--- Commentary: $\textbf{Exercise 11.}$ Suppose that $V$ is finite-dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T = {0} \text{ and } \operatorname{range} T = {Tu : u \in U}.$ $\textbf{Solution 11.}$ There exists a subspace $U$ of $V$ such that $V = \operatorname{null} T \oplus U;$ this follows from 2.33 (with $\operatorname{null} T$ playing the role of $U$ and $U$ playing the role of $W$). From the definition of direct sum, we have $U \cap \operatorname{null} T = {0}$. Clearly $\operatorname{range} T \supseteq {Tu : u \in U}$. To prove the inclusion in the other direction, suppose $v \in V$. Then there exist $w \in \operatorname{null} T$ and $x \in U$ such that $v = w + x.$ Applying $T$ to both sides of this equation, we have $Tv = Tw + Tx = Tx$. Thus $Tv \in {Tu : u \in U}$. Because $v$ was an arbitrary vector in $V$ (and thus $Tv$ is an arbitrary vector in $\operatorname{range} T$), this implies that $\operatorname{range} T \subseteq {Tu : u \in U}.$ Thus $\operatorname{range} T = {Tu : u \in U}$, as desired. $\textit{Commentary:}$ This exercise asks us to prove that for any linear map $T$ from a finite-dimensional space $V$ to another space $W$, there is a subspace $U$ of $V$ that "complements" the null space of $T$ and whose image under $T$ is the range of $T$. The proof relies on the fact that in a finite-dimensional space, any subspace has a complement (a subspace that "completes" it to the whole space). Specifically, we use the result that $V$ is the direct sum of $\operatorname{null} T$ and some subspace $U$. This means that every vector in $V$ can be uniquely written as a sum of a vector in $\operatorname{null} T$ and a vector in $U$. We then show that the range of $T$ is exactly the set of images of vectors in $U$ under $T$. The inclusion $\operatorname{range} T \supseteq {Tu : u \in U}$ is clear, because every $Tu$ with $u \in U$ is in the range of $T$. For the reverse inclusion, we take an arbitrary vector $v \in V$, write it as $w + x$ with $w \in \operatorname{null} T$ and $x \in U$, and then observe that $Tv = Tx$, so $Tv$ is in ${Tu : u \in U}$. Geometrically, this result says that we can understand a linear map $T$ by looking at its behavior on a subspace $U$ that "avoids" the null space. The null space is the part of the domain that $T$ "collapses" to zero, while $U$ is a part that $T$ maps "faithfully" to the range. This result is useful in various contexts. For example, it's used in the proof of the Rank-Nullity Theorem, which relates the dimensions of the domain, null space, and range of a linear map. It's also related to the concept of a "right inverse" of a linear map. $\textit{Examples:}$ 1. Let $T : \mathbb{R}^4 \to \mathbb{R}^3$ be defined by $T(x, y, z, w) = (x, y, z)$. Then $\operatorname{null} T = {(0, 0, 0, w) : w \in \mathbb{R}}$ and we can take $U = {(x, y, z, 0) : x, y, z \in \mathbb{R}}$. Then $U \cap \operatorname{null} T = {0}$ and $\operatorname{range} T = {Tu : u \in U} = \mathbb{R}^3$. 2. Let $T : \mathbb{C}^5 \to \mathbb{C}^4$ be defined by $T(z_1, z_2, z_3, z_4, z_5) = (z_1 + z_2, z_2 + z_3, z_3 + z_4, z_4 + z_5)$. Then $\operatorname{null} T = {(z, -z, z, -z, z) : z \in \mathbb{C}}$ and we can take $U = {(z_1, z_2, z_3, z_4, 0) : z_1, z_2, z_3, z_4 \in \mathbb{C}}$. Then $U \cap \operatorname{null} T = {0}$ and $\operatorname{range} T = {Tu : u \in U}$. 3. Let $T : \mathbb{F}_3^4 \to \mathbb{F}_3^2$ be defined by $T(x, y, z, w) = (x + 2y, z + w)$. Then $\operatorname{null} T = {(x, x, y, 2y) : x, y \in \mathbb{F}_3}$ and we can take $U = {(0, y, 0, w) : y, w \in \mathbb{F}_3}$. Then $U \cap \operatorname{null} T = {0}$ and $\operatorname{range} T = {Tu : u \in U} = \mathbb{F}_3^2$. These examples demonstrate the principle in various vector spaces over different fields. In each case, we have a linear map $T$ and we find a subspace $U$ that has trivial intersection with the null space of $T$ and whose image under $T$ is the range of $T$. The choice of $U$ is not unique, but it always exists and it helps us understand the action of $T$.