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Homework Help: Linear Algebra - Prove that E is not a basis for V.

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Since it's kind of hard to type out, I'll try to post a screenshot:

    [PLAIN]http://img841.imageshack.us/img841/7357/questionq.jpg [Broken]

    2. Relevant equations

    There's the definition of a basis, vector space, and all the axioms.

    3. The attempt at a solution

    I understand part A; it's simple enough, but I'm really stuck on part B. Had they been finite sets, as stated in our textbook E would be the standard basis for that finite set, i.e. {(1,0,0), (0,1,0), (0,0,1)} is the standard basis for R3. But somehow for infinite sequences this is not the case. Since I know by part A the set is linearly independent, it must be the second part of the definition of a basis that is violated, meaning it does NOT generate all infinite sequences. My initial guess is that sequences like {0,0,0,...0,0,1} can't be generated because j cannot reach infinity or something like that, but I don't know if it's correct to say that. Any ideas?

    Thanks in advance.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 26, 2010 #2
    Perhaps you should start with stating to yourself precisely and carefully what your fuzzy term "generate" means.
     
  4. Sep 26, 2010 #3

    Office_Shredder

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    Also {0,0,0,...0,0,1} isn't a sequence in the set, there is no infinity term.
     
  5. Sep 26, 2010 #4
    Alright so maybe {0,...,0,1} doesn't work. By generate I simply mean to span V. As in, when I say {0,0,0,...0,0,1} can't be generated I mean it is not in span(E).
     
  6. Sep 26, 2010 #5
  7. Sep 27, 2010 #6
    Well, what it means "span"? Can you provide the exact definition?
     
  8. Sep 27, 2010 #7

    HallsofIvy

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    That, {0, 0, 0,..., 0, 0, 1} is not in your vector space because it has some unknown but finite number of components. {0, 0, 0, ..., 0, 0, 1, 0, ...} is[\b] in your vector space and is in the span of the given vectors.

    Look at {1, 1, 1, ..., 1, 1, 1, ...}, the vector all of whose components are 1. Is that in the span?

    (You have been asked to state the precise definition of "span". That is not just being picky. In mathematics you use the precise words of definitions in proofs. I can think of one word in the definition of "span" that is crucially important here!)
     
  9. Sep 27, 2010 #8
    Hint: this one important word should start with letter "f".
     
  10. Sep 27, 2010 #9
    I have been thinking about this question and I stumbled upon a similar one: the telescoping series. This has a basis of {1-x, x-x^2, x^2-x^3...}. So 1 = (1-x) + (x-x^2) ... . But 1 is not in the span, I'm guessing for the same reason as in this problem. So, answering your question, no, {1,1,... 1, 1} is not in the span of E. The definition of span (in my textbook, shortened,) is "the set consisting of all linear combinations of the vectors in S." Hmm.
     
  11. Sep 27, 2010 #10
    Ok I looked at the book's definition of linear combination and figured it out. In my book linear combinations need a finite number of vectors to make the sum. Therefore, since {1,1...1,1} requires is an infinite sequence and requires an infinite amount of vectors in E, E does not generate all infinite sequences. Thanks!
     
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