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Showing that a given set of vectors forms a basis

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    show that

    [tex]\left(\begin{array}{cc}2 & -1\\-1 & 1\end{array}\right)[/tex]

    forms a basis for R^2


    2. Relevant equations



    3. The attempt at a solution

    ok...my instructor said he wants me to show that they are linearly independant and to show that they span to form a basis...not just by a theorm

    im trying to show that the c_{1} and c_{2} are zero to say that they are but i get

    2c1-c2=0 and c2=c1 which does not work

    my instructor does not want me to do the determinant way

    i think i can show that it spans R^2 but i just need help showing that it is linear independant
     
  2. jcsd
  3. Nov 3, 2009 #2
    anyone?
     
  4. Nov 3, 2009 #3

    Dick

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    I think you mean to say that you want to show the two column vectors [2,-1] and [-1,1] are linearly independent. You probably don't want to write that as a 2x2 matrix. But are there any solutions to 2c1-c2=0 and c2=c1 that aren't c1=0 and c2=0? BTW don't bump your problem after only 20 minutes.
     
  5. Nov 3, 2009 #4
    sorry bout the bump...

    no...c1 or c2 must be zero to satisfy those equations, which would make them linearly independant?

    pls tell me if next part is right...

    if i say that 2c_1 - c2 = x1 and -c1+c2 = x2.....since this tells me how to get c's....this set spans R^2, thus forming a basis?
     
  6. Nov 3, 2009 #5

    Dick

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    You're forgiven for the bump, just give people time to collect their thoughts. And yes, they are linearly independent and yes, 2c_1 - c2 = x1 and -c1+c2 = x2 can be solved for any x1 and x2. So they span. I think you get it. Happy?
     
  7. Nov 3, 2009 #6
    almost =) perhaps you can take a gander at my other thread about the runga kutta method?

    thank you
     
  8. Nov 3, 2009 #7

    Dick

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    I'm with everyone else on that one. If you are going to do a numerical integration, I think you need a numerical initial value.
     
  9. Nov 3, 2009 #8
    thats the thing...i went to two different professors today and asked for help, one of which was the author of the book. he said its solveable and that you don't need an initial condition to solve, and another professor said that you do, because you can have so many different solutions on that x value that you need the y value to start
     
  10. Nov 3, 2009 #9

    Dick

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    I guess I would pay some attention to the author of the book. Do they just mean to set up the iteration algorithm, or are you actually supposed to run it? You can solve the problem exactly without using numerical techniques at all. Look, I'm really an amateur at numerical techniques, I plug in cookbook methods and use them.
     
  11. Nov 4, 2009 #10
    thats what im trying to do too. actually the instructions say to show work for the iterations, so do u know how i can start? once i get started, i should be able to finish it
     
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