Vector Subspace Proof (Please See Attached File)

[TheWrathOfMath]

Prove or disprove the following claim:
V is a vector space; U and W are vector subspaces of V. Let S be the group of all vectors is V which do not exist in the intersection of U and W, in addition to the zero vector. S is a vector space (over the same field, and with the same operations as in V).

 

[Zermelo]

First of all, 0 included you can write like {v…} U {0}.
Second of all, you need to try to find counterexamples, to try to disprove the claim, or to get a ‘feeling’ why it would be true.
If you don’t have time for that, I would try the following: How can you write v not in A intersect B? What does that mean?

 

[topsquark]

Prove or disprove the following claim:
V is a vector space; U and W are vector subspaces of V. Let S be the group of all vectors is V which do not exist in the intersection of U and W, in addition to the zero vector. S is a vector space (over the same field, and with the same operations as in V).

I am going to assume that V is a finite dimensional vector space over a field k.

Express V as the span of a basis. Then $V = span \{ e, ~ v_1, ~ v_2, \text{ ...} \}$.

Then let U and W be expressed in terms of basis vectors $U = span \{e, ~ v_{i_1}, ~ v_{i_2}, \text{ ...} \}$ and $W = span \{e, ~ v_{j_1}, ~ v_{j_2}, \text{ ...} \}$.

Now for these basis vectors we need:
For V: $a e + b v_1 + c v_2 + \text{ ...} = 0 \implies a = b = c = ... = 0$

For U: $d e + f v_{i_1} + g v_{i_2} + \text{ ...} = 0 \implies d = f = g = ... = 0.$

For W: $h e + m v_{j_1} + n v_{j_2} + \text{ ...} = 0 \implies h = m = n = ... = 0.$

Construct S in terms of the basis vectors of V, $S = span \{e, ~ v_{p_1}, ~ v_{p_2}, \text{ ...} \}$. So you need to prove that there exist q, r, s... in k such that
$q e + r v_{p_1} + s v_{p_2} + \text{ ...} = 0 \implies q = r = s = ... = 0.$

-Dan

Addendum: As Zermelo says, try this with a few (low dimensional) examples. The pattern should be clear.

 

[TheWrathOfMath]

I am going to assume that V is a finite dimensional vector space over a field k.

Express V as the span of a basis. Then $V = span \{ e, ~ v_1, ~ v_2, \text{ ...} \}$.

Then let U and W be expressed in terms of basis vectors $U = span \{e, ~ v_{i_1}, ~ v_{i_2}, \text{ ...} \}$ and $W = span \{e, ~ v_{j_1}, ~ v_{j_2}, \text{ ...} \}$.

Now for these basis vectors we need:
For V: $a e + b v_1 + c v_2 + \text{ ...} = 0 \implies a = b = c = ... = 0$

For U: $d e + f v_{i_1} + g v_{i_2} + \text{ ...} = 0 \implies d = f = g = ... = 0.$

For W: $h e + m v_{j_1} + n v_{j_2} + \text{ ...} = 0 \implies h = m = n = ... = 0.$

Construct S in terms of the basis vectors of V, $S = span \{e, ~ v_{p_1}, ~ v_{p_2}, \text{ ...} \}$. So you need to prove that there exist q, r, s... in k such that
$q e + r v_{p_1} + s v_{p_2} + \text{ ...} = 0 \implies q = r = s = ... = 0.$

-Dan

Addendum: As Zermelo says, try this with a few (low dimensional) examples. The pattern should be clear.

No, it is infinite, but I highly appreciate the time and effort put into this comment.

First of all, 0 included you can write like {v…} U {0}.
Second of all, you need to try to find counterexamples, to try to disprove the claim, or to get a ‘feeling’ why it would be true.
If you don’t have time for that, I would try the following: How can you write v not in A intersect B? What does that mean?

I already wrote: v ∉ U∩W

First of all, 0 included you can write like {v…} U {0}.
Second of all, you need to try to find counterexamples, to try to disprove the claim, or to get a ‘feeling’ why it would be true.
If you don’t have time for that, I would try the following: How can you write v not in A intersect B? What does that mean?

And I do not know whether or not this is true.
I wanted to attempt to prove it, and if I encounter difficulties, then I will probably know that this is a false claim.
But I am not sure about whether or not I am stuck because I do not know how to proceed, or because this is indeed a false claim.

 

[pka]

It is hard to read your handwriting.
If $\mathcal{V}$ is a vector space and $\mathcal{S}\subset\mathcal{V}$ the in order that $\mathcal{S}$ to be a subspace it must be the case that it contains the zero vector and is closed under vector addition and scalar multiplication. Can you verify those three conditions?


$$

 

[TheWrathOfMath]

It is hard to read your handwriting.
If $\mathcal{V}$ is a vector space and $\mathcal{S}\subset\mathcal{V}$ the in order that $\mathcal{S}$ to be a subspace it must be the case that it contains the zero vector and is closed under vector addition and scalar multiplication. Can you verify those three conditions?


$$

I know this. But I do not know how to prove the two last conditions (closed under vector addition and scalar multiplication).

 

[Deleted member 4993]

I know this. But I do not know how to prove the two last conditions (closed under vector addition and scalar multiplication).

Did you try proof by contradiction?

Did you mean verify instead of proof?

 

[TheWrathOfMath]

Did you try proof by contradiction?

Did you mean verify instead of proof?

Yes

 

[Deleted member 4993]

Yes

I asked 2 questions in response #7. I am assuming "yes" answer to both of my queries.

Excellent - please share your work.

 

[TheWrathOfMath]

I asked 2 questions in response #7. I am assuming "yes" answer to both of my queries.

Excellent - please share your work.

I did.
Didn't you see the attached file?

 

[Deleted member 4993]

I did.
Didn't you see the attached file?

You did NOT show any proof by contradiction!

 

[TheWrathOfMath]

You did NOT show any proof by contradiction!

Well, I am stuck.
I attempted to prove it, but I am stuck.
My question is whether or not this claim is even true to begin with.
Perhaps I am stuck because this is a false claim...

 

[Steven G]

Note that the intersection of two v.s. is another vector space. So I don't think that you even need to consider that you are dealing with an intersection of two vector spaces if you are looking for a contradiction.
Also note that if V is a v.s. then V intersect {e} is a v.s.

 

[Steven G]

Look at any v.s. V and find a subspace, S, of V. Is V \ S a v.s.??

 

[TheWrathOfMath]

Note that the intersection of two v.s. is another vector space. So I don't think that you even need to consider that you are dealing with an intersection of two vector spaces if you are looking for a contradiction.
Also note that if V is a v.s. then V intersect {e} is a v.s.

Have you seen the attached file? I wrote on it: " probably need to use the fact that the the vectors u and v which exist in S do not exist in the intersection of U and W, which is a known vector space.

I do not know how to proceed, however.

Note that the intersection of two v.s. is another vector space. So I don't think that you even need to consider that you are dealing with an intersection of two vector spaces if you are looking for a contradiction.
Also note that if V is a v.s. then V intersect {e} is a v.s.

This is a false statement, isn't it?
That is why I struggle to prove it.
Could you please help me find a counterexample?

 

[Steven G]

In the other forum you posted in you were already told the validity of the statement.

 

[TheWrathOfMath]

In the other forum you posted in you were already told the validity of the statement.

Alright, but the problem is that I cannot find a counterexample.

 

[Steven G]

No one here or at the other forum will give you a counter example. This is a help forum where we help students with their problems. We never do their problems for them.
Please show us your work from the big hint from the other website.
Did you pick a vector space, V, and find a two subspaces, U and W for this vector space? Did you look at S\(U int V) U {0}?
Show us one or two example with all your work and then we can give you a hint towards finding your counter example