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Prove that the intersection of subspaces is subspace

  1. Jul 8, 2009 #1
    1. The problem statement, all variables and given/known data
    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?
     
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  3. Jul 8, 2009 #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.
     
  4. Jul 8, 2009 #3
    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?
     
  5. Jul 8, 2009 #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.
     
  6. Jul 8, 2009 #5
    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

    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....
     
  7. Jul 8, 2009 #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?
     
  8. Jul 8, 2009 #7
    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.
     
  9. Jul 8, 2009 #8
    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: )
     
    Last edited by a moderator: Apr 24, 2017
  10. Jul 8, 2009 #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?
     
  11. Jul 8, 2009 #10
    Righto!
     
  12. Jul 8, 2009 #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?
     
  13. Jul 8, 2009 #12
    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 ...
     
  14. Jul 8, 2009 #13

    Dick

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    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.
     
  15. Jul 8, 2009 #14
    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?
     
  16. Jul 8, 2009 #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.
     
  17. Jul 8, 2009 #16
    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?
     
  18. Jul 8, 2009 #17
    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)?
     
  19. Jul 8, 2009 #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.
     
    Last edited: Jul 8, 2009
  20. Jul 8, 2009 #19

    Dick

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    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?
     
  21. Jul 8, 2009 #20
    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.
     
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