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T/F: Subset of a spanning set always forms a basis

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    T/F: If a finite set of vectors spans a vector space, then some subset of the vectors is a basis.

    2. Relevant equations


    3. The attempt at a solution

    It seems that the answer is true, due to the "Spanning Set Theorem," which says that we are allowed to remove vectors in a spanning set until we get a set that is linearly independent and thus forms a basis. However, what if our original set of vectors only spans the trivial vectors space ##\{ 0 \}##? In that case we can't form a basis, so wouldn't the answer to this question be false?
     
  2. jcsd
  3. Oct 25, 2016 #2

    fresh_42

    Staff: Mentor

    This can only be the case, if all vectors in the set are ##\vec{0}##. Then the subset ##\{\}## spans ##\{0\}##.
     
  4. Oct 25, 2016 #3
    So in the problem statement, why can't we interpret that the finite set of vectors is a set of zero vectors?
     
  5. Oct 25, 2016 #4

    fresh_42

    Staff: Mentor

    Who says we can't? A set ##\{\vec{0},\vec{0},\vec{0},\vec{0},\ldots\}=\{\vec{0}\}## spans the zero dimensional vector space ##V=\{0\}## with the basis ##\{\}##. Of course this is on the edge of the definition and not very geometric, but it is allowed. And if you take away ##\vec{0}## from ##\{\vec{0}\}## you are left with the basis ##\{\}## of ##V=\{0\}##.

    One cannot have some non-zero vectors and their (linear) span will turn out to be ##\{0\}##, because with every ##0 \neq \vec{v}## there will be at least all vectors ##\mathbb{R}\cdot \vec{v}## in the linear span of a set, that contains ##\vec{v}##.
    (I chose ##\mathbb{R}## as the corresponding field for simplicity, but any other will work as well.)
     
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