Let $\{u,v\}$ be a linearly independent set in $\Bbb R^3$. Show that the plane $\{su+tv|s,t \in\Bbb R\}$ through the origin in $\Bbb R^3$ is equal to the null space of some element of $(\Bbb R^3)^{*}$.
Attempt: We are supposed to prove that $su+tv=ker(g)=0$ for some $g \in (R^3)^{*}$. And I am stuck now.
I want to argue algebraically. I know that $g= g(s) g_1+ g(t) g_2 $for $g_1, g_2$ being basis for g.
Then from definition of kernel we have
$g(s)=g(s) g_1 (s) +g(s) g_2 (s)=0 \rightarrow g(s)=0$
and $g(t)=g(s) g_1 (t) +g(s) g_2 (t)=0 \rightarrow g(t)=0$
So we have $su+tv=0=g(s)=g(t)$
I am not sure if that makes sense.
As suggested in the comment, we note that we may extend $\{u,v\}$ to a basis for $\mathbb{R}^3$, say $\{u,v,w\}$. We may define a linear map by specifying its image on basis element. Define a linear map $T:\mathbb{R}^3 \to \mathbb{R}$ by $T(u):=0$, $T(v):=0$, and $T(w):= 1$ (or in fact you can replace 1 with any nonzero real number). Then $T \in (\mathbb{R}^3)^*$ and $\ker(T) = \{su + tv: s,t \in \mathbb{R} \}$.
To tackle a problem, sometimes it is useful to start with a simple example. Let $e_1 = (1,0,0)^T, e_2 = (0,1,0)^T, e_3 = (0,0,1)^T$. Then $\{e_1,e_2,e_3\}$ is a basis for $\mathbb{R}^3$. Take (as an example) $u= e_1$ and $v=e_2$. In this case, one may see (geometrically) that $\{su + tv:s,t \in \mathbb{R} \}$ is simply the xy-plane. It is the nullspace of the linear map $T_0:\mathbb{R^3} \to \mathbb{R}$ defined by $T_0(x,y,z):=z$ (in fact you may replace $z$ with $rz$, where $r$ is any nonzero real number). Another way to think about this $T_0$ is: it is the linear map defined by setting $T_0(u):=0$, $T_0(v):=0$, and $T_0(e_3) = 1$ (and extending it linearly).