$\textbf{Exercise 4.}$ Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ such that $U \neq V$. Prove that there exists $\varphi \in V^*$ such that $\varphi(u) = 0$ for every $u \in U$ but $\varphi \neq 0$. $\textbf{Solution 4.}$ Let $u_1, \ldots, u_m$ be a basis of $U$. Extend this list to a basis $u_1, \ldots, u_m, v_1, \ldots, v_n$ of $V$ (using 2.32). Now the linear map lemma (3.4) implies that there exists $\varphi \in V^*$ such that $\begin{align*} \varphi(u_k) &= 0 \text{ for } k = 1, \ldots, m \\\\ and\\\\ \varphi(v_k) &= 1 \text{ for } k = 1, \ldots, n. \end{align*}$ Thus $\varphi(u) = 0$ for every $u \in U$ but $\varphi \neq 0$, as desired. --- ![[sol-5.pdf#page=21]] --- $\textbf{Exercise 4.}$ Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ such that $U \neq V$. Prove that there exists $\varphi \in V^*$ such that $\varphi(u) = 0$ for every $u \in U$ but $\varphi \neq 0$. $\textbf{Solution 4.}$ Let $u_1, \ldots, u_m$ be a basis of $U$. Extend this list to a basis $u_1, \ldots, u_m, v_1, \ldots, v_n$ of $V$ (using 2.32). Now the linear map lemma (3.4) implies that there exists $\varphi \in V^*$ such that $\begin{align*} \varphi(u_k) &= 0 \text{ for } k = 1, \ldots, m \\ \varphi(v_k) &= 1 \text{ for } k = 1, \ldots, n. \end{align*}$ Thus $\varphi(u) = 0$ for every $u \in U$ but $\varphi \neq 0$, as desired. $\textit{Commentary:}$ This exercise demonstrates the existence of a nonzero linear functional that vanishes on a proper subspace. The construction of such a functional relies on extending a basis of the subspace to a basis of the whole space and then defining the functional to be zero on the subspace basis vectors and nonzero on the additional basis vectors. This exercise is important for understanding the concept of the annihilator of a subspace. $\textit{Example:}$ Let $V = \mathbb{R}^3$ and $U = \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$. Extend the basis $\{(1, 0, 0), (0, 1, 0)\}$ of $U$ to the basis $\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$ of $\mathbb{R}^3$. Define $\varphi: \mathbb{R}^3 \to \mathbb{R}$ by $\varphi(1, 0, 0) = \varphi(0, 1, 0) = 0$ and $\varphi(0, 0, 1) = 1$. Then $\varphi$ is a nonzero linear functional that vanishes on $U$.