Prove that the intersection of subspaces is subspace

  • #26
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So. Back to my annoying questions. (this is a continuation of post #20)

If I am given a set S={1,2,3,4,5} and I am told that it is a vector space, then something is flawed here since the sum of many of those elements are not themselves in the set. So what I have I done to cause this conflict.

Does a Vector space have to be infinite to rectify this?

i.e, since from S we could take 5+5=10, then S would have to include 10 to be a vector space...but then the cycle starts again since for example 5+10=15.
 
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  • #27
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Uh, who told you that was a vector space? I don't really understand the problem here. You don't have to know exactly what a vector space is, just that it is a very nice set with lots of properties. The best way for you to understand more about them is to not get obscured by all the details. For now, you should be able to determine when something is or is not a vector space. If a set satisfies the axioms of a vector space, it's a vector space. If not, who cares?

Revisit the definition of a vector space again if you have to. If this is your first encounter with actually proving various things about the axioms of an algebraic structure, you probably need to read the first chapter more than once. But really the initial question involves some basic set theoretical ideas, nothing more.
 
  • #28
Dick
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So. Back to my annoying questions. (this is a continuation of post #20)

If I am given a set S={1,2,3,4,5} and I am told that it is a vector space, then something is flawed here since the sum of many of those elements are not themselves in the set. So what I have I done to cause this conflict.

Does a Vector space have to be infinite to rectify this?

i.e, since from S we could take 5+5=10, then S would have to include 10 to be a vector space...but then the cycle starts again since for example 5+10=15.

Ok. Then this is a detour. It has nothing to do with any proof, right? You are just asking why S={1,2,3,4,5} isn't a one dimensional vector space? Over the real numbers as scalars? You'd better write it as S={(1),(2),(3),(4),(5)} to be clear they are vectors. You've already given an answer. It's not closed under addition (among other things, it also doesn't have an additive identity and isn't closed under scalar multiplication). S is pretty far from being something you would call a vector space. Yes, the set would have an infinite number of elements if you extend it to a vector space over the reals. Are you sure this 'example' isn't more confusing than helpful?
 
  • #29
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Ok. Then this is a detour. It has nothing to do with any proof, right? You are just asking why S={1,2,3,4,5} isn't a one dimensional vector space? Over the real numbers as scalars? You'd better write it as S={(1),(2),(3),(4),(5)} to be clear they are vectors. You've already given an answer. It's not closed under addition (among other things, it also doesn't have an additive identity and isn't closed under scalar multiplication). S is pretty far from being something you would call a vector space. Yes, the set would have an infinite number of elements if you extend it to a vector space over the reals. Are you sure this 'example' isn't more confusing than helpful?

No. I promise :smile: it is not. I learn a lot more about what something is by knowing what it isn't. I know that makes it painfully annoying for others, but that is honestly how I learn a lot.

So do ALL vector spaces have to have infinite elements? To avoid the aforementioned case?

Thanks!
 
  • #30
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Is {0} a vector space? Does {0} have an infinite amount of elements?

Also consider finite fields (which are vector spaces over themselves) such as F_p, where if p is any prime, then F_p = {0, 1, 2, ... , p - 2, p -1} with addition mod(p) and multiplication mod(p) (F_p is obviously not closed under the "regular" addition and multiplication of real numbers)
 
  • #31
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Is {0} a vector space? Does {0} have an infinite amount of elements?

Also consider finite fields (which are vector spaces over themselves) such as F_p, where if p is any prime, then F_p = {0, 1, 2, ... , p - 2, p -1} with addition mod(p) and multiplication mod(p) (F_p is obviously not closed under the "regular" addition and multiplication of real numbers)

Well these are all clearly very special cases. {0} is closed since you will only get (0) no matter what operation is carried out. I don't know anything about that prime field or what "mod()" means.....nor do I understand why the field of primes would be finite? But anyway, I do not want to get ahead of myself.

In general, when we are talking about general vector fields that are not a special case, wouldn't they have to be infinite so that all sums and products will be contained in them?

Thanks :smile:
 
  • #32
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I hope the original poster doesn't mind my suggesting reading one of those books that teaches one how to prove things, basics of set theory, other important ideas used in abstract and proof-oriented math courses. They can provide, in a sense, the "rules of operation". It can be read parallel to reading the L.A. book. I read one of them when I took my first proofs-based course, and many colleges require the math majors to take a course that teaches out of one of them.
 
  • #33
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In general, when we are talking about general vector fields that are not a special case, wouldn't they have to be infinite so that all sums and products will be contained in them?

Most of the time in an introductory course, the non-trivial subspaces will be infinite, because they are usually over the filed of Real numbers or Complex numbers, but there are also lots of finite fields and any finite-dimensional vector space over one of those will be finite, so all of its subspaces will be finite also.
 
  • #34
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I hope the original poster doesn't mind my suggesting reading one of those books that teaches one how to prove things, basics of set theory, other important ideas used in abstract and proof-oriented math courses. They can provide, in a sense, the "rules of operation". It can be read parallel to reading the L.A. book. I read one of them when I took my first proofs-based course, and many colleges require the math majors to take a course that teaches out of one of them.

No not at all. If anyone has any recommendations, by all means suggest away! Like I have said many times in many posts, I don't really know anything about sets, or set theory. My math background is that of an engineer and for some reason I have gotten through all of my math requirements without having seen even these basic symbols like [itex]\in\text{ or }\cap\text{ or }\cup[/itex] so i am having to fill in a lot of the blanks here. (And there are so many :bugeye:)

That's why I have sooooo many annoying questions. :redface: Know of any good basic set theory books? :smile:
 
  • #35
Try downloading a free book in one of the book recommendation sites on this forum.
Look for a discrete math book. I took discrete math last year and learned all
about sets and subsets and such.
 
  • #36
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Okay then! Back to the OP ! I figure I should finish this problem, else Dick might have a conniption on me :smile:

here we go....

Given U1,..,Un are subspaces of V and U is their intersection.

1.)Since U contains all of the elements that are common to U1,...,Un and (0) is in all of them, then (0) is in U.

2.)
 
  • #37
Dick
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Keep going. Or else! Try closure under addition or scalar multiplication.
 
  • #38
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2.) Referring back to post #6. If some vector 'a' and 'b' are in U then a+b must be in U as well. And by definition of U, they are in all U1,...,Un.

3.) Can't I just say for scalar multiplication that since subspaces are closed under multiplication by definition, than the intersection, by definition, must also be closed?
 
  • #39
Dick
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2.) Referring back to post #6. If some vector 'a' and 'b' are in U then a+b must be in U as well. And by definition of U, they are in all U1,...,Un.

3.) Can't I just say for scalar multiplication that since subspaces are closed under multiplication by definition, than the intersection, by definition, must also be closed?

2) You are doing it backward. U ISN'T known to be a subspace. That's what you are trying to prove. U1...Un ARE subspaces. Use that.
 
  • #40
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Allllriiighty then...

So given vectors 'a' and 'b' that are common to U1,...,Un then clearly a+b is common to U1,...,Un and therefore in U.

How am I doing? Does what I said make U closed under addition?
 
  • #41
Dick
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Try and say it more carefully. What does 'common' mean? If you mean 'a' and 'b' are in ALL of the U1...Un, say so. If that's what you mean to say, then why does that mean 'a+b' is in all of the U1...Un? This is a proof, you are supposed to give reasons.
 
  • #42
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If 'a' and 'b' are in ALL of the U1,...,Un then 'a' and 'b' are in U by definition of an intersection.

If 'a' and 'b' are in ALL of the U1,...,Un then 'a+b' are in ALL of the U1,...,Un because it was given that U1,...,Un are subspaces and thus are closed under addition.

Therefore if 'a+b' is in ALL of the U1,...,Un then 'a+b' is in U by definition of an intersection.

That's the best I can do. If I missed anything, I will probably have to take all of these people in Starbucks hostage and hold them until my demands are met (i.e., a paid round-trip flight for you to come here, have a cup of Joe, and explain to me why I can't get this).
 
  • #43
Dick
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No, no. Release the hostages. I LIKE that! But you want to start with 'a' and 'b' are in U and then say that means 'a' and 'b' are in ALL U1...Un (by definition of intersection, of course). Then it's all good.
 
  • #44
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Okay then. I will do that in awhile. A can't believe I left my power chord at home. Stupid Mac.
 
  • #45
HallsofIvy
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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.
No, no! Two year olds are much more annoying than that! (I know, I used to be one!)

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?
Yes, that's what definitions are for!
 

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