Subspace Addition Proof

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.

(edited 11 years ago)

Reply 2

OP

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.

Reply 3

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, ....

(edited 11 years ago)

Reply 4

OP

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?

Reply 5

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

Reply 6

OP

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)

(edited 11 years ago)

Reply 7

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.

Reply 8

OP

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)

Reply 9

Original post by Noble.

...

Sorted

Reply 10

OP

Original post by ghostwalker

Sorted

Many thanks! :smile:

Reply 11

OP

Original post by ghostwalker

Sorted

Sorry, I have another question. :tongue:

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?

Reply 12

Original post by Noble.

Sorry, I have another question. :tongue:

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?

Reply 13

OP

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)}?

(edited 11 years ago)

Reply 14

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...?

Reply 15

OP

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.

Reply 16

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?

Reply 17

OP

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.

Reply 18

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....

Reply 19

OP