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Thread: A basis for C^2

  1. #1
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    A basis for C^2

    I am trying to show that

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

    Linear independence is alright.

    $\displaystyle a(i,1) + b(i,-1) = (0,0)$

    $\displaystyle (ia+ib,a-b)=0$

    $\displaystyle ia + ib = 0$
    $\displaystyle a-b=0$




    where the first equation implies $\displaystyle a=b=0$.

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

    Let $\displaystyle (a,b) \in C^2$

    Find $\displaystyle r,s \in C$ s.t.

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

    so we get the system of equations

    $\displaystyle ri+si = a$

    $\displaystyle r-s= b$

    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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    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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