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--- math121b:ex2 [2020/02/10 20:57]
pzhou created
+++ math121b:ex2 [2026/02/21 14:41] (current)
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 {{ :math121b:ex_of_s14.jpg?600&nolink |}}
-See the
+See the {{ :math121b:dual_space.pdf |chapter on dual vector space}} for notations. My canonical pairing is using pointed bracket $\langle -, - \rangle$, Halmos uses $[-,-]$.
 ====== Solution ======
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 where we define $\alpha_j = F(t^j)$ for any $j \geq 0$.
+. Yes. By definition, $y: V \to \R$ is a function whose image is not just $\{0\}$. Suppose $v \in V$, and $y(v) \neq 0$. We consider the element $w = v / y(v) \in V$. Then
+$y(w) = y(v/y(v)) = y(v)/y(v) = 1$. Hence, for any $\alpha \in \R$, we have $\alpha w \in V$, such that
+$$\langle \alpha w, y \rangle = y(\alpha w) = \alpha y(w) = \alpha. $$

Lecture Notes