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Vector Space Basis

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find a basis for (1, a, a^2) (1, b, b^2) (1, c, c^2)

    2. Relevant equations



    3. The attempt at a solution

    M(1, a, a^2) + N(1, b, b^2) + K(1, c, c^2) = (0, 0, 0)

    M + N + K = 0
    Ma + Nb + Kc = 0
    Ma^2 + Nb^2 + Kc^2 = 0

    This is as far as I got. I tried monkeying around with these 3 equations to no avail.
     
    Last edited by a moderator: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2
    I don't understand the question: you are asked to find a basis for a collection of 3 vectors?
     
  4. Oct 5, 2011 #3

    Mark44

    Staff: Mentor

    I don't understand the question, either. Please give us the exact wording of the problem. Also, tell us what kinds of things a, b, and c are.
     
  5. Oct 5, 2011 #4
    Show that for any three distinct values a,b,c, the vectors (1,a,a2), (1,b,b2), (1,c,c2) are linearly independent.
     
  6. Oct 5, 2011 #5
    THanks for all your help
     
  7. Oct 5, 2011 #6

    Mark44

    Staff: Mentor

    Now, the work you showed in the OP begins to make sense. Your problem statement (shown below) is completely misleading. The objective is not to find a basis at all.
    So here you have a system of three equations in three unknowns, M, N, and K. You must have some experience in solving systems like this.
     
  8. Oct 5, 2011 #7
    Yea sorry that was a different problem...

    I don't know how to solve a system like that above^.
     
  9. Oct 5, 2011 #8

    Mark44

    Staff: Mentor

    I suspect that you actually do.
    Can you solve a system like this?
    x + 2y + z = 3
    2x - y + 3z = 4
    x - 3y + z = 2
    I'm not asking you to solve this system - I'm just asking if you know how.
    For this system, the variables are x, y, and z. For your system, the variables are M, N, and K.

    If you really don't know how to solve a system of equations, then I don't see how you are going to show that the given vectors are linearly independent.
     
  10. Oct 5, 2011 #9
    Yeah I can do that,^ but normally I just monkey around with them, so in that case I would do something like multiply the 1st by -2 and add that to the second and then substitute, etc...

    I tried employing that method to this problem, and it didn't work.
     
  11. Oct 5, 2011 #10

    Mark44

    Staff: Mentor

    Show us what you tried.
     
  12. Oct 5, 2011 #11
    Can someone please just show me how to solve it. I don't know how to solve it when the variables are general and not actual numbers. Normally I just get rid of one variable by mult/subtracting equations... but it won't work in this case. Please help!
     
  13. Oct 5, 2011 #12
    please help...
     
  14. Oct 5, 2011 #13
    OMG someone please!! I'm like crying right now
     
  15. Oct 5, 2011 #14

    Mark44

    Staff: Mentor

    Yes, it will work in this case.

    M + N + K = 0
    Ma + Nb + Kc = 0
    Ma^2 + Nb^2 + Kc^2 = 0

    Remember, the variables are M, N, and K.

    To get rid of the first variable in the 2nd equation, add -a times the first equation to the second.

    M + N + K = 0
    (b-a)N +(c-a)K = 0

    Do something similar to eliminate the M term from the 3rd equation.

    After this, add some multiple of the 2nd equation to the 3rd equation, to get rid of the N term from the third equation. At this point, your 3rd equation will involve only K, so you can solve for it.

    Knowing K, you can substitute into the 2nd equation and solve for N.

    Finally, knowing K and N, you can substitute into the 1st equation to get M.
     
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