Problem: Suppose $V$ is a finite dimensional vector space and $v \in V$ with $v \neq 0$. Let $V^* = \mathcal{L} (V,\mathbb{R} )$ denotes the set of all linear functionals on $V$. Then prove that there exists $\phi \in \mathcal{L} (V,\mathbb{R} )$ such that $\phi (v) = 1$.
Solution: Let $\left\{ v_1, v_2, \dots, v_n \right\} $ be a basis for $V$ and write $v = \sum_{i=1}^{n} \alpha _i v_i$ where $\alpha _i \in \mathbb{R} $ and at least one of the $\alpha _i\neq 0$. Without loss of generality, let us assume that $\alpha _1 \neq 0$. Let $\left\{ \phi _1, \phi _2, \dots, \phi _n \right\} $ be a basis for the dual space $V^*$. Define $\phi = \frac{\phi _1}{\alpha _1}$. Then we have
\[
\phi (v) = \left( \frac{\phi_1}{\alpha _1} \right) (v) = \frac{1}{\alpha _1} \phi _1(v) = \frac{\alpha_1}{\alpha _1} = 1.
\]