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Let $S$ be a subspace of a vector space $V$ so that $V/S$ is finite dimensional. How do we show that $S$ is the intersection of kernels of a finite number of elements in the dual of $V$? If $S$ is finite dimensional, what can we conclude about the dimension of $V$?

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This looks like homework. What have you tried? –  Alex Becker Aug 24 '12 at 21:23
    
Actually I was trying to solve the 1 st part by looking at examples , but I cannot really generalize it –  Ester Aug 24 '12 at 21:25
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Well, first consider the case where $V/S$ is one-dimensional. Can you use that to get a linear functional from somewhere? –  GEdgar Aug 24 '12 at 21:30
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Put $\,S=Span\{v_i\;\;|\;\;i\in I\}\,$ and $\,W:= V\setminus S\Longrightarrow W=Span\{w_1,...,w_n\}\,$ , and we get at once that $$\{w_1,...,w_n,w_k\;\;|\;\;k\in K:=\text{ a set of indexes}\,\,,\,K\cap\{1,2,...,n\}=\emptyset\}$$

is a basis of $\,V\,$ . Define (with $\,\Bbb F:=\,$ the definition field)

$$\forall\,\,i=1,2,...,n\,\,,\,f_i: V\to \Bbb F\,\,,\,f_iw_j=\delta_{ij}=\left\{\begin{array} {}1 &\text{, if}\,\,1\leq i=j\leq n\\0&\text{, if}\,\,1\leq i\neq j\leq n\,\,\text{ or}\,\,j\in K\end{array}\right.$$ and extend this definition by linearity. Clearly $\,f_i\in V^*\,$ , and letting $$v=\sum_{r=1}a_rw_r+\sum_{k\in K}a_kv_k\,\,,\,\,a_k=0\,\,\text{but for a finite number of indexes in} \,K$$ we get that $$\ker f_i:=\{v\in V\;\;|\;\;f_iv=0\}=\{v\in V\;\;|\;\;a_i=0\,,\,1\leq i\leq n\,,\,\,\text{in the above expression for}\,\,v\}$$

And thus $$\bigcap_{i=1}^n\ker f_i=\{v\in V\;\;|\;\;a_1=...=a_n=0\,\,\text{in the above expression for}\,\,v\}$$ Well, now end the argument.

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Thanks to both of you for your help . –  Ester Aug 25 '12 at 7:32
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Let $\{x_i\}_{i\in I}$ be a basis for $S$. Since $V/S$ is finite-dimensional, we have some finite basis $\{y_1+S,\ldots,y_n+S\}$ with each $y_i\in V$. Then $\{x_i\}_{i\in I}\cup \{y_1,\ldots,y_n\}$ is a basis for $V$. Define the linear functionals $f_j\in V^*$ by $f_j(x_i)=0$ for all $i\in I$ and $f_j(y_k)=\delta_{jk}$. What is the kernel of $f_j$? What are the intersection of these kernels?

For the second part, note that $\{x_i\}_{i\in I}\cup \{y_1,\ldots,y_n\}$ is a basis for $V$, and if $I$ is finite then...

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How are you defining a basis in V which may be infinite dimensional? –  Ester Aug 27 '12 at 20:26
    
@Ricky All vector spaces have bases, even if they are infinite dimensional (assuming axiom of choice, which is natural in this setting). Pick an arbitrary one. –  Alex Becker Aug 27 '12 at 21:57
    
Can we solve this problem without using the concept of basis in infinite dimensional vector spaces? –  Ester Aug 28 '12 at 9:00
    
@Ricky Not without using something equivalent I think. Some form of choice is necessary. –  Alex Becker Aug 28 '12 at 21:17
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