# Prove that the intersection of subspaces is subspace

Okay then! Back to the OP ! I figure I should finish this problem, else Dick might have a conniption on me

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.)

Homework Helper
Keep going. Or else! Try closure under addition or scalar multiplication.

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?

Homework Helper
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.

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?

Homework Helper
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.

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).

Homework Helper
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.

Okay then. I will do that in awhile. A can't believe I left my power chord at home. Stupid Mac.