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Spans in Linear Algebra

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose V is a vector space. Let E,F be subsets of V. Show [itex]E \subseteq F \Leftrightarrow L(E) \subseteq L(F)[/itex]

    3. The attempt at a solution
    Let [itex] x \in L(E)[/itex], there are scalars [itex]q_i[/itex] such that [itex]x = \sum_{i} q_i p_i[/itex] where [itex]p_i \in E[/itex]. [itex]p_i \in F[/itex] because [itex]E \subseteq F[/itex]. Thus it is shown that [itex]x \in L(F)[/itex]. From this result, L(E) is a subset of L(F).

    First of all, I'm not sure if this is a convincing proof or if I have notated it correctly. Second, I don't think I understand the proof very well. For example, why is it (if this proof is true) that [itex]p_i \in E[/itex]? Similarly, is [itex]q_i \in E[/itex]? If so, how is this known? Mostly, what I think I'd like to see is a less mathy and more wordy explanation of what's going on here.

    Thanks!
     
  2. jcsd
  3. Jan 13, 2012 #2

    Mark44

    Staff: Mentor

    Other notation I've seen is Span(E). I believe this is what you mean when you write L(F), the set of all linear combinations of vectors in F.
    Your notation is a bit on the hard side to read, partly because there is a lot you're not saying. For example, I assume you mean that the pi's are a basis for E. Also, your notation makes it difficult to tell scalars from vectors. Making the vectors bold would help alleviate that difficulty.

    I would also recommend using different letters for the vectors in the two sets. Instead of x and pi, I would use e as a vector in E, and e1, ..., en as basis vectors, and maybe c1, ..., cn for the scalars.
    It's possible that pi does not belong to E, such as if E = {0}.
    No. The qi's are scalars, so they belong to some field, not to a vector space. That's what I meant about your notation being confusing - you have managed to confuse yourself.
    Don't forget that this is an if and only if proof, so you need to go the other way, as well.
     
  4. Jan 14, 2012 #3
    [itex] p_i [/itex] are not bases for [itex] E [/itex](they can be though) since [itex] E [/itex] is not a vector space, it's a SUBSET of [itex] V [/itex]. [itex] L(E) [/itex] it's a vector space on the other hand but the [itex] p_i [/itex] might be or might be not bases for [itex] L(E) [/itex]

    For examples.

    [itex] V = ℝ^2 [/itex]

    [itex] E = \left\{ (1,0),(0,1),(1,1),(2,1) \right\} \subseteq V[/itex]

    [itex] L(E) = ℝ^2 [/itex] but [itex] E [/itex] is not a basis for[itex] ℝ^2 [/itex] since [itex] E [/itex] is not linearly independent.




    Why is it true that [itex] p_i \in E [/itex] ? Well that comes from the definition of span of a subset ( L(E ), which is:

    [itex] L(E) [/itex] is the set containing all the linear combinations of the elements of E.
    in formula [itex] L(E) = \left\{ r_1e_1 + r_2e_2 + ... + r_ie_i | e_i \in E , r_i \in R \right\} [/itex]

    or more generally if you know what a field K is

    [itex] L(E) = \left\{ r_1e_1 + r_2e_2 + ... + r_ie_i | e_i \in E , r_i \in K \right\} [/itex]

    you can rewrite this in a more compact form, and using [itex] p_i [/itex] instead of [itex] e_i [/itex] and [itex] q_i [/itex] insted of [itex] r_i [/itex]


    [itex] L(E) = \left\{ \sum q_ip_i| p_i \in E , q_i \in K \right\} [/itex]

    so saying that [itex] x \in L(E) [/itex] means that there exist some [itex] p_i \in E, q_i \in K[/itex] for which [itex] x = \sum q_ip_i [/itex]. Does this answer your question?
     
    Last edited: Jan 14, 2012
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