Dual of a vector space

Question

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 $V$?

Answer 1

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...

Answer 2

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.