$\textbf{Exercise 3.}$ Suppose $V$ is finite-dimensional and $v \in V$ with $v \neq 0$. Prove that there exists $\large\varphi \in V^*$ such that $\varphi(v) = 1$. $\textbf{Solution 3.}$ Because $v \neq 0$, we can extend $v$ to a basis $v, v_2, \ldots, v_n$ of $V$ (by 2.32). Now the linear map lemma (3.4) implies that there exists $\varphi \in V^*$ such that $\varphi(v) = 1$ \[and $\varphi(v_k)$ equals whatever we want for each $k = 2, \ldots, n$]. --- ![[sol-5.pdf#page=20]] --- $\textbf{Exercise 3.}$ Suppose $V$ is finite-dimensional and $v \in V$ with $v \neq 0$. Prove that there exists $\varphi \in V^*$ such that $\varphi(v) = 1$. $\textbf{Solution 3.}$ Because $v \neq 0$, we can extend $v$ to a basis $v, v_2, \ldots, v_n$ of $V$ (by 2.32). Now the linear map lemma (3.4) implies that there exists $\varphi \in V^*$ such that $\varphi(v) = 1$ \[and $\varphi(v_k)$ equals whatever we want for each $k = 2, \ldots, n$]. $\textit{Commentary:}$ This exercise shows that for any nonzero vector in a finite-dimensional vector space, there exists a linear functional that maps this vector to 1. This is a consequence of the linear map lemma, which allows us to define a linear map by specifying its values on a basis. The exercise is a step towards understanding the dual space and its relationship to the original vector space. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and $v = (1, 0)$. We can extend $v$ to the standard basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$. Define $\varphi: \mathbb{R}^2 \to \mathbb{R}$ by $\varphi(1, 0) = 1$ and $\varphi(0, 1) = 0$. Then $\varphi$ is a linear functional with $\varphi(v) = 1$.