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Subspace of a Vector Space over Complex Numbers Proof.

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Aug 29, 2011 #2
    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.
     
  4. Aug 29, 2011 #3
    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.
     
  5. Aug 30, 2011 #4

    HallsofIvy

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    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 [itex]ax- ay\ne ax+ ay[/itex].


    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?)
     
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