$\textbf{Exercise 16.}$ Suppose $W$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $\begin{equation*} T^* = 0 \iff T = 0. \end{equation*}$ $\textbf{Solution 16.}$ First suppose $T = 0$. If $\varphi \in W^*$, then $T^*(\varphi) = \varphi \circ T = 0$, and thus $T^* = 0$. To prove the other direction, now suppose $T^* = 0$. Thus $\begin{equation*} 0 = T^*(\varphi) = \varphi \circ T \end{equation*}$ for every $\varphi \in W^*$. Thus $\begin{equation*} 0 = \varphi(Tv) \end{equation*}$ for every $v \in V$ and every $\varphi \in W^*$. Exercise 3 now implies that $Tv = 0$ for every $v \in V$. Thus $T = 0$, as desired. ![[sol-5.pdf#page=29]] --- Extended answer: $\textbf{Exercise 16.}$ Suppose $W$ is finite-dimensional and $T \in \mathcal{L}(V, W)$. Prove that $\begin{equation*} T^* = 0 \iff T = 0. \end{equation*}$ $\textbf{Solution 16.}$ First suppose $T = 0$. If $\varphi \in W^*$, then $T^*(\varphi) = \varphi \circ T = 0$, and thus $T^* = 0$. To prove the other direction, now suppose $T^* = 0$. Thus $\begin{equation*} 0 = T^*(\varphi) = \varphi \circ T \end{equation*}$ for every $\varphi \in W^*$. Thus $\begin{equation*} 0 = \varphi(Tv) \end{equation*}$ for every $v \in V$ and every $\varphi \in W^*$. Exercise 3 now implies that $Tv = 0$ for every $v \in V$. Thus $T = 0$, as desired. $\textit{Commentary:}$ This exercise shows that a linear map is zero if and only if its dual map is zero. The forward direction is straightforward: if $T$ is zero, then composing any functional with $T$ yields the zero functional. The reverse direction relies on a previous exercise (Exercise 3), which stated that if a vector $v$ in a finite-dimensional space satisfies $\varphi(v) = 0$ for all functionals $\varphi$, then $v$ must be zero. Applying this to each vector $Tv$ shows that $T$ must be the zero map. This result demonstrates a strong connection between a map and its dual, and it is a key step in proving that the double dual of a finite-dimensional vector space is isomorphic to the original space. $\textit{Example:}$ Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be the zero map, so $T(x, y) = (0, 0, 0)$ for all $(x, y) \in \mathbb{R}^2$. For any $\varphi = (a, b, c) \in (\mathbb{R}^3)^*$, we have $\begin{align*} T^*(\varphi)(x, y) &= \varphi(T(x, y)) \\ &= \varphi(0, 0, 0) \\ &= 0 \end{align*}$ for all $(x, y) \in \mathbb{R}^2$. Thus $T^*(\varphi) = 0$ for all $\varphi \in (\mathbb{R}^3)^*$, so $T^* = 0$. Conversely, suppose $S: \mathbb{R}^2 \to \mathbb{R}^3$ is such that $S^* = 0$. Then for every $\varphi = (a, b, c) \in (\mathbb{R}^3)^*$ and every $(x, y) \in \mathbb{R}^2$, $\begin{equation*} 0 = S^*(\varphi)(x, y) = \varphi(S(x, y)) = aS_1(x, y) + bS_2(x, y) + cS_3(x, y), \end{equation*}$ where $S(x, y) = (S_1(x, y), S_2(x, y), S_3(x, y))$. Since this holds for all $\varphi$, we must have $S_1(x, y) = S_2(x, y) = S_3(x, y) = 0$ for all $(x, y)$. Thus $S = 0$.