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Linear Algebra Proof (vector spaces and spans)

  1. Jan 31, 2012 #1

    jmm

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    1. The problem statement, all variables and given/known data

    If [itex]ℝ^{n}=span(X_{1},X_{2},...,X_{k})[/itex] and A is a nonzero m x n matrix, show that [itex]AX_{i}≠0[/itex] for some i.

    2. Relevant equations
    ~


    3. The attempt at a solution
    Hey guys, I'm way over my head here. I really don't know how to approach this problem. I would really appreciate it if someone could give a nudge in the right direction as to where to start.
     
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  3. Jan 31, 2012 #2

    tiny-tim

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    hi jmm! :smile:

    hint: suppose every Axi = 0 :wink:

    (and remember the definition of span)
     
  4. Jan 31, 2012 #3

    jmm

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    If every AXi=0 that would mean that Xi=0 for all i, right? And a set of all 0 vectors would not span Rn, right? ...so that would be impossible? Haha I'm really having trouble wrapping my head around this.

    Thanks for your reply by the way!
     
  5. Jan 31, 2012 #4

    tiny-tim

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    nooo :redface:

    what is the definition of span = ℝn? :smile:
     
  6. Jan 31, 2012 #5

    jmm

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    Does it mean that all of the linear combinations of vectors in the span form the space Rn?
     
  7. Jan 31, 2012 #6

    tiny-tim

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    what does it mean about any individual vector? :smile:
     
  8. Jan 31, 2012 #7

    jmm

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    Ummmm, I really don't know :(
     
  9. Jan 31, 2012 #8

    tiny-tim

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    look it up!! :rolleyes:

    (remember, we're talking about vector spaces :wink:)
     
  10. Jan 31, 2012 #9

    jmm

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    Oh, believe me, I've been looking it up for the past 8 hours haha. Everything I've seen has it defined in terms of combinations of many vectors.
     
  11. Jan 31, 2012 #10

    tiny-tim

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    in a vector space, you can express any vector as … ?
     
  12. Jan 31, 2012 #11

    jmm

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    I don't know that one either. I mean I probably do but I can't figure out where you're trying to lead me :)

    And I thought another linear algebra course would be good for me!
     
  13. Jan 31, 2012 #12

    tiny-tim

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    hint: basis :wink:
     
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