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Math Help - Aother subset proof

  1. #1
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    Aother subset proof

    Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 union S2) = Span(S1)+Span(S2).
    Can someone solve this?

    And did I approach this correctly

    1. Let s1 and s2 be arbitrary subsets of a v. space V.
    2. Let x be an arbitrary vector such that x belongs to (S1 union S2). So x belongs to S1 or x belongs to S2. Then Span(S1) IS all the linear combinations containing the vector x such that (for all a belonging to the field F)(a is a scalar) then the summation a*x belongs to Span(S1)
    and then.....

    my wordin prob sounds bad i dont know how u guys use the summation symbol and the for all symbol on the internet... how do u guys do that anyhow
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  2. #2
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    Quote Originally Posted by ruprotein View Post
    Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 union S2) = Span(S1)+Span(S2).
    Can someone solve this?
    \mbox{span}(S_1\cup S_2)\not = \mbox{span}S_1+\mbox{span}S_2
    Because the right hand side is the set of ordered paris of S_1 and S_2.

    Hence your probably mean to say (and you tell me what):
    1) \mbox{span}(S_1\cup S_2)=\mbox{span}(S_1)\cup \mbox{span}(S_2)
    2) \mbox{span}(S_1\cup S_2)\simeq \mbox{span}(S_1)+\mbox{span}(S_2).

    By \simeq I mean, isomorphic.
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  3. #3
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    Actually he does mean span\left( {S_1  \cup S_2 } \right) = span\left( {S_1 } \right) + span\left( {S_2 } \right). The term \left( {S_1  \cup S_2 } \right) is only a union not a set of ordered pairs.

    As for a proof, it is easy to see that span\left( {S_1 } \right) + span\left( {S_2 } \right) \subseteq span\left( {S_1  \cup S_2 } \right).
    To prove span\left( {S_1  \cup S_2 } \right) \subseteq span\left( {S_1 } \right) + span\left( {S_2 } \right) is just a messy exercise. In any linear combination from span\left( {S_1  \cup S_2 } \right) you need to separate the spanning elements into one of those two sets.
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