$\textbf{Exercise 18.}$ Suppose $V$ and $W$ are finite-dimensional. Prove that the map that takes $T \in \mathcal{L}(V, W)$ to $T^* \in \mathcal{L}(W^*, V^*)$ is an isomorphism of $\mathcal{L}(V, W)$ onto $\mathcal{L}(W^*, V^*)$. $\textbf{Solution 18.}$ Define $\Gamma: \mathcal{L}(V, W) \to \mathcal{L}(W^*, V^*)$ by $\Gamma(T) = T^*$. By 3.120, $\Gamma$ is a linear map. Exercise 16 implies that $\Gamma$ is injective. Note that $ \begin{align*} \dim \mathcal{L}(V, W) &= (\dim V)(\dim W) \\ &= (\dim V^_)(\dim W^_) \ &= (\dim W^_)(\dim V^_) \ &= \dim \mathcal{L}(W^_, V^_), \end{align*}$ where the first and last equalities above come from 3.72 and the second equality above comes from 3.111. The fundamental theorem of linear maps (3.21) and the equation above now imply that $\dim \operatorname{range} \Gamma = \dim(\mathcal{L}(W^_, V^_))$. Thus $\Gamma$ is surjective and hence is an isomorphism of $\mathcal{L}(V, W)$ onto $\mathcal{L}(W^_, V^_)$, as desired. --- ![[sol-5.pdf#page=31]] --- Problem 18 (Section 3F): $\textbf{Exercise 18.}$ Suppose $V$ and $W$ are finite-dimensional. Prove that the map that takes $T \in \mathcal{L}(V, W)$ to $T^* \in \mathcal{L}(W^*, V^*)$ is an isomorphism of $\mathcal{L}(V, W)$ onto $\mathcal{L}(W^*, V^*)$. $\textbf{Solution 18.}$ Define $\Gamma: \mathcal{L}(V, W) \to \mathcal{L}(W^*, V^*)$ by $\Gamma(T) = T^*$. By 3.120, $\Gamma$ is a linear map. Exercise 16 implies that $\Gamma$ is injective. Note that $\begin{align*} \dim \mathcal{L}(V, W) &= (\dim V)(\dim W) \\ &= (\dim V^*)(\dim W^*) \\ &= (\dim W^*)(\dim V^*) \\ &= \dim \mathcal{L}(W^*, V^*), \end{align*}$ where the first and last equalities come from 3.72 and the second equality comes from 3.111. The fundamental theorem of linear maps (3.21) and the equation above now imply that $\dim \operatorname{range} \Gamma = \dim(\mathcal{L}(W^*, V^*))$. Thus $\Gamma$ is surjective and hence is an isomorphism of $\mathcal{L}(V, W)$ onto $\mathcal{L}(W^*, V^*)$, as desired. $\textit{Commentary:}$ This exercise shows that the dual operation induces an isomorphism between the space of linear maps from $V$ to $W$ and the space of linear maps from $W^*$ to $V^*$. In other words, the dual operation gives a "natural" way to convert a map $T: V \to W$ into a map $T^*: W^* \to V^*$, and this conversion process is reversible. The proof first uses a previous exercise to show that the dual map $\Gamma$ is injective. It then computes the dimensions of the domain and codomain of $\Gamma$, using the fact that the dual space has the same dimension as the original space. The fundamental theorem of linear maps then implies that $\Gamma$ is surjective, and hence an isomorphism. This result is a cornerstone of the duality principle in linear algebra and functional analysis. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and $W = \mathbb{R}^3$. The space $\mathcal{L}(V, W)$ can be identified with the space $\mathbb{R}^{2 \times 3}$ of $2 \times 3$ real matrices, which has dimension 6. The space $\mathcal{L}(W^*, V^*)$ can be identified with the space $\mathbb{R}^{3 \times 2}$ of $3 \times 2$ real matrices, which also has dimension 6. The dual map $\Gamma: \mathbb{R}^{2 \times 3} \to \mathbb{R}^{3 \times 2}$ is given by matrix transposition: $\begin{equation*} \Gamma\left(\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}\right) = \begin{pmatrix} a & d \\ b & e \\ c & f \end{pmatrix}. \end{equation*}$ This is clearly a linear map, and it is injective and surjective (since a matrix can be recovered from its transpose), so it is an isomorphism.