Let $\{u,v\}$ be a linearly independent set in $\Bbb R^3$. Show that the plane $\{su+tv\mid s,t \in\Bbb R\}$ through the origin in $\Bbb R^3$ is equal to the null space of some element of $(\Bbb R^3)^{*}$. (Where $(R^3)^*$ is the dual space of $R^3$).
Can we define a linear functional, say $L:R^3 \rightarrow R^2$?
Linear functionals will be mappings $L:\mathbb{R}^3 \to \mathbb{R}$. A plane is generated by two linearly independent vectors, in this case $u$ and $v$. We observe that $u\times v$ is orthogonal to the plane. Hence, for any $x\in P$, where $P$ is the plane, we have $$ (u\times v) \cdot x =0. $$ In order to write this as $Ax=0$, make $A$ a $1\times 3$ matrix where the row is just $u\times v$ (or any constant multiple).