Proving that the interesection of subspaces is a subspace

  • Thread starter Freye
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  • #1
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Homework Statement



Let U1; U2 be subspaces of the vector space V . Prove that their intersection U1 \ U2 is
also a subspace of V

Homework Equations



I see how any equations could be used here

The Attempt at a Solution



Well intuitively this seems obvious from the get go. If U1 and U2 are subspaces, then their intersection, which can at most contain all of U1 if U1=U2, and at the very least the 0 vector if U1 and U2 share no common vectors other than the 0 vector. But I don't know how to prove this. It seems like from what I've said, I've neglected all of the intermediate possibilities.
 

Answers and Replies

  • #2
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What is the definition of a subspace? (Post it here)

Pick arbitrary elements in the intersection and show that all the properties of a subspace hold, using the premise.
 
  • #3
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A subspace must be closed under addition and multiplication, use the same addition and scalar multiplication as it's parent vector space, and have the same additive identity as it's parent vector space.

But how can I pick arbitrary elements of the intersection if I don't even know what V is. It could be a vector space of functions, or of complex numbers, or probably of something else that I don't yet know. So how can I write out V so that it doesn't exclude any possibilities, yet I can work with it to prove the conditions for it's subspaces?
 
  • #4
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Like this:

Let a and b be elements of U1 \ U2.

How do we know that a+b is in U1 \ U2?

Use the properties of the premise that U1 and U2 are subspaces of V.
 
  • #5
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I'm sorry, I'm really not giving you much to go on.

Try a direct proof. It won't take more than 4 lines.

What does it mean that a is in the intersection of U1 and U2? Literally - what does that mean? Same for b. Now put that together to say something about a+b in relation to U1, and similarly with relation to U2.
 
  • #6
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Ok thank you, actually that "let a and b be elemnts of U1\U2" was actually quite helpful
 
  • #7
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No problem, glad to be of help.
 

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