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

Let be subspaces of a vector space

Prove that

Give an example of subspaces of where

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

To see that equality doesn't always hold, the only issue being that none of are actually subspaces of - 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!
2. (Original post by Noble.)
...
Hint: For your example, thinking geometrically, what does a subspace of look like. And see what you can do considering those.
3. (Original post by ghostwalker)
Hint: For your example, thinking geometrically, what does a subspace of look like. And see what you can do considering those.
Subspaces of 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.
4. (Original post by Noble.)
Subspaces of 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, ....
5. (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 so

Will be the set of all points on the line

However,

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

Also, in regards to proving the subset, do you have any hints?
6. (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?
7. Ok, I think I may have a proof for the subset.

If we have that

Then we can write

Where

and

and hence

, , and

Since L is a subspace

and

Hence

and we have shown that

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

8. (Original post by Noble.)

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

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.
9. (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.

Then we can write

Where

and

and hence

, , and

Since L is a subspace

and

Hence

and we have shown that

10. (Original post by Noble.)
...
Sorted
11. (Original post by ghostwalker)
Sorted
Many thanks!
12. (Original post by ghostwalker)
Sorted
Sorry, I have another question.

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

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

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

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?
14. (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)}?
15. (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...?
16. (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.
17. (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?
18. (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.
19. (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....
20. (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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