I was wondering if anyone could offer some insight into the following problem:
Let$\mathit{V}$ be a vector space over a field $\mathbb{F}$. Assume that dim$\mathit{V}$ is finite. Let $f_1, \ldots, f_k$ $\in$ $\mathit{V}$ '
Show that $\{f_1, \ldots, f_k\}$ span $\mathit{V}$ ' iff $\bigcap_{i=1}^n \ker(f_i)$ = $\{0\}$
First suppose $f_1,...,f_n$ is a basis for $V^*$ and $b_1,...,b_n$ is a basis for $V$. Then the matrix $C$ defined by $[C]_{ij} = f_i(b_i)$ is invertible, and if we let $A=C^{-1}$ then the points $\beta_i = \sum_k [A]_{ik} b_k$ satisfy $f_j ( \beta_i) = \delta_{ij}$. That is, $f_k$ is the dual basis for $\beta_k$.
Suppose $f_k$ span $V^*$, and $x \in V$. Without loss of generality we can assume that $f_k$ is a basis for $V^*$. Write $x=\sum_k \alpha_k \beta_k$. Since $\alpha_k = f_k(x)$, we see that if $f_k(x) = 0$ for all $k$ then $x=0$. Hence $\cap_k \ker f_k = \{0\}$.
Now suppose the $f_k$ do not span $V^*$. Without loss of generality we can assume that the $f_k$ are linearly independent. Add functionals $g_j$ such that $f_k,g_j$ form a basis. Construct the basis $\beta_k$ of $V$ as above. Suppose $g_j(\beta_j) = 1$, then we have $f_k(\beta_j) = 0$ for all $k$ and so $\beta_j \in \cap_k \ker f_k$.
I don't think that the solution below is any special, but it deserves attention for the way it's formulated. Assuming that $V$ is finite-dimensional, with dimension $\dim V=n$ and we are given n linear functionals : $f_1, f_2, \dots f_n \in V^{*}$ We have to prove the following : $$\langle f_1, f_2, \dots , f_n \rangle = V^* \Leftrightarrow \bigcap_{1\leq i \leq n} f_i ={0} $$ Proof: $$ \langle f_1, \dots, f_n \rangle =V^{*} \Leftrightarrow \langle f_1, \dots, f_n \rangle ^0 =(V^{*})^0={0}. $$ I used the fact that $V^{**}$ can be identified with $V$ and $U$ with $U^{00}$ respectively.
