# Math Help - dual space

1. ## 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 $b\in Ann(W_2)$?

2. Originally Posted by ziggychick
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 $Ann(I)+Ann(J)=Ann(I \cap J).$

3. Originally Posted by NonCommAlg
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...

4. Originally Posted by ziggychick

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! )

5. hmm ok i understand what you've done... thanks!