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Linear Algebra: Direct sum proof

  1. May 31, 2010 #1
    Let U and V be subspaces of a vector space W. If W=U [tex]\oplus[/tex] V, show U [tex]\bigcap[/tex] 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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  3. May 31, 2010 #2

    vela

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    What's the definition of a direct sum you're using?
     
  4. May 31, 2010 #3
    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.
     
  5. May 31, 2010 #4

    Hurkyl

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    So were you able to do the problem?
     
  6. May 31, 2010 #5
    No... I still need help.
     
  7. May 31, 2010 #6

    Hurkyl

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    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?)
     
  8. Jun 1, 2010 #7
    I think you should add the definition as follows:

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

    W is the direct sum of U and V and can be written [tex]W = U \oplus V[/tex]
     
  9. Jun 1, 2010 #8

    HallsofIvy

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    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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