--- COMMENTARY: $\textbf{Exercise 7.}$ Suppose $V$ and $W$ are finite-dimensional with $2 \leq \dim V \leq \dim W$. Show that ${T \in \mathcal{L}(V, W) : T \text{ is not injective}}$ is not a subspace of $\mathcal{L}(V, W)$. $\textbf{Solution 7.}$ Let $v_1, \ldots, v_m$ be a basis of $V$ and let $w_1, \ldots, w_n$ be a basis of $W$. Thus $2 \leq m \leq n$. Using the linear map lemma (3.4), let $S, T \in \mathcal{L}(V, W)$ be such that $Sv_1 = 0 \text{ and } Sv_k = w_k \text{ for } k = 2, \ldots, m$ and $Tv_1 = w_1 \text{ and } Tv_k = 0 \text{ for } k = 2, \ldots, m.$ Then neither $S$ nor $T$ is injective because $Sv_1 = Tv_2 = 0$. However, $(S+T)v_k = w_k$ for each $k = 1, \ldots, m$, and thus $S + T$ is injective (as is easy to see). Hence the set of linear maps from $V$ to $W$ that are not injective is not closed under addition and thus is not a subspace of $\mathcal{L}(V, W)$. $\textit{Commentary:}$ This exercise asks to prove that the set of non-injective linear maps from $V$ to $W$ is not a subspace of $\mathcal{L}(V, W)$, under the condition that $2 \leq \dim V \leq \dim W$. The proof constructs two specific non-injective maps $S$ and $T$, and shows that their sum $S + T$ is injective. This proves that the set is not closed under addition, and therefore not a subspace. The construction of $S$ and $T$ relies on the condition $2 \leq \dim V \leq \dim W$. This ensures that there are enough basis vectors in $W$ to define $S$ and $T$ as needed. $S$ maps $v_1$ to zero and the other basis vectors of $V$ to distinct basis vectors of $W$, while $T$ does the opposite, mapping $v_1$ to a basis vector of $W$ and the other basis vectors of $V$ to zero. Both of these maps are non-injective, but their sum maps each basis vector of $V$ to a distinct basis vector of $W$, making it injective. Geometrically, $S$ and $T$ can be thought of as projections followed by embeddings. $S$ projects $V$ onto the subspace spanned by $v_2, \ldots, v_m$ (mapping $v_1$ to zero), and then embeds this subspace into $W$. $T$ does the opposite, projecting $V$ onto the subspace spanned by $v_1$ and then embedding this into $W$. The sum $S + T$ effectively cancels out these projections, resulting in an injective map. This result shows that the set of non-injective maps, while it contains the zero map and is closed under scalar multiplication, is not closed under addition, and therefore does not form a subspace. This is a common situation in linear algebra: many important sets of linear maps (such as the set of all maps, the set of surjective maps, etc.) are not subspaces. $\textit{Examples:}$ 1. Let $V = \mathbb{R}^2$ and $W = \mathbb{R}^3$. Define $S, T \in \mathcal{L}(V, W)$ by $S(x, y) = (0, x, 0)$ and $T(x, y) = (x, 0, 0)$. Then $S$ and $T$ are non-injective, but $S + T$ is injective. 2. Let $V = \mathbb{C}^3$ and $W = \mathbb{C}^4$. Define $S, T \in \mathcal{L}(V, W)$ by $S(z_1, z_2, z_3) = (0, z_2, z_3, 0)$ and $T(z_1, z_2, z_3) = (z_1, 0, 0, 0)$. Then $S$ and $T$ are non-injective, but $S + T$ is injective. 3. Let $V = \mathbb{F}_2^2$ and $W = \mathbb{F}_2^3$. Define $S, T \in \mathcal{L}(V, W)$ by $S(x, y) = (0, x, 0)$ and $T(x, y) = (x, 0, y)$. Then $S$ and $T$ are non-injective, but $S + T$ is injective. These examples demonstrate the principle in various vector spaces over different fields. In each case, we have two non-injective maps whose sum is injective, showing that the set of non-injective maps is not a subspace. The geometric interpretation is the same in each case: the non-injective maps are projections followed by embeddings, and their sum cancels the projections, resulting in an injective map.