# Linear Algebra: Direct sum proof

1. May 31, 2010

### loesch.19

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!

2. May 31, 2010

### vela

Staff Emeritus
What's the definition of a direct sum you're using?

3. May 31, 2010

### loesch.19

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.

4. May 31, 2010

### Hurkyl

Staff Emeritus
So were you able to do the problem?

5. May 31, 2010

### loesch.19

No... I still need help.

6. May 31, 2010

### Hurkyl

Staff Emeritus
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?)

7. Jun 1, 2010

### Susanne217

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 $$W = U \oplus V$$

8. Jun 1, 2010

### HallsofIvy

Staff Emeritus
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.