![[sol-6.pdf#page=3]] $\textbf{Exercise 22.}$ Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$. (a) Show that $(U + W)^0 = U^0 \cap W^0$. (b) Show that $(U \cap W)^0 = U^0 + W^0$. $\textbf{Solution 22.}$ (a) First suppose $\varphi \in (U + W)^0$. Then $\varphi(u + w) = 0$ for all $u \in U$ and all $w \in W$. Taking $w = 0$ and then taking $u = 0$, in particular we see that $\varphi(u) = 0$ for all $u \in U$ and $\varphi(w) = 0$ for all $w \in W$. Thus $\varphi \in U^0$ and $\varphi \in W^0$. In other words, $\varphi \in U^0 \cap W^0$. Thus $(U + W)^0 \subseteq U^0 \cap W^0$. To prove the inclusion in the other direction, suppose $\varphi \in U^0 \cap W^0$. If $u \in U$ and $w \in W$, then $\begin{equation*} \varphi(u + w) = \varphi(u) + \varphi(w) = 0 + 0 = 0. \end{equation*}$ Hence $\varphi \in (U + W)^0$. Thus $(U + W)^0 \supseteq U^0 \cap W^0$. Thus $(U + W)^0 = U^0 \cap W^0$, as desired. (b) First suppose $\varphi \in U^0 + W^0$. Thus $\varphi = \varphi_1 + \varphi_2$, where $\varphi_1 \in U^0$ and $\varphi_2 \in W^0$. If $v \in U \cap W$, then $\begin{equation*} \varphi(v) = \varphi_1(v) + \varphi_2(v) = 0 + 0 = 0. \end{equation*}$ Hence $\varphi \in (U \cap W)^0$. Thus $\begin{equation*} U^0 + W^0 \subseteq (U \cap W)^0. \end{equation*}$ To show that the inclusion above is an equality, we will show that both sides have the same dimension. We have $\begin{align*} \dim(U \cap W)^0 &= \dim V - \dim(U \cap W) \\ &= \dim V - (\dim U + \dim W - \dim(U + W)) \\ &= \dim U^0 + \dim W^0 - \dim(U + W)^0 \\ &= \dim U^0 + \dim W^0 - \dim(U^0 \cap W^0) \\ &= \dim(U^0 + W^0), \end{align*}$ where the first equality comes from 3.125, the second equality comes from 2.43, the third equality comes from 3.125, the fourth equality comes from Exercise 22(a), and the fifth equality comes from 2.43. The equality of dimensions along with the inclusion above show that $\begin{equation*} (U \cap W)^0 = U^0 + W^0, \end{equation*}$ as desired. --- Commentary $\textbf{Exercise 22.}$ Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$. (a) Show that $(U + W)^0 = U^0 \cap W^0$. (b) Show that $(U \cap W)^0 = U^0 + W^0$. $\textbf{Solution 22.}$ (a) First suppose $\varphi \in (U + W)^0$. Then $\varphi(u + w) = 0$ for all $u \in U$ and all $w \in W$. Taking $w = 0$ and then taking $u = 0$, in particular we see that $\varphi(u) = 0$ for all $u \in U$ and $\varphi(w) = 0$ for all $w \in W$. Thus $\varphi \in U^0$ and $\varphi \in W^0$. In other words, $\varphi \in U^0 \cap W^0$. Thus $(U + W)^0 \subseteq U^0 \cap W^0$. To prove the inclusion in the other direction, suppose $\varphi \in U^0 \cap W^0$. If $u \in U$ and $w \in W$, then $\begin{equation*} \varphi(u + w) = \varphi(u) + \varphi(w) = 0 + 0 = 0. \end{equation*}$ Hence $\varphi \in (U + W)^0$. Thus $(U + W)^0 \supseteq U^0 \cap W^0$. Thus $(U + W)^0 = U^0 \cap W^0$, as desired. (b) First suppose $\varphi \in U ^0 + W^0$. Thus $\varphi = \varphi_1 + \varphi_2$, where $\varphi_1 \in U^0$ and $\varphi_2 \in W^0$. If $v \in U \cap W$, then $\begin{equation*} \varphi(v) = \varphi_1(v) + \varphi_2(v) = 0 + 0 = 0. \end{equation*}$ Hence $\varphi \in (U \cap W)^0$. Thus $\begin{equation*} U^0 + W^0 \subseteq (U \cap W)^0. \end{equation*}$ To show that the inclusion above is an equality, we will show that both sides have the same dimension. We have $\begin{align*} \dim(U \cap W)^0 &= \dim V - \dim(U \cap W) \ &= \dim V - (\dim U + \dim W - \dim(U + W)) \ &= \dim U^0 + \dim W^0 - \dim(U + W)^0 \ &= \dim U^0 + \dim W^0 - \dim(U^0 \cap W^0) \ &= \dim(U^0 + W^0), \end{align*}$ where the first equality comes from 3.125, the second equality comes from 2.43, the third equality comes from 3.125, the fourth equality comes from Exercise 22(a), and the fifth equality comes from 2.43. The equality of dimensions along with the inclusion above show that $\begin{equation*} (U \cap W)^0 = U^0 + W^0, \end{equation*}$ as desired. $\textit{Commentary:}$ This exercise establishes two important identities involving annihilators and sums and intersections of subspaces. Part (a) shows that the annihilator of the sum of two subspaces is the intersection of their annihilators. This is because a functional vanishes on the sum $U + W$ if and only if it vanishes on both $U$ and $W$ separately. Part (b) is the dual statement: the annihilator of the intersection of two subspaces is the sum of their annihilators. The proof of part (b) uses part (a) and a dimension argument. These identities are part of a larger theory of dual operations on subspaces and quotient spaces, which has numerous applications in linear algebra and functional analysis. $\textit{Example:}$ Let $V = \mathbb{R}^3$, $U = \operatorname{span}{(1, 0, 0)}$ (the $x$-axis), and $W = \operatorname{span}{(0, 1, 0)}$ (the $y$-axis). Then $U + W = \operatorname{span}{(1, 0, 0), (0, 1, 0)}$ (the $xy$-plane) and $U \cap W = {(0, 0, 0)}$ (the origin). The annihilator $(U + W)^0$ consists of functionals of the form $\varphi(x, y, z) = cz$, while $U^0$ consists of functionals of the form $\varphi(x, y, z) = by + cz$ and $W^0$ consists of functionals of the form $\varphi(x, y, z) = ax + cz$. Clearly, $(U + W)^0 = U^0 \cap W^0$, as predicted by the theorem. On the other hand, $(U \cap W)^0 = V^* = (\mathbb{R}^3)^_$ (since the only functional that vanishes at the origin is the zero functional), while $U^0 + W^0$ consists of all functionals of the form $\varphi(x, y, z) = ax + by + cz$, which is indeed all of $(\mathbb{R}^3)^_$. ---