How to show that the reciprocal set is also a basis of $\mathbb{R}^3$

Question

If three vectors (say $\vec v_1$, $\vec v_2$ and $\vec v_3$) are a basis of $\mathbb{R}^3$, their reciprocal vectors ($\vec u_1$, $\vec u_2$ and $\vec u_3$, defined by $\vec{u}_i \cdot \vec{v}_j = \delta_{ij}$) must also be a basis of $\mathbb{R}^3$.

How do I go about proving this? I have tried using $\alpha_{1}\vec v_{1} + \alpha_{2}\vec v_{2} + \alpha_{3}\vec v_{3} = 0$ means that $\alpha_1 = \alpha_2 = \alpha_3 = 0$, and dotting different things through it, but I'm not getting very far.

Answer 1

Hint: If $\alpha_1\vec{u_1}+\alpha_2\vec{u_2}+\alpha_3\vec{u_3}=0$, then $\vec{v_1}.\left(\alpha_1\vec{u_1}+\alpha_2\vec{u_2}+\alpha_3\vec{u_3}\right)=0$ too. In other words, $\alpha_1=0$.