--- Commentary: $\textbf{Section 2A}$ $\textbf{Exercise 1.}$ Find a list of four distinct vectors in $\mathbb{F}^3$ whose span equals ${(x, y, z) \in \mathbb{F}^3 : x + y + z = 0}.$ $\textbf{Solution 1.}$ Let $U = {(x, y, z) \in \mathbb{F}^3 : x + y + z = 0}$. The list $(1, -1, 0)$, $(0, -1, 1)$, $(0, 0, 0)$, $(2, -2, 0)$ consists of four distinct vectors, each of which is in $U$. Thus the span of these four vectors is contained in $U$. Conversely, if $(x, y, z) \in U$, then $y = -x - z$ and thus $(x, y, z) = x(1, -1, 0) + z(0, -1, 1) + 0(0, 0, 0) + 0(2, -2, 0).$ Thus $(x, y, z) \in \operatorname{span}((1, -1, 0), (0, -1, 1), (0, 0, 0), (2, -2, 0)).$ This shows that $U$ is contained in the span of our list of four vectors. Hence $U$ equals the span of our list of four vectors. $\textit{Commentary:}$ This exercise asks to find a spanning set for a specific subspace of $\mathbb{F}^3$. The subspace $U$ is defined by the equation $x + y + z = 0$, which is a linear constraint. The solution provides a list of four vectors in $U$ and shows that their span is exactly $U$. The forward inclusion (that the span is contained in $U$) follows from the fact that each vector in the list is in $U$, and $U$ is a subspace, so it must contain all linear combinations of these vectors. The reverse inclusion (that $U$ is contained in the span) is shown by taking an arbitrary vector $(x, y, z)$ in $U$, expressing $y$ in terms of $x$ and $z$ using the defining equation of $U$, and then writing $(x, y, z)$ as a linear combination of the vectors in the list. This exercise demonstrates the concept of a spanning set and provides practice in proving that a given set spans a particular subspace. It also illustrates the connection between linear equations and subspaces. $\textit{Additional Examples:}$ In $\mathbb{R}^3$, the subspace ${(x, y, z) : 2x - y + 3z = 0}$ is spanned by the vectors $(1, 2, 0)$, $(0, 3, 1)$, $(1, 5, 1)$, $(2, 4, 0)$. In $\mathbb{C}^3$, the subspace ${(z_1, z_2, z_3) : z_1 + iz_2 - z_3 = 0}$ is spanned by the vectors $(1, i, 1)$, $(i, -1, i)$, $(0, 0, 0)$, $(2, 2i, 2)$. In $\mathbb{F}_2^3$, the subspace ${(a, b, c) : a + b + c = 0}$ is spanned by the vectors $(1, 1, 0)$, $(1, 0, 1)$, $(0, 0, 0)$, $(0, 1, 1)$. In the space of polynomials $\mathbb{Q}[x]$, the subspace of polynomials $p(x)$ satisfying $p(1) = 0$ is spanned by the polynomials $x-1$, $x^2-1$, $0$, $x^3-1$. These examples illustrate spanning sets for subspaces defined by linear equations in various contexts, including over the real and complex numbers, over a finite field, and in a polynomial space. In each case, a set of vectors is provided that generates the entire subspace through linear combinations. ---