Proving V = U ⊕ W: A Subspaces Problem

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In summary, the conversation discusses an exercise that involves proving the direct sum of two subspaces is equal to the vector space if and only if the intersection of the two subspaces is only the trivial element. The conversation also provides an explanation and proof for the exercise.
  • #1
Evalesco
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Hi, I've being going over some course material in preparation for my exams and I've come across an exercise which I can't seem to work out.

The exercise:
Let V be a vector space over a field F and let U and W be two subspaces of V.
Suppose V = U + W. Prove that V = U ⊕ W iff {(u,w)∈U×W : u + w = 0} = {(0,0)}

What I've done so far:
At first I was unsure of the operator ⊕ but from Wikipedia I discovered the following:
V = U ⊕ W ⇔ (V = U + W) ∧ (U ∩ W = ∅)
Which I guess is a consequence of the following: The dimension of V ⊕ W is equal to the sum of the dimensions of V and W.

So I thought ok that identity will make it quite easy to solve, I just need to make sure U and W share no common elements.

We are given {(u,w)∈U×W : u + w = 0} = {(0,0)}
Ok I thought, that uses the property that U and W are both closed under addition to ensure the two subspaces U and W have no non-zero elements in common.

But this is where I'm stuck I thought in order to solve it I would need U and V to share no common elements but 0 can be both an element of U and V.

I know this isn't a proof but I was just trying to formulate an idea of how I would go about solving it. Any insight would be welcome, Thanks.
 
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  • #2
OK, let V be the direct sum of U and W. It follows that U[itex]\cap[/itex]W = {0}, so there cannot exist an element of (u, v) from U x W such that u + w = 0, because then the intersection of U and W would not be trivial.

For the other direction, assume {(0, 0)} = { (u, v) from U x W : u + w = 0}. Take an element from U[itex]\cap[/itex]W, let's say x. Then x is in U and x is in W. Since the intersection is a subspace, there exists an inverse for the element x. It's easy to finish the proof now.
 
  • #3
Thanks for the clarification, it helped me heaps, it looks like I had my definitions a little off lol.
 

1. What does it mean to prove V = U ⊕ W?

Proving V = U ⊕ W means showing that the vector space V can be decomposed into the direct sum of two subspaces, U and W. This means that every vector in V can be written as a unique combination of vectors from U and W, and that U and W only have the zero vector in common.

2. How do you prove a subspace problem?

To prove V = U ⊕ W, you need to show that (1) every vector in V can be written as a unique combination of vectors from U and W, and (2) U and W only have the zero vector in common. This can be done by showing that U and W span V, are linearly independent, and their intersection is only the zero vector.

3. What is the difference between a direct sum and a sum of subspaces?

A direct sum, denoted by ⊕, is a special type of sum where the subspaces involved have no vectors in common except for the zero vector. This means that the sum is "direct" in the sense that there is no overlap between the subspaces. In contrast, a sum of subspaces, denoted by +, may have non-zero vectors in common.

4. Can V = U ⊕ W if U and W are not subspaces of V?

No, V = U ⊕ W can only be true if U and W are subspaces of V. This is because the definition of a direct sum requires U and W to be subspaces of the larger vector space V.

5. Why is proving V = U ⊕ W important in linear algebra?

Proving V = U ⊕ W is important in linear algebra because it allows us to understand the structure of a vector space V by breaking it down into simpler subspaces, U and W. This decomposition can make it easier to solve problems and understand the properties of V. Additionally, the concept of direct sums is used in many other areas of mathematics, making it a fundamental concept to understand.

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