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$\textbf{Exercise 20.}$ Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Show that
$\begin{equation*}
U = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}.
\end{equation*}$
$\textbf{Solution 20.}$ If $u \in U$, then $\varphi(u) = 0$ for every $\varphi \in U^0$. Thus
$\begin{equation*}
U \subseteq \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}.
\end{equation*}$
To prove the inclusion in the other direction, suppose $w \in V$ but $w \notin U$. Let $u_1, \ldots, u_m$ be a basis of $U$. Because $w \notin U$, the list $u_1, \ldots, u_m, w$ is linearly independent and hence can be extended to a basis of $V$. Thus by the linear map lemma (3.4), there exists $\psi \in V^*$ such that
$\begin{align*}
\psi(u_k) &= 0 \text{ for } k = 1, \ldots, m \text{ and} \\
\psi(w) &= 1.
\end{align*}$
Thus $\psi \in U^0$ but $\psi(w) \neq 0$. Hence
$\begin{equation*}
w \notin \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}.
\end{equation*}$
Thus
$\begin{equation*}
U \supseteq \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\},
\end{equation*}$
which implies that
$\begin{equation*}
U = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\},
\end{equation*}$
as desired.
$\textit{Commentary:}$ This exercise gives a characterization of a subspace $U$ in terms of the functionals that vanish on it.
The set $U^0$, called the annihilator of $U$, consists of all functionals that map every vector in $U$ to zero.
The exercise shows that a vector $v$ is in $U$ if and only if $\varphi(v) = 0$ for every $\varphi \in U^0$.
The forward direction is straightforward: if $v \in U$, then every functional that vanishes on $U$ must vanish on $v$.
The reverse direction is proved by contradiction: if $v \notin U$, then one can construct a functional that vanishes on $U$ but not on $v$, using the linear map lemma.
This result is a key tool in the study of dual spaces and has many applications, such as in the theory of Banach spaces and operator algebras.
$\textit{Example:}$ Let $V = \mathbb{R}^3$ and $U = \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$, the $xy$-plane in $\mathbb{R}^3$. The annihilator $U^0$ consists of all functionals of the form $\varphi(x, y, z) = cz$ for some $c \in \mathbb{R}$, since these are precisely the functionals that vanish on the $xy$-plane.
The set $\{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}$ consists of all vectors of the form $(x, y, 0)$, which is exactly $U$.
This illustrates the theorem.