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Linear Algebra: Subspace Proof

  1. Sep 6, 2007 #1
    Prove that the intersection of any collection of subspaces of V is a subspace of V.

    Ok, I know I need to show that:

    1. For all u and v in the intersection, it must imply that u+v is in the intersection, and

    2. For all u in the intersection and c in some field, cu must be in the intersection.

    I can show both 1. and 2. for the trivial case where the intersection is zero, but I'm not sure what I need to do for the arbitrary case.

    Any suggestions?
     
  2. jcsd
  3. Sep 6, 2007 #2

    Hurkyl

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    Well, I'm not sure where you're stuck, so I'll throw out two random hints.


    (a) Let W be one of the spaces in the intersection...

    (b) What is the definition of "x is in the intersection"?
     
  4. Sep 6, 2007 #3
    Since u and v are elements of the intersection, u and v will also be elements of any subspace W that is in the intersection. And since u and v are in W and W is a subspace, this guarantees that u+v will also be in W. This same argument would apply to scalar multiplication.

    Is that the right idea?
     
  5. Sep 6, 2007 #4

    Hurkyl

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    Yes, that is one of the essential points of the proof.
     
  6. Sep 6, 2007 #5
    Thanks!
     
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