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I need an authoritative answer

  1. May 2, 2012 #1
    I thought the definition of a field was the set of all real numbers plus addition and multiplication (or whatever the particular set of operations are) and since its elements have no direction, by definition, they are not vectors; thus cannot be a vector space.

    (1) Am I wrong?

    (2) Can a field be a vector space?

    (3) Does the following statement make sense?

    A field is a vector space over itself with dimension 1.

    (4) Can a field have a subspace? (I thought subfields are to fields as subspaces are to vector spaces.

    Thanks!

    Joe
     
  2. jcsd
  3. May 2, 2012 #2
    Yes. [itex]\mathbb{R}[/itex] is an example of a field, but it is much more general than that. As far as direction goes...

    A vector space has a very specific definition that is much more general than the idea of magnitude and direction.
    http://mathworld.wolfram.com/VectorSpace.html
    The set of all real numbers fulfills all of those requirements and is therefore a vector space. Yes, it is a silly vector space, but it has to be one. Think of it this way
    [itex]\mathbb{R}^3[/itex] is a VS with 3 dimensions
    [itex]\mathbb{R}^2[/itex] is a VS with 2 dimensions
    [itex]\mathbb{R}^1=\mathbb{R}[/itex] is a VS with 1 dimension

    yes. The basis is [itex]\{a\}[/itex] where [itex]a[/itex] is any single element in the field.

    I think that properly, you shouldn't say this, but saying that the reals have no proper nontrivial subspace means

    a one dimensional vector space over the reals has no proper nontrivial subspace
     
  4. May 2, 2012 #3
    Thank you! Thank you! Thank you! Thank you!
     
  5. May 3, 2012 #4

    micromass

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    One thing you should know is that, formally, there is no such thing as a vector space. That is: if you want to have a vector space then you should always give what the underlying field is. If you do not, then the question is ambiguous.

    For example, [itex]\mathbb{R}[/itex] gives rise to a [itex]\mathbb{R}[/itex]-vector space, but also to a [itex]\mathbb{Q}[/itex]-vector space and several other vector spaces. These vector spaces are very different from each other.

    So, if you ask: is [itex]\mathbb{R}[/itex] a vector space, then this question is incomplete. You should ask: is [itex]\mathbb{R}[/itex] an [itex]\mathbb{R}[/itex]-vector space. The answer to this is yes.

    Every field F gives rise to an F-vector space, and as F-vector space it has dimension 1. Should a field F be a G-vector space over another field G, then it might be that its dimension is no longer 1.

    A field F, as an F-vector space, has exactly 2 subspaces: {0} and F itself. On the other hand, it might have many subfields. So the notion of subspace and subfield are not equal.
     
  6. May 5, 2012 #5
    ]I think you're having this confusion because you're using the definition of a vector as "a quantity with magnitude and direction". Unfortunately a different, more general definition of vector is used in higher mathematics. See this article for the axioms used to define a vector space.
     
  7. May 6, 2012 #6
    Thank you very much!!!
     
  8. May 6, 2012 #7
    Thank you very much, as well!
     
  9. May 8, 2012 #8
    Thank you!

    By the skin of my teeth, some help from you, and the grace of God, I received the best grade I could have expected in Linear Algebra.

    Thanks, again!

    Joe
     
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