If $\alpha_{1},\ldots,\alpha_{m}$ are vectors different from zero vector, then there is a linear functional $f$ such that $f(\alpha_{i})\neq 0$ I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise 14 from page 106.

Let $\mathbb{F}$ be a field of characteristic zero and let $V$ be a
  finite-dimensional vector space over $\mathbb{F}$. If
  $\alpha_{1},\ldots,\alpha_{m}$ are finitely many vectors vectors in
  $V$, each different from zero vector, prove that there is a linear
  functional $f$ such that $f(\alpha_{i})\neq 0, i=0,\ldots,m.$

I thought that prove by induction on $m$ would work. I showed the base case, but I got stuck on inductive step.
I would appreciate your help.
 A: The following should work.
Suppose the statement holds for $m-1$. Let $f$ be a functional such that $f(\alpha_i)\neq 0$ for $1\le i\le m-1$. If $f(\alpha_m)\neq 0$, we are done, otherwise $f(\alpha_m)=0$. Using the inductive hypothesis again, there must also be a functional $g$ such that $g(\alpha_m)\neq 0$. Using the fact that $\mathbb F$ has characteristic $0$, we can see $(*)$ that there exists a natural number $n\in\mathbb N$ such that $ng(\alpha_i)\neq -f(\alpha_i)$ for all $1\le i\le m$. But then $f+ng$ is the functional we wanted.

$(*)$: To see this, define $A_i:=\lbrace n\in\mathbb N|\text{ }ng(\alpha_i)=-f(\alpha_i)\rbrace$ for each $i\in\lbrace 1,2,\ldots,m\rbrace$. Each $A_i$ may contain at most one element.
(Otherwise, we would have $n_1g(\alpha_i)=-f(\alpha_i)=n_2g(\alpha_i)$ for some $i$ and some $n_1\neq n_2$. Since the characteristic of $\mathbb F$ is $0$ this would imply $g(\alpha_i)=0$ which would be impossible for $i<m$ because then $f(\alpha_i)=-n_1 g(\alpha_i)=0$ and also impossible for $i=m$ since $g(\alpha_m)\neq 0$ by the choice of $g$.)
Therefore $\bigcup\limits_{i=1}^m A_i$ is a finite subset of $\mathbb N$, which proves the claim above i.e. we can choose a natural number $n$ such that $ng(\alpha_i)\neq -f(\alpha_i)$ for all $i$.
A: Suppose you write $V = F^n$. Then since $F$ contains $\mathbb{Q}$ it suffices to find a linear functional $f$ on $E = \mathbb{Q}^n$ such that its prolongation $\tilde f$ on $V$ satisfies $\tilde f(\alpha_i)\ne 0$. For each $i$ the equation $\tilde f(\alpha_i)=0$ defines a subspace $F_i$ of $E^* \simeq E$ that is of dimension at most $n-1$. The closure of each $F_i$ in $\mathbb{R}^n$ is a negligible closed set, therefore their union is negligible and closed so the complement is a non-empty open set. This implies there is a point in $\mathbb{Q}^n$ not in the union of the $F_i$.
