# Correct reasoning about direct sums proof?

1. Sep 18, 2007

### quasar_4

Howdy everyone. I'm not very good at writing proofs, so I am wondering if someone can tell me if I'm even on the right page with this. I am not sure if I understand the idea correctly. The theorem goes as follows: Suppose V is a finite dim. vector space with subspaces U and W such that V is the direct sum of U and W. Let Z also be a subspace of V. Then Z cannot equal the direct sums of (i) the intersection of Z and U and (ii) the intersection of Z and W unless Z is a subspace of U or W.

Here's my thoughts:

Let "C" denote the direct sum of the intersections (i) and (ii) above. Then two conditions must hold:
1) The intersection of the intersections (i) and (ii) must be the zero vector; and
2) The sum of the intersections must equal V.

The elements of C are given as {x|x is in Z and x is in U and x is in W}={0v}. Thus every x in Z that intersects U and W is in C. But we have that C={0v} since it is a direct sum. The only way that {0v}=(ZintersectU)intersect(ZintersectW) is if Z contains only the zero vector. Thus Z is the zero subspace, and it follows that Z must be a subspace of U and W.

Yeah, let me know if that's right... I'm not sure that I am correct in asserting that Z contains only the zero vector.

2. Sep 18, 2007

### morphism

Why?

3. Sep 19, 2007

### quasar_4

well it sounded reasonable to me... if C is the set of {(Z intersects U) intersects (Z intersects W)}. So for (Z intersects U) we have the set containing {x|x is in Z and x is in U} and similarly for (Z intersects W) we have the set {x|x is in Z and x is in W}.
The direct sum would have all of these x too... wouldn't it? Although looking above it seems that I mixed up an intersection and direct sum. But even so, the intersection of the two intersections has to be the zero vector.

4. Sep 19, 2007

### morphism

Yes, which means your second sentence is irrelevant! So what if the intersection of those two things is {0}? C (the direct sum) will certainly contain this, but it definitely does not have to equal it.

5. Sep 21, 2007

### fopc

Maybe you've already proved it. If not, here's a suggestion.

Put the statement into a form that'll shed some more light on the different options you have for proving it.
The first thing I'd do is get rid of the "unless"; then you'll have a positive statement. It'll be an implication.
Move its antecedent into your list of hypotheses (giving you more stuff to work with).
Notice the antecedent's a disjunction (which suggests one proof method).
Of course, there are other options too.

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