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Linear Algebra, spans

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    I think I get the basic concept of spans (all possible combinations of vectors with all possible scalers).


    3. The attempt at a solution:blushing:
    It seems obvious that the span of S would have to be in the span of T, I don't understand what is left to "prove". I have not done a lot of proofs. I don't know where to begin with this problem.
     
    Last edited: Mar 26, 2008
  2. jcsd
  3. Mar 26, 2008 #2

    quasar987

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    To prove a set A is a subset of some other set B, you must show that every element of A is also an element of B.

    In your case, this will follow very easily from the definitions of span(S) and span(T).
     
  4. Mar 26, 2008 #3
    [​IMG]
    Thank you quasar987,
    I don't know how I could better state what seems obvious.

    Maybe it should say 'All possible linear combinations...'

    I won't be including the definition (or my misspelling, the definition was copied from my text book). Sorry if this seems dumb; maybe I'm not clear what proofs are about; the premise of the question almost seems like proof.
     
    Last edited: Mar 26, 2008
  5. Mar 26, 2008 #4

    quasar987

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    I understand how you feel perfectly. Feels like I was there yesterday!

    But stick to it... as you read more proofs and attempt to write some yourself, you will eventually see a pattern in the techniques used and you will pick up the proper vocabulary for writing proofs.

    In the meantime, I give you this to feed on. Compare my proof to yours.

    Proof: Let [tex]s=c_1v_1+...+c_kv_k[/tex] be an element of span(S).

    We want to show that s is an element of span(T) also.

    Recall that span(T) is the set of all elements of the form [tex]t=d_1v_1+...+d_mv_m[/tex]. In particular, for [tex]d_1=c_1,...,d_k=c_k[/tex] and [tex]d_{k+1}=...=d_m=0[/tex], we get that [tex]c_1v_1+...+c_kv_k+0v_{k+1}+...+0v_m=c_1v_1+...+c_kv_k=s[/tex] is an element of span(T).

    Since the element s is arbitrary, it follows that all elements of span(S) are in span(T); that is to say, span(S) is a subset of span(T).
     
    Last edited: Mar 26, 2008
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