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Proof with intersection of subspaces

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose L, M, and N are subspaces of a vector space.

    (a)
    Show that the equation
    [tex] L \cap (M+N) = (L \cap M)+(L \cap N) [/tex]
    is not necessarily true.

    (b)
    Prove that
    [tex] L \cap (M+(L \cap N))=(L \cap M) + (L \cap N) [/tex]

    2. Relevant equations
    N/A

    3. The attempt at a solution

    (a)
    I let
    [tex]
    \begin{aligned}
    M=&\;span\{(0,0),(1,0)\}\\
    N=&\;span\{(0,0),(0,1)\}\\
    L=&\;span\{(0,0),(1,1)\}
    \end{aligned}
    [/tex]

    Then,

    [tex]
    \begin{aligned}
    M+N=&\;span\{(0,0),(1,0),(0,1),(1,1)\}\\
    L \cap (M+N)=&\;span\{(0,0),(1,1)\}\\
    L \cap M=&\;span\{(0,0)\}\\
    L \cap N=&\;span\{(0,0)\}
    \end{aligned}
    [/tex]

    and the equation is not true.

    This in fact leads me to believe that the equation does not hold when [tex]L \subset (M+N) [/tex], because then [tex]L \cap (M+N) = L[/tex] and [tex]L \cap M[/tex] and [tex]L \cap N[/tex] are something else.
    I would guess they turn out to be something like [tex]L-L \cap N[/tex] and [tex]L-L \cap M[/tex], respectively..

    (b)

    [tex] L \cap M + L \cap (L \cap N) = (L \cap M) + (L \cap N) [/tex]

    That's all I can come up with on my own.
    Any suggestions are appreciated, thanks!
     
  2. jcsd
  3. Feb 6, 2010 #2

    vela

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    For (b), show that [itex]L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)[/itex] and [itex](L \cap M) + (L \cap N) \subset L \cap (M+(L \cap N))[/itex].

    For example, assume [itex]x \in L \cap (M+(L \cap N))[/itex]. Then [itex]x \in M+(L \cap N)[/itex], so you can write [itex]x = m + n[/itex] where [itex]m \in M[/itex] and [itex]n \in L \cap N[/itex]. If you can show that [itex]m \in L[/itex], then you'll have that [itex]m \in (L \cap M)[/itex] and consequently that [itex]x \in (L \cap M)+(L \cap N)[/itex], so [itex]L \cap (M+(L \cap N))\subset (L \cap M) + (L \cap N)[/itex].
     
  4. Feb 6, 2010 #3
    From our assumption that [tex]x\in L \cap (M+(L\cap N)) [/tex], we have that [tex]x\in L[/tex].
    Since [tex]x=m+n[/tex] we have that [tex]m\in L[/tex] and [tex]n \in L [/tex], so
    [tex] L \cap (M+(L\cap N)) \subset (L\cap M)+(L\cap N) [/tex].

    Is that it or do I have to show that [tex] (L\cap M)+(L\cap N)\subset L\cap (M+(L\cap N)) [/tex]?

    Suppose [tex] x \in (L\cap M)+(L\cap N) [/tex] and [tex]x=m+n[/tex] where [tex]m\in (L\cap M)[/tex] and [tex]n\in (L\cap N)[/tex].
    Then [tex]m\in L[/tex] and [tex]n\in L[/tex].
    I do not know how to proceed..

    Thank you!
     
  5. Feb 6, 2010 #4

    vela

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    Wait, how do you know that m is in L from knowing x is in L? M is not necessarily contained in L, so isn't it possible that m is in M but not in L. You need to explain why m must also be in L.

    You have to show both directions to prove equality.

    You know that [itex]m \in M[/itex] and [itex]n \in L\cap N[/itex], so it follows that [itex]x=m+n \in M+(L \cap N)[/itex]. If you can show that x is in L, you can conclude it's an element of [itex]L\cap (M+(L\cap N))[/itex].

    Hint: Remember that L, M, and N are subspaces.
     
  6. Feb 7, 2010 #5
    Since [tex]m+n \in L,\;n\in L[/tex] and [tex]L[/tex] is a subspace (closed under vector addition), we know that [tex]m \in L[/tex]?

    From [tex](L \cap N)[/tex] I know that [tex]n \in L[/tex], and from [tex](L \cap M)[/tex] I know that [tex]m \in L[/tex].
    [tex]m+n[/tex] must also be in [tex]L[/tex] since it is a subspace.
    Now, [tex]m+n \in (M+(L \cap N))[/tex] and [tex]m+n \in L[/tex] and so [tex]x[/tex] is an element of [tex]L \cap (M+(L\cap N))[/tex].

    Is this any good?

    Thanks!
     
  7. Feb 7, 2010 #6

    vela

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    Yes.
    Perfect.
     
  8. Feb 7, 2010 #7
    Thank you very much :)
     
  9. Feb 8, 2010 #8

    Mark44

    Staff: Mentor

    One other thing that I didn't notice coming up in this thread.
    There is no need to include (0,0) as a vector in your span{} sets. It's not incorrect to include it, but it needlessly clutters things up. You could have more simply said that
    M = span{(1, 0)}
    N = span{(0, 1)}
    L = span{(1, 1)}
    All of these are one-dimensional subspaces of R2, and for each the entire subspace is generated by scalar multiples of the given vector. Choose the scalar to be 0 for each, and you get the zero vector.
     
  10. Feb 8, 2010 #9
    That makes sense and looks prettier. Thanks.
     
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