--- $\textbf{Exercise 21.}$ Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$. (a) Prove that $W^0 \subseteq U^0$ if and only if $U \subseteq W$. (b) Prove that $W^0 = U^0$ if and only if $U = W$. $\textbf{Solution 21.}$ (a) First suppose $U \subseteq W$. Suppose $\varphi \in W^0$. If $u \in U$, then $u \in W$, and hence $\varphi(u) = 0$. Thus $\varphi \in U^0$. Hence $W^0 \subseteq U^0$. To prove the implication in the other direction, now suppose $W^0 \subseteq U^0$. Suppose $u \in U$. Thus $\varphi(u) = 0$ for every $\varphi \in U^0$. Hence $\varphi(u) = 0$ for every $\varphi \in W^0$. Now Exercise 20 implies that $u \in W$. Thus $U \subseteq W$, as desired. (b) If $U = W$, then clearly $W^0 = U^0$. To prove the implication in the other direction, now suppose $W^0 = U^0$. Then $W^0 \subseteq U^0$ and $U^0 \subseteq W^0$. Now (a) implies that $U \subseteq W$ and $W \subseteq U$. Thus $U = W$, as desired. $\textit{Commentary:}$ This exercise explores the relationship between subspaces and their annihilators. Part (a) shows that the annihilator reverses inclusions: if $U$ is a subspace of $W$, then the annihilator of $W$ is a subspace of the annihilator of $U$. This is because any functional that vanishes on the larger space $W$ must also vanish on the smaller space $U$. The converse is also true and is proved using the result of Exercise 20. Part (b) is a simple consequence of part (a): two subspaces are equal if and only if their annihilators are equal. These results are fundamental in the study of dual spaces and have many applications in functional analysis and operator theory. $\textit{Example:}$ Let $V = \mathbb{R}^3$, $U = \operatorname{span}\{(1, 0, 0)\}$ (the $x$-axis), and $W = \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$ (the $xy$-plane). Then $U \subseteq W$. The annihilator $W^0$ consists of all functionals of the form $\varphi(x, y, z) = cz$ for some $c \in \mathbb{R}$, while $U^0$ consists of all functionals of the form $\varphi(x, y, z) = by + cz$ for some $b, c \in \mathbb{R}$. Clearly, $W^0 \subseteq U^0$, as predicted by the theorem. On the other hand, if $U = W$, then their annihilators are also equal, consisting of functionals of the form $\varphi(x, y, z) = cz$.