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Thread: dual space

  1. #1
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    dual space

    ok..

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

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

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

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

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

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

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

    For instance if $\displaystyle w\in Ann(W_1\cap W_2)$ then f(w)=0 for all $\displaystyle w\in W_1\cap W_2$. But how can I show this implies $\displaystyle w=a+b$ for some $\displaystyle a\in Ann(W_1)$ and $\displaystyle b\in Ann(W_2)$?
    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 $\displaystyle V$ is a finite dimensional vector space over $\displaystyle F$ and $\displaystyle W$ is a subspace of $\displaystyle V^*.$ then for any linear map $\displaystyle \varphi: W \to F,$ there exists $\displaystyle v \in V$

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

    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 $\displaystyle R$ is self-injective and $\displaystyle I,J$ are two ideals

    of $\displaystyle R,$ then $\displaystyle Ann(I)+Ann(J)=Ann(I \cap J).$
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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 $\displaystyle V$ is a finite dimensional vector space over $\displaystyle F$ and $\displaystyle W$ is a subspace of $\displaystyle V^*.$ then for any linear map $\displaystyle \varphi: W \to F,$ there exists $\displaystyle v \in V$

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

    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 $\displaystyle R$ is self-injective and $\displaystyle I,J$ are two ideals

    of $\displaystyle R,$ then $\displaystyle Ann(I)+Ann(J)=Ann(I \cap J).$
    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, $\displaystyle V \cong V^{**}$ is good enough to prove "my" theorem! so, let $\displaystyle w \in Ann(W_1 \cap W_2).$ define the linear map $\displaystyle \varphi: W_1 + W_2 \longrightarrow F$ by $\displaystyle \varphi(f_1 + f_2)=f_1(w).$ note that $\displaystyle \varphi$ is well-defined because

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

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

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

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

    that $\displaystyle Ann(W_1 \cap W_2) \subseteq Ann(W_1) + Ann(W_2).$ 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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