--- COMMENTARY: $\textbf{Exercise 4.}$ Show that ${T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) : \dim \operatorname{null} T > 2}$ is not a subspace of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$. $\textbf{Solution 4.}$ Define $S, T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$ by $S(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, 0, 0)$ and $T(x_1, x_2, x_3, x_4, x_5) = (0, 0, x_3, x_4).$ Then $\operatorname{null} S = {(0, 0, x_3, x_4, x_5) : x_3, x_4, x_5 \in \mathbb{R}}$ and $\operatorname{null} T = {(x_1, x_2, 0, 0, x_5) : x_1, x_2, x_5 \in \mathbb{R}}.$ Thus $\dim \operatorname{null} S = \dim \operatorname{null} T = 3$. However, $(S + T)(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, x_3, x_4).$ Thus $\operatorname{null}(S + T) = {(0, 0, 0, 0, x_5) : x_5 \in \mathbb{R}}$, which has dimension one. Thus ${T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^4) : \dim \operatorname{null} T > 2}$ is not closed under addition and hence is not a subspace of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$. $\textit{Commentary:}$ This exercise asks to prove that a certain subset of $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$ is not a subspace. The subset consists of all linear maps whose null space has dimension greater than 2. The proof constructs two specific linear maps $S$ and $T$ in this set, and shows that their sum $S + T$ is not in the set. This proves that the set is not closed under addition, and therefore cannot be a subspace. Geometrically, $S$ is a projection onto the first two coordinates, and $T$ is a projection onto the third and fourth coordinates. Both of these projections have 3-dimensional null spaces (the subspaces where the projection is zero). However, their sum $S + T$ is a projection onto the first four coordinates, which has only a 1-dimensional null space (the fifth coordinate axis). This example illustrates that the dimension of the null space, while linear in the sense that $\dim(\operatorname{null} S + \operatorname{null} T) \leq \dim \operatorname{null} S + \dim \operatorname{null} T$, is not linear in the stronger sense required for the set of maps with high-dimensional null spaces to be a subspace. $\textit{Examples:}$ 1. In $\mathcal{L}(\mathbb{R}^4, \mathbb{R}^3)$, the set ${T : \dim \operatorname{null} T > 1}$ is not a subspace. For example, if $S(x, y, z, w) = (x, 0, 0)$ and $T(x, y, z, w) = (0, y, 0)$, then $\dim \operatorname{null} S = \dim \operatorname{null} T = 3$, but $\dim \operatorname{null}(S + T) = 2$. 2. In $\mathcal{L}(\mathcal{P}_3(\mathbb{F}_2), \mathbb{F}_2^2)$, the set ${T : \dim \operatorname{null} T > 2}$ is not a subspace. For example, if $S(a + bx + cx^2 + dx^3) = (a, 0)$ and $T(a + bx + cx^2 + dx^3) = (0, b)$, then $\dim \operatorname{null} S = \dim \operatorname{null} T = 3$, but $\dim \operatorname{null}(S + T) = 2$. 3. In $\mathcal{L}(\mathbb{C}^6, \mathbb{C}^5)$, the set ${T : \dim \operatorname{null} T > 3}$ is not a subspace. For example, if $S(z_1, \ldots, z_6) = (z_1, z_2, 0, 0, 0)$ and $T(z_1, \ldots, z_6) = (0, 0, z_3, z_4, 0)$, then $\dim \operatorname{null} S = \dim \operatorname{null} T = 4$, but $\dim \operatorname{null}(S + T) = 2$. These examples show that the non-subspace property of the set ${T : \dim \operatorname{null} T > k}$ is not specific to $\mathcal{L}(\mathbb{R}^5, \mathbb{R}^4)$, but is a general phenomenon in spaces of linear maps. The geometric reason is the same in each case: the sum of two maps with high-dimensional null spaces can have a lower-dimensional null space, due to the null spaces "overlapping" in a certain sense.