Show that there exists $0 \neq v \in V$ such that $\alpha (v) = 0$ for all $\alpha \in R.$

Question

Let $V$ be a finite dimensional vector space and $R \subseteq V^{\ast}$ be such that $\text {span}\ R \subsetneq V^{\ast}.$ Then there exists $0 \neq v \in V$ such that $\alpha (v) = 0$ for all $\alpha \in R.$

Contrapositively, I have to show the following $:$

If for any $0 \neq v \in V$ there exists $\alpha_{v} \in R$ (depending upon $v$) such that $\alpha_{v} (v) \neq 0$ then $\text {span}\ R = V^{\ast}.$

$\textbf {Guess} :$ Let $\{v_1, \cdots, v_n \}$ be a basis of $V$ then $\left \{\alpha_{v_1}, \cdots, \alpha_{v_n} \right \}$ is a basis of $V^{\ast}$ contained in $R.$

But I can't prove it. Any help in this regard would be greatly appreciated.

Thanks for your time.

Answer 1

Hint: Start with a basis of $R$, and extend it to $V^*$. Notice that since $V$ is finite-dimensional, this basis of $V^*$ must be the dual to some basis of $V$. Can we find a non-zero $V$ that satisfies the given condition?