I'm trying to solve below exercise, i.e.,
Let $X$ be a vector space. Fix $n$ linear functionals $f_1, \ldots, f_n$ and $n$ real numbers $\alpha_1, \ldots, \alpha_n$. Then the following statements are equivalent.
- (a) There is $x \in E$ such that $f_i(x) = \alpha_i$ for all $i=1, \ldots,n$.
- (b) For any $(\beta_1, \ldots, \beta_n) \in \mathbb R^n$ such that $\sum_{i=1}^n \beta_i f_i=0$, we have $\sum_{i=1}^n \beta_i \alpha_i = 0$.
Could you please have a check on my attempt?
Proof: The direction (a) $\implies$ (b) is trivial. Let's prove the converse. WLOG, we assume $\{f_1, \ldots, f_m\}$ is a maximal linearly independent subset of $\{f_1, \ldots, f_n\}$ for some $m \le n$. Consider $$ F:X \to \mathbb R^m, x \mapsto (f_1(x), \ldots, f_m(x)). $$
Then $F$ is surjective, so there is $a \in X$ such that $F(a) = (\alpha_1, \ldots, \alpha_m)$. Let $p \in \{m+1, \ldots, n\}$. We assume $f_p = \lambda_1 f_1 +\cdots +\lambda_m f_m$ for some $(\lambda_1, \ldots, \lambda_m) \in \mathbb R^m \setminus \{0\}$. Then $f_p - \lambda_1 f_1 - \cdots -\lambda_m f_m=0$. By (b), we get $\alpha_p - \lambda_1 \alpha_1 - \cdots -\lambda_m \alpha_m=0$. On the other hand, $f_p (a) = \lambda_1 f_1 (a)+\cdots +\lambda_m f_m (a) = \lambda_1 \alpha_1 + \cdots + \lambda_m \alpha_m$. It follows that $f_p(a) = \alpha_p$. This completes the proof.
