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Am I the only one that finds this funny?

  1. Sep 23, 2013 #1
    Is this really worthy of being pointed out as a corollary??
     

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  3. Sep 23, 2013 #2
    Oops. It's just the contrapositive. Then no, it's not worthy of being a corollary.
     
  4. Sep 23, 2013 #3

    fluidistic

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    If it follows from the theorem 1.6 then why not? :) Not really funny to me.
     
  5. Sep 23, 2013 #4
    Are you sure? :wink:
     
  6. Sep 23, 2013 #5
    I misread it, thought that the corollary was about dependance too. :redface:
     
  7. Sep 23, 2013 #6
    Hah, I don't blame you. I had to do a double (or triple) take.

    Though, interestingly enough, if the corollary was about dependence, then that would mean (S1 dependent) <=> (S2 dependent), which would mean that any subset or superset of any linearly dependent set would be linearly dependent. Which would mean that all subsets of a vector space would be linearly dependent, because all vector spaces must contain some linearly dependent subset (namely itself) (unless I'm just confusing myself here!)
     
    Last edited: Sep 23, 2013
  8. Sep 23, 2013 #7

    Office_Shredder

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    That's true, but not for the reason that you seem to be thinking. Every vector space contains 0, and therefore is a linearly dependent set, but if you remove that point you could have a linearly independent set. For example (and probably the only example), F2 the field of two elements, as a one dimensional vector space over itself.
     
  9. Sep 23, 2013 #8
    I'm saying - if the corollary was talking about dependence, a consequence would be that any subset of a vector space is linearly dependent because the vector space itself is.
     
  10. Sep 24, 2013 #9

    AlephZero

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    I think it's not just to "point it out", but ask you to prove both results from first principles, to get used to making proofs about linear dependence and independence.
     
  11. Sep 24, 2013 #10

    FlexGunship

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    Wait. Why? Because there's two extra commas in there?
     
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