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Linear Algebra spanning set

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine whether the set S spans R3. If the set does not span R3, then give a geometric description of the subspace that it does span

    S = [ (2,0,3) , (2,0,-1) , (6,0,5) , (4,0,6) ]

    2. Relevant equations


    3. The attempt at a solution
    I know S does not span R3 because the system of equations has 3 equations but 4 variables. But how do I give a geometric description of the subspace it does span?
     
  2. jcsd
  3. Nov 5, 2014 #2

    RUber

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    A spanning set must be able to be combined by addition or scalar multiplication to reach anything in the space it spans.
    In this case, you have 4 vectors, so the largest space it could span is R^4.
    Edit:
    But since each vector only has 3 entries, it can only span a 3-D space.
     
  4. Nov 5, 2014 #3
    but because the system of equations has no unique solution that means it does not span R3. so does it span a subspace at all?
     
  5. Nov 5, 2014 #4

    LCKurtz

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    First you say it spans R3 then you say it doesn't. In neither case have you shown us any equations to back up your opinion.
     
  6. Nov 5, 2014 #5

    RUber

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    No unique solution implies the set in not linearly independent.
    4 vectors must be linearly independent to span 4-D space.

    You should be looking to see if there is anywhere in R3 that they can not reach, i.e. if the null space of the set of vectors is more than just the zero vector.
     
  7. Nov 5, 2014 #6

    Mark44

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    No. The fact that the system doesn't have a unique solution is not a guarantee that the set doesn't span R3.

    For example, the set of vectors {(1, 0), (0, 1), (1, 2)} spans R2.
    @RUber, you have mentioned R4 in two of your posts in this thread. Since the vectors all have only three components, they couldn't possibly belong to R4.
     
  8. Nov 6, 2014 #7
    I never said the vectors spanned R3. so basically the subspace the vectors span is R3 minus its nullspace?
     
  9. Nov 6, 2014 #8

    HallsofIvy

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    No. Your very first statement "I know S does not span R3 because the system of equations has 3 equations but 4 variables" is wrong. In the first place, you have 4 vectors, not variables! More importantly, because you have four vectors in three dimensional space they cannot be independent and so cannot be a basis. Perhaps you are confusing "span" and "independent". A set of fewer vectors than the dimension of the space cannot span it but a larger set certainly can.
     
  10. Nov 6, 2014 #9

    LCKurtz

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    @nate9519: So far you have stated questions but given no definitions and shown no arguments or effort at actually trying to work this problem. This violates forum guidelines and I'm surprised you haven't been dinged by the Mentors. Maybe you have, I wouldn't know. But you need to show some effort. Have you tried row reduction or anything else? Show us.
     
  11. Nov 6, 2014 #10

    pasmith

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    It is immediately obvious on looking at the second component of each vector in S that S spans a subset of the plane [itex]\{(x,0,z) : (x,z) \in \mathbb{R}^2\} \varsubsetneq \mathbb{R}^3 [/itex]. What is not immediately obvious, and I leave for you to determine, is whether S spans all of that plane or only a line lying in that plane and passing through the origin.
     
  12. Nov 6, 2014 #11

    Mark44

    Staff: Mentor

    Before adding more replies, let's wait and see if the OP addresses the comments already made.
     
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