, 2004
  7. Dec 9, 2004 #6
    Honestly, that's the first thing I did. Here's what I did and where I got stuck.

    We know that:
    [tex]\alpha V_1 + \beta V_2 + \gamma V_3 = 0[/tex]
    And not all of the scalars are zero.

    We need to show that:
    [tex]x(V_1 + V_2) + y(V_2 + V_3) + z(V_1 + V_3) = 0[/tex]
    [tex](x + z)V_1 + (x + y)V_2 + (y + z)V_3 = 0[/tex]
    When not all scalars are zero.

    So what I did was create this system of equations:
    [tex]\alpha = x + z[/tex]
    [tex]\beta = x + y[/tex]
    [tex]\gamma = y + z[/tex]
    And then I got stuck.

    I also tried saying that [tex]V3 = aV_1 + bV_2[/tex] and didn't get much further.
     
  8. Dec 9, 2004 #7

    Hurkyl

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    Ok, so you already figured out the part I was hinting. :tongue: You're just blocking on the easy part -- all you have to do is show whether or not that system of equations has a solution, right?
     
  9. Dec 9, 2004 #8

    Hurkyl

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    3. It might help to look for bases for the row spaces of A and A'.
     
  10. Dec 9, 2004 #9
    Ok, I solved and got:
    [tex]x = \frac{\alpha - \beta + \gamma}{2}[/tex]
    [tex]y = \frac{-\alpha + \beta + \gamma}{2}[/tex]
    [tex]z = \frac{\alpha + \beta - \gamma}{2}[/tex]
    And now how do I use the fact that at least one of the scalars is different than zero? :cry:

    Oh wait... I'm done, aren't I?

    Yeah, but we didn't study bases yet, so I can't use it to prove anything in the homework... not that I would know how to find them anyway. :tongue2: What I did try to do is show that the span of the rows of each matrix are equivalent, but that road got too complicated as well.
     
  11. Dec 9, 2004 #10

    Hurkyl

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    1. Sure looks like it!

    3. Okay... maybe try starting simple. You know a few vectors in the row space of A... how can you make them out of the rows of A'?
     
  12. Dec 9, 2004 #11
    Hmm... Let me think about that a little while (i.e tomorrow when I'm more awake!). :smile:

    In the mean time, I solved another question but I'm not sure the solution is correct. Would you be so kind to verify it?

    Let A and B be non-zero matrices so that AB = 0. Show that if C is row equivalent to A, then there exists a non-zero matrix D so that CD = 0.

    What I did was say that since C is row equivalent to A we can write it as:
    [tex]C = E_n...E_2E_1A[/tex]
    (where E is an elementary matrix)
    So if we multiply both sides of the equations by B we get:
    [tex]CB = E_n...E_2E_1AB = E_n...E_2E_1(0) = 0[/tex]
    So I found a D... and it even equals B! :tongue: Is this correct?
     
  13. Dec 9, 2004 #12

    Hurkyl

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    4. seems reasonable.
     
  14. Dec 9, 2004 #13
    Right, I see what you mean. I need to be able to express every row in A as a linear combination of the rows of A', yes?

    So for example:
    [a b c] = x[1 1 a] + y[1 1 b] + z[1 1 c]
    Which means that a=b, right? But what about c?
     
  15. Dec 10, 2004 #14
    Ok so I know that a=b. Now I did it with rows from A' and the row space of A:

    [1 1 c] = x[1 1 1] + y[a b c] = x[1 1 1] + y[a a c]

    And I got a system of equations:
    1 = x + ya
    1 = x + ya
    c = x + yc

    The ranked system matrix is:
    [tex]A = \left(\begin{array}{cc|c}1&a&1\\0&c-a&c-1\end{array}\right)[/tex]
    Can you please help with that? I need to find when the system to has at least one solution. I think that it's:
    c!=1: c!=a
    c=1: any a
    Am I correct?
     
  16. Dec 10, 2004 #15

    Hurkyl

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    Sounds reasonable.
     
  17. Dec 10, 2004 #16
    Of course it does, I wouldn't do anything unreasonbale now would I? :approve: I just want to make sure it's also correct.
     
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