$V$ $= \text{span}\{(1,2,3),(1,1,1)\}$ $\subseteq \mathbb{R}^3$. Find the vectors spanning $V^0$ in terms of the usual basis for $(\mathbb{R}^3)^*$.
So we want linear functionals $f \in V^*$ such that $f(v)=0$ for every $f \in V$. We know that $v=\alpha_1 v_1 + \alpha_2 v_2$, so
$0=f(v)=\alpha_1 f(v_1)+\alpha_2 f(v_2)$. But I cannot see where to go from here.#
Best Answer
So far you've shown $0 = \alpha_1f(v_1) + \alpha_2f(v_2)$. It follows that $f \in V^0$ if and only if $f(v_1) = 0$ and $f(v_2) = 0$.
For any $f \in V^*$, there is $w\in \mathbb{R}^3$ such that $f(v) = v\cdot w$. Let $w = (w_1, w_2, w_3)$ and suppose $f \in V^0$, then $f(1, 2, 3) = w_1 + 2w_2 + 3w_3 = 0$ and $f(1, 1, 1) = w_1 + w_2 + w_3 = 0$. So if $f \in V^0$, the corresponding $w = (w_1, w_2, w_3)$ satisfies the equation
\begin{align*} w_1 + 2w_2 + 3w_3 &= 0\\ w_1 + w_2 + w_3 &= 0. \end{align*}
Solving this system of linear equations, we see that $(w_1, w_2, w_3) = t(-1, -2, 1)$ where $t \in \mathbb{R}$. So $V^0 = \{tf \mid t \in \mathbb{R}\}$ where $f(v_1, v_2, v_3) = -v_1 - 2v_2 + v_3$. If $\{e_1, e_2, e_3\}$ denotes the standard basis for $\mathbb{R}^3$ and $\{e^1, e^2, e^3\}$ denotes the dual basis (i.e. $e^i(e_j) = \delta^i_j$), then $f = -e^1-2e^2+e^3$.
More generally, given a subspace $V$ of $\mathbb{R}^n$ with basis $\{v_1, \dots, v_m\}$, $f \in V^0$ if and only if $f(v_i) = 0$ for $i = 1, \dots, m$. As above, writing $f(v) = v\cdot w$ and $w = (w_1, \dots, w_n)$ we see that $f \in V^0$ if and only if
$$\left[\begin{array}\ v_1^T\\ \ \ \vdots\\ v_m^T\end{array}\right]\left[\begin{array}\ w_1\\ \ \ \vdots\\ w_n\end{array}\right] = \left[\begin{array}\ 0\\ \, \vdots\\ 0\end{array}\right].$$
As $\{v_1, \dots, v_m\}$ is a basis, the first matrix has rank $m$, so by the rank-nullity theorem, the dimension of its nullspace is $n - m$. So $\dim V^0 = n - m = n - \dim V$.