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Math Help - null-space and annihilators

  1. #1
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    null-space and annihilators

    I tried to work the following two rules, if someone can please confirm to me that they are indeed correct and guide how can I prove/counter-prove them

    V - n dim vector space
    V* dual space of V i.e. Hom(V,F)

    Let S be a sub-set of V. A(S) is annihilator of S (thus is a sub-space of V*)
    Let F be a sub-set of V*. N(F) is null-space of F (which means all v in V, such that for all f in F f(v)=0). N(F) can be shown to be a sub-space of V.

    L(X) is linear span of any sub-set of any vector space.

    Under the above notations, I want to establish/dis-prove the following:

    1. N(A(S)) = L(S)

    2. A(N(F)) = L(F)

    Will really appreciate help here. I am just getting too confused here.
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    I tried to work the following two rules, if someone can please confirm to me that they are indeed correct and guide how can I prove/counter-prove them

    V - n dim vector space
    V* dual space of V i.e. Hom(V,F)

    Let S be a sub-set of V. A(S) is annihilator of S (thus is a sub-space of V*)
    Let F be a sub-set of V*. N(F) is null-space of F (which means all v in V, such that for all f in F f(v)=0). N(F) can be shown to be a sub-space of V.

    L(X) is linear span of any sub-set of any vector space.

    Under the above notations, I want to establish/dis-prove the following:

    1. N(A(S)) = L(S)

    2. A(N(F)) = L(F)
    For 1., N(A(S)) is a subspace containing S, so it contains L(S). For the reverse inclusion, suppose that x∉L(S). Let \{e_1,\ldots,e_k\} be a basis for S. Then the set \{e_1,\ldots,e_k,x\} is linearly independent, so we can extend it to a basis \{e_1,\ldots,e_n\} for V, where e_{k+1} = x. Define an element f of V* by f\bigl(\textstyle\sum \alpha_ie_i\bigr) = \alpha_{k+1}. Then f∈A(S). But f(x)≠0, so x∉N(A(S)).

    2. is also true, and it should be provable in a similar way.
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  3. #3
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    @Opalg - Thanks. I follow your proof.
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