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Please help me on this proof. Span!

  1. Jun 27, 2009 #1
    The Problem:
    Let [tex]S_{1}[/tex] and [tex]S_{2}[/tex] be subsets of a vector space [tex]V[/tex]. Prove that [tex]span\left(S_{1}\cap S_{2}\right)\subseteq span\left(S_{1}\right)\cap span\left(S_{2}\right)[/tex].
    Give an example in which [tex]span\left(S_{1}\cap S_{2}\right)[/tex] and [tex]span\left(S_{1}\right)\cap span\left(S_{2}\right)[/tex] are equal and one in which they are unequal.

    Solution:
    I could do the proof, so that is not a problem. I found an example when they are equal to each other, but I can't think of an example that those two are not equal. It'd be nice if you could explain it in general case, but it is okay if you just give me an example. Please help me on this!
     
    Last edited: Jun 27, 2009
  2. jcsd
  3. Jun 27, 2009 #2
    Can you find two sets S_1, S_2 so that their intersection is empty, but they each span the whole space V ?

    By the way, doesn't [tex]S_1\cap S_2[/tex] look better than S_1[tex]\bigcap[/tex]S_2 ?
     
  4. Jun 27, 2009 #3
    test
     
    Last edited: Jun 27, 2009
  5. Jun 27, 2009 #4
    fixed
     
  6. Jun 27, 2009 #5
    Yes. so that means [tex]span\left(S_1\cap S_2)[/tex] equal to a zero set [tex]\left\{0\right\}[/tex]?

    And since both [tex]span(s_1)[/tex] and [tex]span(s_2)[/tex] are subspaces of V, they both have a zero set as well, so [tex]span(s_1)\cap span(s_2)[/tex] also should be a zero set. That meas they are equal. I didn't get your answer... Sorry
     
  7. Jun 27, 2009 #6
    Linear Algebra. spans!

    1. The problem statement, all variables and given/known data
    The Problem:
    Let [tex]S_{1}[/tex] and [tex]S_{2}[/tex] be subsets of a vector space [tex]V[/tex]. Prove that [tex]span\left(S_{1}\cap S_{2}\right)\subseteq span\left(S_{1}\right)\cap span\left(S_{2}\right)[/tex].
    Give an example in which [tex]span\left(S_{1}\cap S_{2}\right)[/tex] and [tex]span\left(S_{1}\right)\cap span\left(S_{2}\right)[/tex] are equal and one in which they are unequal.

    2. Relevant equations
    Nothing.


    3. The attempt at a solution
    Solution:
    I could do the proof, so that is not a problem. I found an example when they are equal to each other, but I can't think of an example that those two are not equal. It'd be nice if you could explain it in general case, but it is okay if you just give me an example. Please help me on this!
     
  8. Jun 27, 2009 #7
    Find [tex]S_1[/tex] and [tex]S_2[/tex] so that [tex]S_1 \cap S_2 = \emptyset[/tex], [tex]\text{span}(S_1) = V[/tex] and [tex]\text{span}(S_2) = V[/tex] .
     
  9. Jun 27, 2009 #8

    quasar987

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    Re: Linear Algebra. spans!

    Take V=R, S_1={1}, S_2={2}. Then S_1 n S_2 is empty, so span(S_1 n S_2)=0. And span(S_1)=R=span(S_2) so span(S_1) n span(S_2)=R.

    More generally, let S_1={v_1,...,v_k} be a basis for a subspace W of V and let S_2={u_1,...,u_k} be another basis for that subspace such that {v_1,...,v_k} n {u_1,...,u_k} is empty (for instance, u_i=2v_i). Then span(S_1 n S_2)=0, while span(S_1) n span(S_2)=W.

    The idea behind the example is of course that given a subspace, there are many distinct basis for it.
     
  10. Jun 27, 2009 #9
    Yeah. That's what i meant.

    Well, thanks.
     
  11. Jun 27, 2009 #10
    Re: Linear Algebra. spans!

    Thank you! =)
     
  12. Jun 27, 2009 #11

    Fredrik

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    I suggest that you consider subsets S1 and S2 of [itex]\mathbb R^3[/itex] that each contain exactly two vectors.
     
  13. Jun 27, 2009 #12

    Fredrik

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    Re: Linear Algebra. spans!

    Don't post the same question in two forums.
     
  14. Jun 28, 2009 #13

    HallsofIvy

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    I have merged the other thread with this one.
     
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