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Subspace Addition Proof

  1. Jan 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Let [itex]L,M,N[/itex] be subspaces of a vector space [itex]V[/itex]

    Prove that

    [itex](L \cap M) + (L \cap N) \subseteq L \cap (M + N)[/itex]

    Give an example of subspaces [itex]L,M,N[/itex] of [itex]\mathbb{R}^2[/itex] where

    [itex](L \cap M) + (L \cap N) \neq L \cap (M + N)[/itex]


    2. Relevant equations



    3. The attempt at a solution

    Ok so, I can see how the LHS is a subset of the RHS I'm just having trouble showing that applying the intersection of the subspace L before adding the two subspaces M and N limits the resulting set. Also, I have shown the last part, that they're not equal, by using

    L = {(-3,2),(-1,1),(-2,3)}
    M = {(-1,1),(-4,3),(0,2)}
    N = {(-3,2),(-2,3),(8,0)}

    Which shows that they're not equal, but I don't know if this can be used because then L,M and N aren't subspaces of R2. So some guidance into what L,M and N can be used would be appreciated also.
     
  2. jcsd
  3. Jan 4, 2013 #2

    Dick

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    I'd be interesting in seeing how you proved the first part. And for the second part you are right that the sets you have shown aren't subspaces. You generally describe a subspace as an span of a set of vectors. Try that. You can pick L, M and N to all be 1-dimensional subspaces of R^2.
     
    Last edited: Jan 5, 2013
  4. Jan 5, 2013 #3
    Thanks, I've solved this now.

    One good counter example is to

    Let [itex]M = \left\{(x,0) | x \in \mathbb{R}\right\}[/itex]

    Let [itex]N = \left\{(0,y) | y \in \mathbb{R}\right\}[/itex]

    and let L be any line through the origin. Which gives you that [itex]L \cap (M + N)[/itex] is the set of all points on the line [itex]L[/itex]

    However,

    [itex](L \cap M) + (L \cap N) = \left\{(0,0)\right\} [/itex]


    As for the proving of the subspace.

    Let [itex]t \in (L \cap M) + (L \cap N)[/itex]

    Then we can write

    [itex]t = r + s[/itex]

    Where

    [itex]r \in (L \cap M)[/itex]

    and

    [itex]s \in (L \cap N)[/itex]

    and hence

    [itex]r \in L[/itex], [itex]r \in M[/itex], [itex]s \in L[/itex] and [itex]s \in N[/itex]

    Since L is a subspace

    [itex]r + s = t \in L[/itex]

    and

    [itex]r + s = t \in (M + N)[/itex]

    Hence

    [itex]t \in L \cap (M+N)[/itex]

    and we have shown that

    [itex](L \cap M) + (L \cap N) \subseteq L \cap (M + N)[/itex]
     
  5. Jan 5, 2013 #4

    Dick

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    Well done!
     
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