Prove that the intersection of subspaces is subspace

  • #36
Saladsamurai
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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
Saladsamurai
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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
Saladsamurai
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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
Saladsamurai
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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
Saladsamurai
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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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