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About spanning sets

  1. Mar 21, 2012 #1
    I have some questions about spanning sets

    1. Why does empty set spans the zero subspace?

    2. Why is this true: Since any vector u in A is dependent on A, A⊆<A>? (<A> is the set of all vecotrs in R^n that are dependent on A)
     
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  3. Mar 21, 2012 #2

    mathwonk

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    2. u = 1.u

    1. the intersection of all subspaces containing the empty set equals {0}.

    second answer: if you want the summation symbol to be additive over disjoint decomposition of the index set, i.e. if you want the sum of the vectors in SuT to equal the sum of the vectors in S plus the sum of the vectors in T, when S and T are disjoint, you have to agree that the sum of the vectors in the empty set is zero.

    third answer: because i said so (i.e. it is defined that way).
     
  4. Mar 21, 2012 #3

    Fredrik

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    For non-empty sets S, "the smallest vector space that contains S", "the intersection of all vector spaces that contain S" and "the set of all linear combinations of members of S" are all the same. But if S=∅, the last one is ∅ and the first two are {0} (if your definition of vector space requires the set to be non-empty).

    So it looks like your book uses one of the first two as the definition of span S, and requires vector spaces to be non-empty.
     
  5. Mar 21, 2012 #4

    mathwonk

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    I am arguing that the empty linear combination is zero. see above mumbo jumbo about empty index sets. for the same reason the empty product is 1. this is a pretty standard convention.
     
  6. Mar 21, 2012 #5

    Fredrik

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    Ah, I always forget to consider empty index sets. :smile: Yes, if we define ##\sum_{k=1}^0 a_k s_k=0##, then span S = ∅, even if the left-hand side is defined as the set of all linear combinations of members of S.
     
  7. Mar 21, 2012 #6

    mathwonk

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    this bothered me for a long time in connection with euclid's proof of infinitude of primes. i.e. to construct as many primes as desired start from any set of primes, multiply them together and add 1. then we claim that number has anew prime factor. but for this to work, it needs to equal 2 or more.

    i thought this was a gap, and that one should exhibit at least one prime to begin the induction.

    fortunately however, by this convention, even if we begin with the empty set of primes, their product is 1, and so the number constructed is 2.
     
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