Linear Algebra: Direct sum proof

loesch.19
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Let U and V be subspaces of a vector space W. If W=U \oplus V, show U \bigcap V={0}.

I'm a bit lost on this one... as I thought this was essentially the definition of direct sum. I'm unsure where to start. Any help would be great!
 
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What's the definition of a direct sum you're using?
 
vela said:
What's the definition of a direct sum you're using?

I wasn't sure if that was necessary info or not... looks like I was wrong :)

If U and V are subspaces of vector space W, and each w in W can be written uniquely as a sum u+v where u is in U and v is in V then W is a direct sum of U and V.
 
So were you able to do the problem?
 
Hurkyl said:
So were you able to do the problem?

No... I still need help.
 
Well, surely you can do something on it -- even if it's just rewriting the problem in a less opaque form.

e.g. do you know anything about proving two subspaces equal? (Or two sets?)
 
loesch.19 said:
Let U and V be subspaces of a vector space W. If W=U \oplus V, show U \bigcap V={0}.

I'm a bit lost on this one... as I thought this was essentially the definition of direct sum. I'm unsure where to start. Any help would be great!

I think you should add the definition as follows:

If U and V are subspaces of a Vector space W and each w \in W can be written as the unique sum as u+v where u \in U and v \in V then

W is the direct sum of U and V and can be written W = U \oplus V
 
Suppose there were a non-zero vector, w, in both U and V and let u be any vector in U. Now, write u as two different sums of vectors in U and V.
 
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