1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

LA - Proof Involving Linear Independence and Spanning

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data

    I was trying to prove a theorem from Axler's Linear Algebra text and my proof is different from the one in the book, and I'm wondering if someone can check whether or not my proof works, since I'm just starting to write proofs.

    Theorem 2.6 (pg. 25): 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.

    2. Relevant equations



    3. The attempt at a solution

    Let [itex]V[/itex] be a finite-dimensional vector space. Suppose [itex](u_1,...,u_m)[/itex], where [itex]u_1,...,u_m \in V[/itex] is a list of linearly independent vectors and [itex] (w_1,...,w_n)[/itex] where [itex]w_1,...,w_n \in V[/itex] spans [itex]V[/itex]. We need to prove that [itex]m \leq n[/itex]. Suppose [itex]m>n[/itex]. Because the list [itex] (w_1,...,w_n)[/itex] spans V, there exist scalars [itex]a_{i,j}[/itex] where [itex]i \in [1,m], j \in [1,n][/itex] such that [itex]u_i=a_{i,1}w_1+...+a_{i,n}w_n[/itex]. Thus, the vectors in [itex](u_1,...,u_m)[/itex] can be replaced by these linear combinations. However, since [itex]m>n[/itex], there exists some [itex]k[/itex] where [itex]m<k\leq n[/itex] such that [itex]u_k[/itex] can be written in terms of [itex]u_1,...,u_{k-1}[/itex], and thus [itex](u_1,...,u_m)[/itex] is a linearly dependent list, and we have reached a contradiction. Therefore, [itex]m\leq n[/itex].

    If someone could please look this over that would be great. Also, I would really like any criticism on my logic or proof writing skills in general since I have learned proofs on my own and no one has really seen my proofs before.

    Thanks for your help :)
     
  2. jcsd
  3. Jul 8, 2011 #2

    lanedance

    User Avatar
    Homework Helper

    i think that's looking good and so is the logic ;)
     
  4. Jul 8, 2011 #3
    That's good to hear I'm on the right track. Thanks for taking a look at it :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: LA - Proof Involving Linear Independence and Spanning
Loading...