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Vector spaces

  1. Jul 25, 2008 #1
    Would the space C(a,b) (where any element of the space is a continuous complex function) also be a space over the field R of real numbers since the field C has a subfield that is isomorphic to R?

    EDIT: I am thinking yes because all of the axioms that have to be satisfied in order for a set to be a vector space is satisfied if you have C(a,b) being a space over R.
    Last edited: Jul 25, 2008
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  3. Jul 25, 2008 #2


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    Yes, you certainly can multiply complex valued functions by real numbers and so you can define C(a,b) as a vector space over the real numbers.
  4. Jul 25, 2008 #3


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    good observation. now can you see how the dimension over R relates to the dimension over C?
  5. Jul 26, 2008 #4
    They're the same, roy: twice any infinite cardinal is still the same infinite cardinal. If I did smilies I'd put one in to show I was pulling your leg a little. But I don't do them.
  6. Jul 26, 2008 #5
    Looks like the question has already been answered. I was going to say that since a real space is infinite-dimensional, and that a complex space contains a real space, then the dimension of a complex space must also be infinite.

    EDIT: More specifically, since you can find an infinite amount of linearly independent vectors in a real space, then any amount of linearly independent vectors found in a complex space will be added to the amount of linearly independent vectors in the real space, which is infinite. So infinite plus anything is also infinite.
  7. Jul 27, 2008 #6
    There are distinct notions of infinite cardinals, so you shouldn't be so blase.
  8. Jul 27, 2008 #7
    I've been learning Linear Algebra for about a week now...I just answered the question with what I knew
  9. Jul 27, 2008 #8


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    good points all. now let me try to continue to make what is an interesting point for a learner. if V is a complex vector space with basis{xj}, what is a real basis for V?
  10. Jul 29, 2008 #9

    I am assuming by you asking "what is a real basis for v", you are asking what is a basis for V regarded as a real space. So, the dimension of a vector space is equal to the amount of basis vectors in that space. So, in a complex vector space with basis {xj} where I am assuming J is the amount of basis vectors, then J is also the dimension of the complex vector space, V. Now, the dimension of a real space is twice that of a complex space. So the dimension of the complex space regarded as a real space is 2j, so the basis vector would be {x2j}.
  11. Jul 29, 2008 #10


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    that does not answer the question. you have not said what ARE the extra vectors xJ+1,....,x2J.

    you are only saying how MANY vectors are in the new basis. to prove you are right you need to produce those vectors from the old ones explicitly.
  12. Jul 29, 2008 #11

    Just thinking out loud here...

    A complex space should be a subspace of a real space, so since the vectors for the complex basis is x1, x2,...,xj, then there are certain vectors, belonging only to the real space, that when combined with the complex basis form a new basis: x1,x2,...,xj, xj+1,...,x2j

    So now I have to explain what the new vectors, xj+1,...,x2j are. Doesn't that depend on what an element of that certain space is? For example, in a space Kn, an element of that space is any ordered n-tuple, whereas an element of the space R(a,b) is any continuous real function.

    If the element of the spaces you are asking about is any ordered n-tuple, then any element x of the real space can be represented by:

    x = c1(1,0,....,0) + c2(0,1,...0) +...+ cj (0,0,...,1) + cj+1(i,0,...,0) + ... + c2j(0,0,...i)

    where the numbers c1,c2,...,c2j are components of the vector x with respect to the basis.

    The basis for the complex space was x1,x2,...,xj. So, the basis for the real space would be x1,x2,...,xj, i(x1), i(x2),...,i(xj), where i(x1) = xj+1, i(x2) = xj+2 and so on
  13. Jul 29, 2008 #12
    That just isn't the way to say it: a subspace of a vector space V is by definition taken to be over the same field as V.
  14. Jul 29, 2008 #13

    Thanks for the correction. How would I say what I was trying to say?
    Last edited: Jul 29, 2008
  15. Jul 30, 2008 #14


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    well i agree with your answer: namely if x1,x2,.. is the complex basis then x1,ix1,x2,ix2,... is a real basis.

    and thats how you prove the real dimension is twice the complex dimension.
  16. Jul 31, 2008 #15
    Thanks for taking the time to help me learn.
  17. Jul 31, 2008 #16


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    thanks for the appreciation. my pleasure.
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