Prove that the intersection of subspaces is subspace

  • #1
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Homework Statement


Prove that the intersection of any collection of subspaces of V is a subspace of V.

Okay, so I had to look up on wiki what an intersection is. To my understanding, it is basically the 'place' where sets or spaces 'overlap.'

I am not sure how to construct the problem in the language of math.

If U1,...,Un are subspaces of V then I have show that their intersection includes the additive ID and is closed under addition and scalar multiplication.

can someone give me a kick in the right direction?
 

Answers and Replies

  • #2
Dick
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If an element 'x' is in all of the U1...Un, then it is in the intersection (call it U). And if 'x' is in U then it is in all of the U1...Un. Suppose 'x' is the additive identity. What can you conclude? The other properties of the subspace U follow from similar arguments.
 
  • #3
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If an element 'x' is in all of the U1...Un, then it is in the intersection (call it U). And if 'x' is in U then it is in all of the U1...Un. Suppose 'x' is the additive identity. What can you conclude? The other properties of the subspace U follow from similar arguments.

Well, then the additive ID is in U. But I am not seeing something. Let's just take the example of two subspaces intersecting, U1 and U2. This might be a silly question: how do we know that the 'place' where these two subspaces 'overlap' does indeed include the additive ID?
 
  • #4
Dick
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Let's call the identity '0', ok? 0 is in U1 (since it's a subspace). 0 is in U2 (since it's a subspace). If 0 is in U1 and U2, then it is in their intersection U. You are either overthinking this or you might want to look up the definition of 'intersection' in set theory.
 
  • #5
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Okay. I just looked it up again and read it more carefully (the 1st time I only looked at the Venn Diagram). So, the definition says

the intersection of two sets A and B is the set that contains all elements of A that also belong to B (or equivalently, all elements of B that also belong to A), but no other elements.

Your statement now makes mucho more sense.

So since the intersection U contains all of the elements that are common to U1,...,Un then U must contain the 0, since by def, U1,..,Un contain 0.

Now for the other properties....
 
  • #6
Dick
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Ok, if 'a' is in U, then it's in U1...Un, right? Ditto for 'b'. If 'a' is in U1 and 'b' is in U1 then 'a+b' is in U1 (because it's a subspace). Similarly, 'a+b' is in ALL of the U's. Hence?
 
  • #7
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So now I need to show that the intersection is closed under addition. Let U be the intersection of the subspaces U1,...,Un and let u1,...,u_n be their elements.

I have the feeling I should use the definition of a Sum of Subspaces here, but I am not sure.

EDIT: I just saw your last post. I am again getting a little lost by the definitions here. I am not sure how to express what it is that confuses me. Give me a minute to try and word it.
 
  • #8
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Let me digress a little from this particular problem and talk about sets and spaces in general. To make sure that I understand some of the basics here. Please correct any misconceptions that I might convey.

If K is a field, then K is also a set. If V is a vector space over K, then some element of V named v is an ordered list of coordinates which are comprised of elements of K right?

Now when we say that V is closed under addition, we mean that when two (or any number of) elements from V are added we get an other element that is in V.

Now, I get a little confused there. What does it mean in V.

I some of the other problems that I have done, it has been given that some condition has to be met like https://www.physicsforums.com/showthread.php?t=323907". Look at case (d). It says that x1=5*x3. So when I add two arbitrary vectors, I can then look and check that the resulting vector takes the same form as the given condition.

But when we speak so generally about a vector space like "when elements from V are added we get an other element that is in V" but we are not given a condition that I can test, I get lost.

I am not trying to make this more complex than it is, I am just trying to fill in the blanks (in my head :smile: )
 
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  • #9
Dick
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You are right. There is nothing to calculate here. It's all just logic. Go back to having just U1 and U2 and their intersection U. If a vector 'a' and a vector 'b' are in U, then they must be in U1 AND U2. Ok so far?
 
  • #10
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You are right. There is nothing to calculate here. It's all just logic. Go back to having just U1 and U2 and their intersection U. If a vector 'a' and a vector 'b' are in U, then they must be in U1 AND U2. Ok so far?

Righto!
 
  • #11
Dick
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Ok, then if 'a' and 'b' are in U1, then 'a+b' is in U1 (since it's a subspace). Ditto for U2. Still ok? Then 'a+b' is in U1 and U2. Is it in their intersection U?
 
  • #12
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No wait! That's what my last post was about :smile: if 'a' and 'b' are in U1, then why does that necessitate that 'a+b' also be in U1 ? I know I am missing a detail of a definition somewhere.



Side question: It might have to do with a field. Can a field be finite? Like can the scalars 1-100 be considered a field? Or ...
 
  • #13
Dick
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No wait! That's what my last post was about :smile: if 'a' and 'b' are in U1, then why does that necessitate that 'a+b' also be in U1 ? I know I am missing a detail of a definition somewhere.



Side question: It might have to do with a field. Can a field be finite? Like can the scalars 1-100 be considered a field? Or ...

I said it. You are GIVEN U1 is a subspace. It's the premise of the problem. It's a SUBSPACE. If 'a' is in U1 and 'b' is in U1 then 'a+b' is in U1. Otherwise it wouldn't be a subspace. A field can be finite. But that doesn't have anything to do with this question.
 
  • #14
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Okay. But what you are saying is that because it is a subspace, a+b must be in U1.

But what I am trying to ask is: why? I know I sound like an annoying 2-year-old right now, but it is something fundamental that I do not understand.

Is it just because of the definition? That is, we are saying that in order for a set to even be called a subspace, if 'a' and 'b' are in the set, then 'a+b' is also in the set by definition?
 
  • #15
Dick
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Yes, annoying two year old, it's the definition of a subspace. :) It's 'just because'. There's nothing to calculate here. It's a definition. No numbers, no nothing. All you have are definitions.
 
  • #16
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Okay. Because when I look at the definitions of subspace, it doesn't click that we are defining it like that.

My interpretation of the definition was this: if a set is closed under addition, scalar mult, etc, then it is a subspace. I know it's basically the same, but in the context of this problem, oh forget it!!! It will take all day to explain what I am trying to say!:mad:

Can we talk about vector spaces for a second? Do we define a vector space the same way? I mean I know what he def of a vector space is from my book: addition commutes, associates, is distributive and scalar multiplication distributes. and some other stuff.

BUT, do we also say that if 'a' and 'b' are in V, then a+b must be in V? Of course we do; that's a stupid question (now I am talking to myself:bugeye:).....right?
 
  • #17
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Okay. Here we go. Silly example that will give me some good insight. Pretend that I really am a 2-year-old.

I am given that a=(1,1) and b=(2,2) and that both are in V. I add them up for a cookie and get a+b=(3,3) mmmmm chocolate chip, my favorite.

You then tell me that (3,3) is also in V. I say "why"? How do we know that V doesn't "only go up to" (2,2) ? How do we know it did not 'end' at (2,2)?
 
  • #18
Dick
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Right. You are talking to yourself. If the problem says V is a vector space, then if a and b are in V, then a+b. Period. It's CLOSED UNDER ADDITION. It's in the definition. Are you saying they could be lying when the say V is a vector space? This is getting too existential for me.
 
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  • #19
Dick
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Okay. Here we go. Silly example that will give me some good insight. Pretend that I really am a 2-year-old.

I am given that a=(1,1) and b=(2,2) and that both are in V. I add them up for a cookie and get a+b=(3,3) mmmmm chocolate chip, my favorite.

You then tell me that (3,3) is also in V. I say "why"? How do we know that V doesn't "only go up to" (2,2) ? How do we know it did not 'end' at (2,2)?

Because they told you V is a vector space, you silly duck. If (1,1) is in V and (2,2) is in V and V IS A VECTOR SPACE then (3,3) is in V. If it's not then they lied about V being a vector space. And what would be the point to that?
 
  • #20
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Okay. So we back to "because I said so" ('I' being the math gods).

So similar to the definition of a subspace we could say: Hey, check out this set! S={1,2,3,4,5} is it a vector space?

(before I move on ... is this question even OK to ask?)

If it is OK I would answer NO. Since I could add two elements like 5+4=9 which is not in S.
 
  • #21
Dick
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It's fine to ask but it's not going anywhere. This is just silly. When you are asked to prove something, you can't prove the premise. That's what you are given. If I ask you to prove the sum of the angles of a triangle T sum to 180 degrees are you going to say I didn't prove T has three sides?? That's what you are doing, you know. What's this 'math gods' crap? Definitions are 'just because'.
 
  • #22
Dick
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Okay. So we back to "because I said so" ('I' being the math gods).

So similar to the definition of a subspace we could say: Hey, check out this set! S={1,2,3,4,5} is it a vector space?

(before I move on ... is this question even OK to ask?)

If it is OK I would answer NO. Since I could add two elements like 5+4=9 which is not in S.

Ok, Mr Question Authority. If you prove a theorem about vector spaces and S={1,2,3,4,5} then S is not a vector space. So the theorem doesn't apply to S.
 
  • #23
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Now now. Don't be miffed. I am not insulting the math gods, I am trying to get clarification from them.

There are a lot of details that are not clear to me when I read the definition of a Vector Space. And I sweat the details. You might not be getting anything out of this conversation, but I can assure you that I am (and I appreciate it). :smile:
 
  • #24
Dick
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Now now. Don't be miffed. I am not insulting the math gods, I am trying to get clarification from them.

There are a lot of details that are not clear to me when I read the definition of a Vector Space. And I sweat the details. You might not be getting anything out of this conversation, but I can assure you that I am (and I appreciate it). :smile:

NIce of you to say. But I am getting miffed. A proof says if A then B. If A isn't true then who cares? You are allowed to assume A. Please do so, before I assume I'm wasting my time here.
 
  • #25
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I am not sure why you would assume that you are wasting your time. I am asking questions and you are answering them. That's kind of the point, right?

If my questions are a waste of your time, I apologize. But, just as a reminder, at post #8 and then again at #16 I asked if we could detour from the Proof (the OP) and into a discussion on the definition of a Vector space. So when you keep saying that I am "assuming that A is not true" that is only because you have forgotten that I am no longer talking about the original problem. I am talking about the definition of a Vector space.

I am not trying to "prove the premise" (or at least I did not mean try to), I just want to understand the premise fully before I move on.
 

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