$\textbf{Exercise 6.}$ Suppose $\varphi, \beta \in V^*$. Prove that $\operatorname{null} \varphi \subseteq \operatorname{null} \beta$ if and only if there exists $c \in \mathbb{F}$ such that $\beta = c\varphi$. $\textbf{Solution 6.}$ If $\beta = c\varphi$ for some $c \in \mathbb{F}$, then clearly $\operatorname{null} \varphi \subseteq \operatorname{null} \beta$. The implication in the other direction follows from Exercise 25 in Section 3B (with $W = \mathbb{F}$) and from Exercise 7 in Section 3A. --- ![[sol-5.pdf#page=22]] --- $\textbf{Exercise 6.}$ Suppose $\varphi, \beta \in V^*$. Prove that $\operatorname{null} \varphi \subseteq \operatorname{null} \beta$ if and only if there exists $c \in \mathbb{F}$ such that $\beta = c\varphi$. $\textbf{Solution 6.}$ If $\beta = c\varphi$ for some $c \in \mathbb{F}$, then clearly $\operatorname{null} \varphi \subseteq \operatorname{null} \beta$. The implication in the other direction follows from Exercise 25 in Section 3B (with $W = \mathbb{F}$) and from Exercise 7 in Section 3A. $\textit{Commentary:}$ This exercise characterizes when the null space of one linear functional is contained in the null space of another. It turns out that this happens precisely when the functionals are scalar multiples of each other. The proof uses earlier exercises about the relationship between the null spaces of linear maps and the existence of a linear map between them. This result is important for understanding the structure of the dual space. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $\varphi, \beta: \mathbb{R}^2 \to \mathbb{R}$ by $\varphi(x, y) = x$ and $\beta(x, y) = 2x$. Then $\operatorname{null} \varphi = \operatorname{null} \beta = \{(0, y) : y \in \mathbb{R}\}$, and indeed, $\beta = 2\varphi$. On the other hand, if we define $\gamma: \mathbb{R}^2 \to \mathbb{R}$ by $\gamma(x, y) = y$, then $\operatorname{null} \varphi \not\subseteq \operatorname{null} \gamma$ and $\operatorname{null} \gamma \not\subseteq \operatorname{null} \varphi$, and there is no scalar $c$ such that $\gamma = c\varphi$.