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Linear algebra proof and vectors

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    a) Prove that if u1, ... um are vectors in Rn , S = {u1,u2,...uk} and T = {u1,...uk, uk+1,...um} then span(S) [tex]\subseteq[/tex]span(T).

    b) deduce also that if Rn = span(S), then Rn=span(T)


    2. Relevant equations



    3. The attempt at a solution
    I think I got part a:

    the span S is represented by linear combination c1u1 + c2u2 + ... ckuk
    and the span T is represented by the linear combination
    c1u1 + c2u2 + ... ckuk +... cmum
    and since span(S) is contained in span(T)

    span(S)[tex]\subseteq[/tex]span(T)

    Does this look alright?


    b) part b is giving me trouble

    since span(S) = Rn, the entire set of linear combinations of the vectors {u1,u2,...uk} in set S forms a plane.

    since span(S) is a subset of (T) which spans Rn,
    span(T) = Rn.

    I don't feel like this right at all, and I really can't visualize in my head what's going on here. Could somebody please help me? Thanks.
     
  2. jcsd
  3. Mar 7, 2009 #2

    lurflurf

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    Homework Helper

    for b)
    We know span(T) contains R^n.
    How do we know it can not contain anything else?
     
  4. Mar 7, 2009 #3
    Algebraically you are on the right track. Just remember these definitions:
    1) Span(u1, u2, ..., ui) = {a1u1 + a2u2 + ... + aiui | a1, a2, ..., ai in R}

    2) A is a subset of B if every x in A is also in B.

    3) Sets A and B are equal if A is a subset of B and B is a subset of A.

    For some intuition, it may be helpful to consider what this looks like in R3. Here are some pictures:

    http://img242.imageshack.us/img242/1087/spanliner3.png [Broken]

    http://img243.imageshack.us/img243/8680/spanplaner3.png [Broken]

    http://img243.imageshack.us/img243/9002/spanblockr3.png [Broken]
     
    Last edited by a moderator: May 4, 2017
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