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Math Help - A basis for C^2

  1. #1
    Junior Member
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    A basis for C^2

    I am trying to show that

    (i,1) , (i,-1) is a basis for C^2

    Linear independence is alright.

    a(i,1) + b(i,-1) = (0,0)

    (ia+ib,a-b)=0

    ia + ib = 0
    a-b=0




    where the first equation implies a=b=0.

    I can't figure out how to show it spans C^2

    Let (a,b) \in C^2

    Find r,s \in C s.t.

    r(i,1)+s(i,-1) = (a,b)

    so we get the system of equations

    ri+si = a

    r-s= b

    I can kind of see how r,s could exist. But I don't know how to express them explicitly.

    Thank you for your help.



    OR EDIT:

    Since they are linearly independent, and dim C^2 = 2, it forms a basis.
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  2. #2
    MHF Contributor

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    Tejas
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    Re: A basis for C^2

    the first equation does NOT imply a = b = 0, it implies a+b = 0, that is b = -a.

    the second equation implies b = a, which TOGETHER with the first, implies a = -a, so 2a = 0, so a = 0/2 = 0.

    since b = a, b = 0 as well.

    your edit is perfectly good reasoning. but we can solve for r,s explicitly, as well:

    ri + si = a --> r + s = -ai. adding that to the second equation gives:

    2r = b-ai --> r = (b-ai)/2.

    then (b-ai)/2 - s = b, so

    s = (b-ai)/2 - b = (b-ai)/2 - 2b/2 = (-b-ai)/2
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