# Thread: A basis for C^2

1. ## 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.

OR EDIT:

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

2. ## 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