See commentary, below --- $\textbf{Exercise 32.}$ The double dual space of $V$, denoted by $V^{**}$, is defined to be the dual space of $V^*$. In other words, $V^{**} = (V^*)^*$. Define $\Lambda: V \to V^{**}$ by $\begin{equation*} (\Lambda v)(\varphi) = \varphi(v) \end{equation*}$ for each $v \in V$ and each $\varphi \in V^*$. (a) Show that $\Lambda$ is a linear map from $V$ to $V^{**}$. (b) Show that if $T \in \mathcal{L}(V)$, then $T^{**} \circ \Lambda = \Lambda \circ T$, where $T^{**} = (T^*)^*$. (c) Show that if $V$ is finite-dimensional, then $\Lambda$ is an isomorphism from $V$ onto $V^{**}$. Suppose $V$ is finite-dimensional. Then $V$ and $V^*$ are isomorphic, but finding an isomorphism from $V$ onto $V^*$ generally requires choosing a basis of $V$. In contrast, the isomorphism $\Lambda$ from $V$ onto $V^{**}$ does not require a choice of basis and thus is considered more natural. $\textbf{Solution 32.}$ (a) A straightforward application of the definitions shows that $\Lambda$ is a linear map from $V$ to $V^{**}$. (b) Suppose $T \in \mathcal{L}(V)$. Suppose $v \in V$ and $\varphi \in V^*$. Then $\begin{align*} ((T^{**} \circ \Lambda)(v))(\varphi) &= (T^{**}(\Lambda v))(\varphi) \\ &= (\Lambda v \circ T^*)(\varphi) \\ &= (\Lambda v)(T^*(\varphi)) \\ &= (\Lambda v)(\varphi \circ T) \\ &= (\varphi \circ T)(v) \\ &= (\varphi \circ T)(v) \\ &= \varphi(Tv). \end{align*}$ Also, $\begin{align*} ((\Lambda \circ T)(v))(\varphi) &= (\Lambda(Tv))(\varphi) \\ &= \varphi(Tv). \end{align*}$ Thus $\begin{equation*} ((T^{**} \circ \Lambda)(v))(\varphi) = ((\Lambda \circ T)(v))(\varphi), \end{equation*}$ for all $\varphi \in V^*$, which implies that $\begin{equation*} (T^{**} \circ \Lambda)(v) = (\Lambda \circ T)(v), \end{equation*}$ for all $v \in V$, which implies that $T^{**} \circ \Lambda = \Lambda \circ T$, as desired. (c) Suppose $V$ is finite-dimensional. Suppose $v \in V$ and $\Lambda v = 0$. Thus $\varphi(v) = 0$ for every $\varphi \in V^*$. Now Exercise 3 implies that $v = 0$. Thus $\Lambda$ is injective. Because $\dim V = \dim V^* = \dim V^{**}$ (by 3.111), we can conclude that $\Lambda$ is an isomorphism of $V$ onto $V^{**}$. ---  --- Commentary Problem 32 (Section 3F): $\textbf{Exercise 32.}$ The double dual space of $V$, denoted by $V^{**}$, is defined to be the dual space of $V^*$. In other words, $V^{**} = (V^*)^*$. Define $\Lambda: V \to V^{**}$ by $\begin{equation*} (\Lambda v)(\varphi) = \varphi(v) \end{equation*}$ for each $v \in V$ and each $\varphi \in V^*$. (a) Show that $\Lambda$ is a linear map from $V$ to $V^{**}$. (b) Show that if $T \in \mathcal{L}(V)$, then $T^{**} \circ \Lambda = \Lambda \circ T$, where $T^{**} = (T^*)^*$. (c) Show that if $V$ is finite-dimensional, then $\Lambda$ is an isomorphism from $V$ onto $V^{**}$. Suppose $V$ is finite-dimensional. Then $V$ and $V^*$ are isomorphic, but finding an isomorphism from $V$ onto $V^*$ generally requires choosing a basis of $V$. In contrast, the isomorphism $\Lambda$ from $V$ onto $V^{**}$ does not require a choice of basis and thus is considered more natural. $\textbf{Solution 32.}$ (a) A straightforward application of the definitions shows that $\Lambda$ is a linear map from $V$ to $V^{**}$. (b) Suppose $T \in \mathcal{L}(V)$. Suppose $v \in V$ and $\varphi \in V^*$. Then $\begin{align*} ((T^{**} \circ \Lambda)(v))(\varphi) &= (T^{**}(\Lambda v))(\varphi) \\ &= (\Lambda v \circ T^*)(\varphi) \\ &= (\Lambda v)(T^*(\varphi)) \\ &= (\Lambda v)(\varphi \circ T) \\ &= (\varphi \circ T)(v) \\ &= \varphi(Tv). \end{align*}$ Also, $\begin{align*} ((\Lambda \circ T)(v))(\varphi) &= (\Lambda(Tv))(\varphi) \\ &= \varphi(Tv). \end{align*}$ Thus $\begin{equation*} ((T^{**} \circ \Lambda)(v))(\varphi) = ((\Lambda \circ T)(v))(\varphi), \end{equation*}$ for all $\varphi \in V^*$, which implies that $\begin{equation*} (T^{**} \circ \Lambda)(v) = (\Lambda \circ T)(v), \end{equation*}$ for all $v \in V$, which implies that $T^{**} \circ \Lambda = \Lambda \circ T$, as desired. (c) Suppose $V$ is finite-dimensional. Suppose $v \in V$ and $\Lambda v = 0$. Thus $\varphi(v) = 0$ for every $\varphi \in V^*$. Now Exercise 3 implies that $v = 0$. Thus $\Lambda$ is injective. Because $\dim V = \dim V^* = \dim V^{**}$ (by 3.111), we can conclude that $\Lambda$ is an isomorphism of $V$ onto $V^{**}$. $\textit{Commentary:}$ This exercise introduces the double dual space $V^{**}$ and the natural map $\Lambda: V \to V^{**}$. Part (a) verifies that $\Lambda$ is indeed a linear map. Part (b) shows that $\Lambda$ "intertwines" the action of an operator $T$ on $V$ with the action of the double dual operator $T^{**}$ on $V^{**}$. This property is crucial for many applications of the double dual, such as in the study of reflexive spaces and the weak* topology. Part (c) proves that if $V$ is finite-dimensional, then $\Lambda$ is an isomorphism. This is a special case of a more general result: $\Lambda$ is always injective, and it is surjective if and only if $V$ is reflexive (which is true for all finite-dimensional spaces). The commentary notes that while $V$ and $V^*$ are isomorphic for finite-dimensional $V$, the isomorphism $\Lambda$ is more "natural" because it doesn't require choosing a basis. $\textit{Example:}$ Let $V = \mathbb{R}^2$. For $v = (a, b) \in \mathbb{R}^2$ and $\varphi = (c, d) \in (\mathbb{R}^2)^*$, we have $\begin{equation*} (\Lambda v)(\varphi) = \varphi(v) = ac + bd. \end{equation*}$ Thus, $\Lambda(a, b)$ is the functional on $(\mathbb{R}^2)^*$ given by $\begin{equation*} (\Lambda(a, b))(c, d) = ac + bd. \end{equation*}$ This is clearly a linear functional on $(\mathbb{R}^2)^*$, so $\Lambda(a, b) \in (\mathbb{R}^2)^{**}$. In fact, the map $\Lambda: \mathbb{R}^2 \to (\mathbb{R}^2)^{**}$ is an isomorphism. It's injective because if $(\Lambda(a, b))(c, d) = 0$ for all $(c, d)$, then taking $(c, d) = (1, 0)$ and $(0, 1)$ shows that $a = b = 0$. It's surjective because $\dim \mathbb{R}^2 = \dim (\mathbb{R}^2)^{**} = 2$.