Subspace of a Vector Space over Complex Numbers Proof.

[ahsanxr]

Homework Statement

Let V = C (complex numbers). Prove that the only C-subspaces of V are V itself and {0}.

Homework Equations

The Attempt at a Solution

Well this problem has me confused since I have clearly found a complex subspace for example all the complex numbers of the form

{a+ib : a=b}

are closed under addition and scalar multiplication, hence it is a subspace. So if I've found another subspace in the complex numbers how can its only subspace be itself and the empty set?

 

[Timo]

I would assume that "C-subspace" is supposed to mean that the multiplication with a scalar in this case means multiplication with a complex number. None of the elements in your suggested subspace (except zero) map back on it when multiplied by i, for example.

 

[ahsanxr]

I see, that would make sense. And yes a complex number can be represented by R(cos(x) + i*sin(x)) but how will that allow me to prove that there are no other subspaces? I'm familiar with proving whether something is a subspace, but haven't done any where I'm to show that there cannot be any more subspaces.

 

[HallsofIvy]

a+ ai is NOT a subspace of the complex numbers, as a vector space over the complex numbers since it is not closed under scalar multiplication. (x+ iy)(a+ ai)= (ax-ay)+ (ax+ ay)i and $ax- ay\ne ax+ ay$.


What is a basis for the complex numbers as a vector space over the complex numbers? What is its dimension?

Remember that if V has dimension n, all subspaces must have dimension between 0 and n.


("a+ ai" is a subspace of the complex numbers as a vector space over the real numbers. What is the dimension of the complex numbers as a vector space over the real numbers?)