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Dimension of the span of a set of vectors

  1. Oct 20, 2015 #1
    My linear algebra is a bit rusty.

    Let ##A=\{\bar{v}_1, \dots, \bar{v}_1\}## be a set of vectors in ##R^n##. Can dim(span##(A))=n## without spanning ##R^n##?

    I guess I'm unclear on how to interpret the dimension of the span of a set of vectors.
     
  2. jcsd
  3. Oct 20, 2015 #2

    andrewkirk

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    The span of a set of vectors is a vector space. There cannot be a proper n-dimensional subspace of an n-dimensional vector space. Any n-dimensional subspace must be the whole thing.

    That is one area where vector spaces differ from modules.
     
  4. Oct 20, 2015 #3

    Mark44

    Staff: Mentor

    Are there n vectors in A? Your subscripts all appear to be 1. Presumably you meant ##\{\bar{v}_1, \dots, \bar{v}_n\}##.
     
  5. Oct 21, 2015 #4
    Yeah that second, 1 should have been an m. Not necessarily an n. (at least I don't think it should necessarily be an n).
     
  6. Oct 21, 2015 #5
    Ok, this is exactly what I needed to know!

    I've not gotten to modules yet. I'm taking a second graduate Algebra class next semester though. I'm told we will finally get into them then.
     
  7. Oct 21, 2015 #6

    Mark44

    Staff: Mentor

    Then you should specify some condition on m. If m < n, then your set of vectors could not possibly span ##\mathbb{R}^n##, since dim(span(A)) ##\le m < n##. If ##m \ge n##, the set of vectors might span ##\mathbb{R}^n##, or might not.
     
  8. Oct 21, 2015 #7
    The only condition I wanted on m is that it was not necessarily equal to n. So that it was truly an arbitrary set of vectors in ##R^n##. That way the question was just: "Given an arbitrary set of vectors, A, such that dim(span(A)) = n. Does A necessarily span ##R^n##?" This is a more concise way of asking the question.. I believe the two questions are equivalent though. If not, this is definitely the question I mean to ask.

    Which the answer appears to be yes... as per andrewkirk.

    Edit: I guess the condition that dim(span(A))=n forces the condition that ##m\geq n##.
     
  9. Oct 21, 2015 #8

    Mark44

    Staff: Mentor

    Yes
     
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