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Subspace proof

  1. May 1, 2009 #1
    Prove that if L and P are three dimentional subspaces of R5, then L and P must have a nonzero constant in common.

    i am just stuck now on how to get this proof started... any thoughts on how to start?
     
  2. jcsd
  3. May 1, 2009 #2

    matt grime

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    L is thee dimensional: let a,b,c be a basis. P is three dimensional: let x,y,z be a basis. What can you say about a,b,c,x,y,z given that they live in a 5 dimensional space?
     
  4. May 2, 2009 #3
    okay, if so, they will form a six dimentinal subspace, which dont span the vector spcce.


    thanks a lot!
     
  5. May 2, 2009 #4

    matt grime

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    How do you know that these 6 vectors do not span the space? They may or may not span the space, but that is immaterial: you have 6 vectors in a 5 dimensional space, therefore they are .....?
     
  6. May 2, 2009 #5
    they are not linearly independent? and thus not a basis for such space
     
  7. May 2, 2009 #6

    Matterwave

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    The part that they are "not linearly independent" is what is pertinent to the question. You don't care whether they make a basis for a 6 dimensional space.

    If your 6 basis vectors are not linearly independent, then they share...

    What does it mean if the vectors are not linearly independent?
     
  8. May 2, 2009 #7
    then they dont share common subspaces...
     
  9. May 3, 2009 #8

    matt grime

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    I think you're just guessing and hoping now. I don't know what Matterwave means, but all I want you to do is look at the definition of linearly dependent (or the negation of linear independence).
     
  10. May 3, 2009 #9

    HallsofIvy

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    Notice that you did not use the fact that the two subspaces have no non-zero vector (you said "constant" but I presume you meant vector) in common. That's the important thing.
     
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