--- COMMENTARY: $\textbf{Exercise 6.}$ Prove that there does not exist $T \in \mathcal{L}(\mathbb{R}^5)$ such that $\operatorname{range} T = \operatorname{null} T$. $\textbf{Solution 6.}$ Suppose $T \in \mathcal{L}(\mathbb{R}^5, \mathbb{R}^5)$. By the fundamental theorem of linear maps (3.21), we have $5 = \dim \operatorname{null} T + \dim \operatorname{range} T.$ If $\operatorname{range} T = \operatorname{null} T$, then the right side of the equation above would be an even number, which is a contradiction to the value on the left side. Hence $\operatorname{range} T \neq \operatorname{null} T$. $\textit{Commentary:}$ This exercise asks to prove that there is no linear operator on $\mathbb{R}^5$ whose range and null space are equal. The proof uses the fundamental theorem of linear maps, which states that for a linear map $T$ on a finite-dimensional space $V$, we have $\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T.$ If $\operatorname{range} T = \operatorname{null} T$, then the right side of this equation would be $2 \dim \operatorname{null} T$, which is always even. But the left side is $\dim \mathbb{R}^5 = 5$, which is odd. This contradiction proves that no such $T$ can exist. Geometrically, this result is saying that a linear operator on an odd-dimensional space cannot map the space onto a subspace and also map the complement of that subspace to zero. This is because such a map would effectively "compress" the space into a subspace of even dimension, which is impossible in an odd-dimensional space. This result illustrates the interplay between the algebraic properties of linear maps and the geometric properties of the spaces they act on. The parity of the dimension of the space places a constraint on the possible relationships between the range and null space of a linear map. $\textit{Examples:}$ 1. There is no $T \in \mathcal{L}(\mathbb{R}^3)$ such that $\operatorname{range} T = \operatorname{null} T$, because $\dim \mathbb{R}^3 = 3$ is odd. 2. There is no $T \in \mathcal{L}(\mathbb{C}^7)$ such that $\operatorname{range} T = \operatorname{null} T$, because $\dim \mathbb{C}^7 = 7$ is odd. 3. There is no $T \in \mathcal{L}(\mathbb{F}_2^5)$ such that $\operatorname{range} T = \operatorname{null} T$, because $\dim \mathbb{F}_2^5 = 5$ is odd. These examples show that the nonexistence of a linear operator with equal range and null space on an odd-dimensional space is a general phenomenon, not specific to $\mathbb{R}^5$. It holds in any odd-dimensional vector space over any field. The geometric reason is the same in each case: a linear map cannot compress an odd-dimensional space into an even-dimensional subspace. $\textbf{Exercise 7.}$ 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 $\mathcal{L}(V, W)$.