![[sol-5.pdf#page=30]] --- $\textbf{Exercise 17.}$ Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T$ is invertible if and only if $T^* \in \mathcal{L}(W^*, V^*)$ is invertible. $\textbf{Solution 17.}$ A linear map is invertible if and only if it is injective and surjective. Thus the desired result follows from 3.129 and 3.131. --- $\textbf{Exercise 17.}$ Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T$ is invertible if and only if $T^* \in \mathcal{L}(W^_, V^_)$ is invertible. $\textbf{Solution 17.}$ A linear map is invertible if and only if it is injective and surjective. Thus the desired result follows from 3.129 and 3.131. $\textit{Commentary:}$ This exercise establishes a fundamental connection between a linear map and its dual: a map is invertible if and only if its dual map is invertible. The proof relies on two previous results (3.129 and 3.131), which state that a map is injective (respectively, surjective) if and only if its dual map is surjective (respectively, injective). Combining these results yields the desired equivalence. This property is a manifestation of the duality principle in linear algebra, which asserts that many concepts and results have dual counterparts obtained by reversing arrows and interchanging "injective" and "surjective". This exercise is a key step in establishing the double dual isomorphism for finite-dimensional spaces. $\textit{Example:}$ Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the rotation by $90^\circ$ counterclockwise, so $\begin{equation*} T(x, y) = (-y, x). \end{equation*}$ The matrix of $T$ with respect to the standard basis is $\begin{equation*} [T] = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}. \end{equation*}$ Since $\det([T]) = 1 \neq 0$, $T$ is invertible. The dual map $T^_: (\mathbb{R}^2)^_ \to (\mathbb{R}^2)^_$ is given by $\begin{equation_} T^_(\varphi)(x, y) = \varphi(T(x, y)) = \varphi(-y, x) = -b\varphi_1(x, y) + a\varphi_2(x, y) \end{equation_}$ for $\varphi = (a, b) \in (\mathbb{R}^2)^_$, where $\varphi_1(x, y) = x$ and $\varphi_2(x, y) = y$ form the dual basis. Thus, the matrix of $T^_$ with respect to the dual basis is $\begin{equation*} [T^_] = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}. \end{equation_}$ Since $\det([T^_]) = 1 \neq 0$, $T^_$ is also invertible, as predicted by the theorem.