Let \(F\) be a field of characteristic zero and let \(V\) be a finite-dimensional vector space over \(F\). If \(\alpha_1, \ldots, \alpha_m\) are finitely many vectors in \(V\), each different from the zero vector, prove that there is a linear functional \(f\) on \(V\) such that \[ f(\alpha_i) \neq 0, \quad i = 1, \ldots, m. \]
Solution
We prove the statement by induction on \(m\). For \(m=1\), extend \(\alpha_1\) to a basis of \(V\) and choose a coordinate functional taking value \(1\) on \(\alpha_1\).
Assume the result for \(m-1\). Choose \(f\in V^*\) such that \[ f(\alpha_i)\ne0\qquad(1\le i<m). \] Choose \(g\in V^*\) with \(g(\alpha_m)\ne0\); such a \(g\) exists by extending \(\alpha_m\) to a basis.
For \(t\in F\), put \(f_t=f+t g\). For each fixed \(i\), the equation \[ f_t(\alpha_i)=0 \] excludes at most one value of \(t\); for \(i=m\) this is true because \(g(\alpha_m)\ne0\), and for \(i<m\) it is either one excluded value or none. Thus only finitely many scalars \(t\) are forbidden.
Since \(F\) has characteristic zero, it is infinite. Choose \(t\) outside this finite set. Then \[ f_t(\alpha_i)\ne0\qquad(1\le i\le m), \] as required.