Proving Subspace Addition: LHS vs RHS

[harvesl]

Homework Statement

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)

Homework Equations

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.

 

[Dick]

Homework Statement

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)

Homework Equations

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.

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.

 

[harvesl]

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.

Thanks, I've solved this now.

One good counter example is to

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

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

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

However,

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


As for the proving of the subspace.

Let 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)

 

[Dick]

Well done!