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Finite Dimensional Vector Space & Span Proof

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data

    So basically, I'm studying the proof for this:

    "In a finite dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors."

    What the book (Axler's Linear Algebra Done Right 2e) does to prove it is that:

    Suppose the list (u1, ..., um) is linaerly independent in V and (w1, ..., wn) spans V. We need to prove that m ≤ n.

    So what they do is they state that the list (w1, ..., wn) spans V, and so adding any vector to it makes it linearly dependent, and when they add one, they remove one of the w's (because in a lemma they proved that you can express any of the w's in terms of the previous vectors in the list).

    My question is, why do they do that to prove this? What is it about having more u's that creates a contradiction?

    I have a two hunches:

    1. If there are more u's than w's, then once you add all the u's (and remove the w's), you'll have no more w's, and so the list will no longer necessarily span V?
    2. If there are more u's than w's, you added a dimension, which doesn't make sense?

    Thank you so much ^_^
     
  2. jcsd
  3. Oct 10, 2012 #2

    Dick

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    Once you have that (u1, ...,un) spans V, then if there is a un+1 then it must be a linear combination of (u1, ...,un). But the u's were supposed to be linearly independent. You've created a contradiction, not "added a dimension".
     
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