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    Ok, I've had this question bugging me for two days now so help would be appreciated.

    Let L,M,N be subspaces of a vector space V

    Prove that

    (L \cap M) + (L \cap N) \subseteq L \cap (M + N)

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

    (L \cap M) + (L \cap N) \neq L \cap (M + N)



    Firstly, I can see how the equation is easily true and used

    L = \left\{(-3,2),(-1,1),(-2,3)\right\}

M = \left\{(-1,1),(-4,3),(-7,2)\right\}

N = \left\{(-3,2),(-2,3),(6,-1)\right\}

    To see that equality doesn't always hold, the only issue being that none of L,M,N are actually subspaces of \mathbb{R}^2 - so I'm pretty sure that can't be used to show equality doesn't always hold.

    Also I need some help proving the LHS is a subspace of the RHS.

    Thanks!
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    (Original post by Noble.)
    ...
    Hint: For your example, thinking geometrically, what does a subspace of \mathbb{R}^2 look like. And see what you can do considering those.
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    (Original post by ghostwalker)
    Hint: For your example, thinking geometrically, what does a subspace of \mathbb{R}^2 look like. And see what you can do considering those.
    Subspaces of \mathbb{R}^2 will be lines going through the origin? I tried using a similar idea yesterday, but I'm pretty sure I'm confusing the idea since lines through the origin are only going to have the origin as a point common to them all so couldn't see how it could be used to show LHS != RHS.
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    (Original post by Noble.)
    Subspaces of \mathbb{R}^2 will be lines going through the origin? I tried using a similar idea yesterday, but I'm pretty sure I'm confusing the idea since lines through the origin are only going to have the origin as a point common to them all so couldn't see how it could be used to show LHS != RHS.
    Let M={(x,0)| x real}

    Let N={(0,y)| y real}

    Let L be any other line through the origin.

    Consider, what's M+N, ....
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    (Original post by ghostwalker)
    Let M={(x,0)| x real}

    Let N={(0,y)| y real}

    Let L be any other line through the origin.

    Consider, what's M+N, ....
    Ok, so M+N will be the whole of \mathbb{R}^2 so

    L \cap (M + N)

    Will be the set of all points on the line L

    However,

    (L \cap M) + (L \cap N) = \left\{(0,0)\right\}

    I think that's correct, let me know if it's not.

    Also, in regards to proving the subset, do you have any hints?
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    (Original post by Noble.)
    Also, in regards to proving the subset, do you have any hints?
    Standard method for showing A is a subset of B. Let x by an element of the LHS, and show that it is an element of the RHS.

    Further hint:
    Spoiler:
    Show

    If s is an element of the LHS, what form can it be written in?
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    Ok, I think I may have a proof for the subset.

    If we have that

    (a,b) \in (L \cap M) + (L \cap N)

    Then we can write

    (a,b) = (c,d) + (e,f)

    Where

    (c,d) \in (L \cap M)

    and

    (e,f) \in (L \cap N)

    and hence

    (c,d) \in L, (c,d) \in M, (e,f) \in L and (e,f) \in N

    Since L is a subspace

    (c,d) + (e,f) = (a,b) \in L

    and

    (c,d) + (e,f) = (a,b) \in (M + N)

    Hence

    (a,b) \in L \cap (M+N)

    and we have shown that

    (L \cap M) + (L \cap N) \subseteq L \cap (M + N)



    Which could be expanded to show it's true in any dimension, would that be better?

    i.e. instead of using (a,b) use (a_1, a_2, ..., a_n)
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    (Original post by Noble.)

    Which could be expanded to show it's true in any dimension, would that be better?

    i.e. instead of using (a,b) use (a_1, a_2, ..., a_n)
    In essence your proof is correct. BUT, V is any old vector space, not necessarily even finite dimensional, or over the field R.

    You don't need to specify your chosen vector in terms of co-ordinates, you can just say. Let t be an element of LHS, then t can be written in the form r+s, where r...., s.... etc.
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    (Original post by ghostwalker)
    In essence your proof is correct. BUT, V is any old vector space, not necessarily even finite dimensional, or over the field R.

    You don't need to specify your chosen vector in terms of co-ordinates, you can just say. Let t be an element of LHS, then t can be written in the form r+s, where r...., s.... etc.
    Yeah, so this instead

    t \in (L \cap M) + (L \cap N)

    Then we can write

    t = r + s

    Where

    r \in (L \cap M)

    and

    s \in (L \cap N)

    and hence

    r \in L, r \in M, s \in L and s \in N

    Since L is a subspace

    r + s = t \in L

    and

    r + s =  t \in (M + N)

    Hence

    t \in L \cap (M+N)

    and we have shown that

    (L \cap M) + (L \cap N) \subseteq L \cap (M + N)
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    (Original post by Noble.)
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    (Original post by ghostwalker)
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    Many thanks!
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    (Original post by ghostwalker)
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    Sorry, I have another question.

    Can the same counter example for L,M,N be used to disprove

    L + (M \cap N) = (L + M) \cap (L + N)

    Which I'm thinking gives L on the LHS and only the set with (0,0) on the RHS?
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    (Original post by Noble.)
    Sorry, I have another question.

    Can the same counter example for L,M,N be used to disprove

    L + (M \cap N) = (L + M) \cap (L + N)

    Which I'm thinking gives L on the LHS and only the set with (0,0) on the RHS?
    You can use it as a counter-example to that equation, however your interpretation is not correct.

    For the LHS, you're correct, but the right does not equal {(0,0)}, assuming you're still operating in R^2. Have another think. What's L+M?
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    (Original post by ghostwalker)
    You can use it as a counter-example to that equation, however your interpretation is not correct.

    For the LHS, you're correct, but the right does not equal {(0,0)}, assuming you're still operating in R^2. Have another think. What's L+M?
    With M being the x-axis and L some random line, is

    L+M = {(x,0) + (x,ux)} = {(x,ux)}?
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    (Original post by Noble.)
    With M being the x-axis and L some random line, is

    L+M = {(x,0) + (x,ux)} = {(x,ux)}?
    The problem is that you've used the same variable for the point in L and the point in M, which artifically restricts your choices.

    L+M = {(x,0) + (y,uy)} = {(x+y,uy)} which equals...?
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    (Original post by ghostwalker)
    The problem is that you've used the same variable for the point in L and the point in M, which artifically restricts your choices.

    L+M = {(x,0) + (y,uy)} = {(x+y,uy)} which equals...?
    Not sure of the direction you're going in to be honest. So,

    L+N = {(y,uy) + (0,q)} = {(y,uy+q)}

    But I'm not really sure of what the set is when you intersect these two.
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    (Original post by Noble.)
    Not sure of the direction you're going in to be honest. So,

    L+N = {(y,uy) + (0,q)} = {(y,uy+q)}

    But I'm not really sure of what the set is when you intersect these two.
    You need to be clear what L+M is first. Choose a value for u, and plot some points if necessary. What points are feasible?
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    (Original post by ghostwalker)
    You need to be clear what L+M is first. Choose a value for u, and plot some points if necessary. What points are feasible?
    For a fixed y it's going to be a horizontal line. For a fixed x it will be lines through x with gradient u. Sorry if this isn't the direction you're trying to push me towards.
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    (Original post by Noble.)
    For a fixed y it's going to be a horizontal line. For a fixed x it will be lines through x with gradient u. Sorry if this isn't the direction you're trying to push me towards.
    It's a good start.

    Both of those are correct, and define points in L+M.

    Let's work with the first one - "For a fixed y it's going to be a horizontal line".
    OK, but y isn't fixed, it could be anything. So, say it takes one value, then a different value, then another, ..., and all these lines are in L+M, hence L+M is....
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    (Original post by ghostwalker)
    It's a good start.

    Both of those are correct, and define points in L+M.

    Let's work with the first one - "For a fixed y it's going to be a horizontal line".
    OK, but y isn't fixed, it could be anything. So, say it takes one value, then a different value, then another, ..., and all these lines are in L+M, hence L+M is....
    Ah right, so is L+M = R^2?

    Am I right in saying L+N is also R^2? So the RHS is R^2 with L on the LHS giving a counter-example?
 
 
 
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