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A subspaces problem.

  1. Oct 30, 2007 #1
    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.
     
  2. jcsd
  3. Oct 30, 2007 #2

    radou

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    Homework Helper

    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.
     
  4. Oct 30, 2007 #3
    Thanks for the clarification, it helped me heaps, it looks like I had my definitions a little off lol.
     
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