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

  • $\begingroup$ 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 $V^*$? $\endgroup$ Jun 18, 2019 at 18:31
  • $\begingroup$ Sorry I should have mentioned, that direction is simple but I was wondering about the if and only if condition $\endgroup$
    – samlu1999
    Jun 18, 2019 at 18:59

1 Answer 1


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


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