# Set of linear functionals span the dual space iff intersection of their kernels is $\{0\}$.

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 $$\bigcap_{i=1}^n \ker(f_i)$$ = $$\{0\}$$

• Suppose the intersections of their kernels is not empty. Then there exists a $v \in V$ which gets mapped to $0$ by all $f_i$. Why does this prove that $\{f_1, \dots, f_k \}$ does not span $V^*$? Jun 18, 2019 at 18:31
• Sorry I should have mentioned, that direction is simple but I was wondering about the if and only if condition Jun 18, 2019 at 18:59

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 $$\beta_j \in \cap_k \ker f_k$$.