Vector Subspace Proof (Please See Attached File)

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TheWrathOfMath

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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).
 

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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?
 
TheWrathOfMath said:
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).
Click to expand...
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 [imath]V = span \{ e, ~ v_1, ~ v_2, \text{ ...} \}[/imath].

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

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

For U: [imath]d e + f v_{i_1} + g v_{i_2} + \text{ ...} = 0 \implies d = f = g = ... = 0.[/imath]

For W: [imath]h e + m v_{j_1} + n v_{j_2} + \text{ ...} = 0 \implies h = m = n = ... = 0.[/imath]

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

-Dan

Addendum: As Zermelo says, try this with a few (low dimensional) examples. The pattern should be clear.
 
topsquark said:
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 [imath]V = span \{ e, ~ v_1, ~ v_2, \text{ ...} \}[/imath].

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

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

For U: [imath]d e + f v_{i_1} + g v_{i_2} + \text{ ...} = 0 \implies d = f = g = ... = 0.[/imath]

For W: [imath]h e + m v_{j_1} + n v_{j_2} + \text{ ...} = 0 \implies h = m = n = ... = 0.[/imath]

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

-Dan

Addendum: As Zermelo says, try this with a few (low dimensional) examples. The pattern should be clear.
Click to expand...
No, it is infinite, but I highly appreciate the time and effort put into this comment.

Zermelo said:
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?
Click to expand...
I already wrote: v ∉ U∩W

Zermelo said:
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?
Click to expand...
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.
 
It is hard to read your handwriting.
If [imath]\mathcal{V}[/imath] is a vector space and [imath]\mathcal{S}\subset\mathcal{V}[/imath] the in order that [imath]\mathcal{S}[/imath] 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?


[imath][/imath]
 
pka said:
It is hard to read your handwriting.
If [imath]\mathcal{V}[/imath] is a vector space and [imath]\mathcal{S}\subset\mathcal{V}[/imath] the in order that [imath]\mathcal{S}[/imath] 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?


[imath][/imath]
Click to expand...
I know this. But I do not know how to prove the two last conditions (closed under vector addition and scalar multiplication).
 
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.
 
Look at any v.s. V and find a subspace, S, of V. Is V \ S a v.s.??
 
Steven G said:
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.
Click to expand...
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.

Steven G said:
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.
Click to expand...
This is a false statement, isn't it?
That is why I struggle to prove it.
Could you please help me find a counterexample?
 
In the other forum you posted in you were already told the validity of the statement.
 
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
 
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