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Show this is a basis

  1. Nov 17, 2013 #1
    Let β be a basis for ℝ over Q (the set of all rational numbers) and let a[itex]\in[/itex]ℝ, a≠1.

    Show that aβ={ay|y[itex]\in[/itex]β} is a basis for ℝ over Q for all a≠0.

    So I need to show (1) Linear independence, and (2) spanning. I am a little confused, especially because the dimension for the vector space is uncountably infinite.
    Here is what I have for linear independence:

    I took a finite subset of aβ, {az|z[itex]\in[/itex]S[itex]\subset[/itex]β, |S|<|ℝ|} (not sure if I wrote that correctly), so if

    [itex]\sum_{z\in S}[/itex]az=0

    then z must be zero since a≠0 (initial assumption). Therefore, it follows that aβ is linearly independent.

    Is this correct?
    I don't know how to approach spanning.

    Thanks in advance.
     
  2. jcsd
  3. Nov 18, 2013 #2

    Mark44

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    This seems like armwaving to me.
    Do you have a sense of what this problem is about? In particular, what a basis over Q that spans R might look like? For example, can you list a few of the basis elements of such a basis?

    Given that β is some basis, you need to show that aβ is also a basis, where a ≠ 0.

    To show that any set is a basis for some space, you need to do two things:
    1. Show that the set of basis elements is linearly independent.
    2. Show that the set of basis elements spans the space.

    The latter requirement means that if x ##\in## R, then x can be written as a linear combination of the basis elements.
     
  4. Nov 18, 2013 #3
    Honestly, I can't fully grasp it intuitively. I know how a basis is for ℝ with n-dimension, but not ℝ.
    I just can't see how something can span ℝ, is the standard basis applicable here?

    If I can't show linear independence/spanning the usual way, how should I do this?


    Thanks.
     
  5. Nov 18, 2013 #4

    vela

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    You do show independence the usual way. You need to show that for any finite subset ##\{v_1,v_2,\dots,v_n\} \subset a\beta## that
    $$q_1 v_1 + q_2 v_2 + \cdots + q_n v_n = 0$$ implies ##q_1=q_2=\cdots=q_n=0##, where ##q_i \in \mathbb{Q}##. What you wrote above doesn't show this. For one thing, there's no mention of the ##q_i##'s.
     
  6. Nov 18, 2013 #5
    But a basis for ℝ over Q is uncountable, so how can I take a finite, countable subset?
     
  7. Nov 18, 2013 #6

    Dick

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    Because that's the definition of linear independence. To show it's linearly independent you need to show every finite subset is linearly independent.
     
  8. Nov 18, 2013 #7
    Okay, correct me if I'm wrong, but if I take a finite subset, I thought it won't be countable? Like I can't order the terms...
     
  9. Nov 18, 2013 #8

    vela

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    Finite implies countable.
     
  10. Nov 18, 2013 #9
    So if I take a finite subset of ℝ, then I can order the terms? Sorry, I just want to clarify.
     
  11. Nov 18, 2013 #10

    vela

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    What's the precise definition of finite? If you know this basic definition, you should be able to answer your own question.
     
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