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Math Help - dual space

  1. #1
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    dual space

    ok..

    let V be a finite dimensional vector space and let V* be the dual space.

    Let W_1 and W_2 be subspaces of V*. SHOW Ann(W_1\cap W_2)=Ann(W_1)+ Ann(W_2) where Ann(W)=\{v\in V| f(v)=0 for all f\in W\}.

    I can get Ann(W_1)+ Ann(W_2)\subset Ann(W_1\cap W_2). But i cannot get containment in the other direction.

    For instance if w\in Ann(W_1\cap W_2) then f(w)=0 for all w\in W_1\cap W_2. But how can I show this implies w=a+b for some a\in Ann(W_1) and b\in Ann(W_2)?
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  2. #2
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    Quote Originally Posted by ziggychick View Post
    ok..

    let V be a finite dimensional vector space and let V* be the dual space.

    Let W_1 and W_2 be subspaces of V*. SHOW Ann(W_1\cap W_2)=Ann(W_1)+ Ann(W_2) where Ann(W)=\{v\in V| f(v)=0 for all f\in W\}.

    I can get Ann(W_1)+ Ann(W_2)\subset Ann(W_1\cap W_2). But i cannot get containment in the other direction.

    For instance if w\in Ann(W_1\cap W_2) then f(w)=0 for all w\in W_1\cap W_2. But how can I show this implies w=a+b for some a\in Ann(W_1) and b\in Ann(W_2)?
    i can solve the problem provided that the following is true (so you need to tell me if such a thing basically exists in your textbook or not):

    Theorem (i just made it up!): suppose V is a finite dimensional vector space over F and W is a subspace of V^*. then for any linear map \varphi: W \to F, there exists v \in V

    and a linear map \tilde{\varphi} : V^* \to F such that \tilde{\varphi} |_W=\varphi and \tilde{\varphi}(g)=g(v), for all g \in V^*.

    the reason that i think there must be such a theorem in linear algebra is that we have the same situation in ring theory, i.e. if R is self-injective and I,J are two ideals

    of R, then Ann(I)+Ann(J)=Ann(I \cap J).
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    i can solve the problem provided that the following is true (so you need to tell me if such a thing basically exists in your textbook or not):

    Theorem (i just made it up!): suppose V is a finite dimensional vector space over F and W is a subspace of V^*. then for any linear map \varphi: W \to F, there exists v \in V

    and a linear map \tilde{\varphi} : V^* \to F such that \tilde{\varphi} |_W=\varphi and \tilde{\varphi}(g)=g(v), for all g \in V^*.

    the reason that i think there must be such a theorem in linear algebra is that we have the same situation in ring theory, i.e. if R is self-injective and I,J are two ideals

    of R, then Ann(I)+Ann(J)=Ann(I \cap J).
    unfortunately my book only has a couple pages on the subject. i know dim(V)=dim(V*), and V is isomorphic to its double dual. there's nothing else very enlightening in the chapter...
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  4. #4
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    Quote Originally Posted by ziggychick View Post

    unfortunately my book only has a couple pages on the subject. i know dim(V)=dim(V*), and V is isomorphic to its double dual. there's nothing else very enlightening in the chapter...
    ok, V \cong V^{**} is good enough to prove "my" theorem! so, let w \in Ann(W_1 \cap W_2). define the linear map \varphi: W_1 + W_2 \longrightarrow F by \varphi(f_1 + f_2)=f_1(w). note that \varphi is well-defined because

    if f_1+f_2=0, then f_1=-f_2 \in W_2 and hence f_1 \in W_1 \cap W_2. thus f_1(w)=0. we know that any linear map from a vector subspace can be extended to the entire vector space, i.e. there

    exists \tilde{\varphi}: V^* \longrightarrow F such that \tilde{\varphi}|_{W_1+W_2}=\varphi. now we know that there's an isomorphism \Phi: V \longrightarrow V^{**} defined by \Phi(v)(g)=g(v), for all v \in V, \ g \in V^*. thus, since \tilde{\varphi} \in V^{**}, there must

    exist v \in V such that \Phi(v)=\tilde{\varphi}, i.e. \tilde{\varphi}(g)=\Phi(v)(g)=g(v), for all g \in V^*. (this proves the theorem that i mentioned!) therefore for any f_1 \in W_1: \ f_1(w)=\varphi(f_1)=\tilde{\varphi}(f_1)=f_1(v). thus

    f_1(w - v)=0, \ \forall \ f_1 \in W_1. hence w - v \in Ann(W_1). call this result (1). on the other hand for any f_2 \in W_2: \ 0=\varphi(f_2)=\tilde{\varphi}(f_2)=f_2(v). hence v \in Ann(W_2). this result with (1) proves

    that Ann(W_1 \cap W_2) \subseteq Ann(W_1) + Ann(W_2). the other direction of the inclusion is trivial. (that was a nice problem! )
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  5. #5
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    hmm ok i understand what you've done... thanks!
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