--- $ \begin{align} &\text{1. For each of the following subsets of } 𝐅^3, \text{ determine whether it is a subspace of } 𝐅^3 \\ \\ (a)\ &\{(𝑥,𝑥,𝑥)∈ 𝐅^3 \mid 𝑥 +2𝑥 +3𝑥 =0\} \\ (b)\ &\{(𝑥 , 𝑥 , 𝑥 ) ∈ 𝐅^3\mid 𝑥 + 2𝑥 + 3𝑥 = 4\} \\ (c)\ &\{(𝑥,𝑥,𝑥)∈ 𝐅^3\mid 𝑥𝑥𝑥 =0\} \\ (d)\ &\{(𝑥,𝑥,𝑥)∈ 𝐅^3\mid 𝑥 =5𝑥\} \\ \\ &\text{SOLUTION} \\ \\ (a)\ &\text{Let } 𝑈=\{(𝑥,𝑥,𝑥)∈ 𝐅^3 \mid 𝑥 +2𝑥 +3𝑥 =0\}.\\ &\text{To show that } 𝑈 \text{ is a subspace of } 𝐅^3, \text{ first note that } (0,0,0) \in 𝑈, \text{ so } 𝑈 \neq \emptyset.\\ &\text{Next, suppose } (𝑥_1, 𝑥_2, 𝑥_3) \in 𝑈 \text{ and } (𝑦_1, 𝑦_2, 𝑦_3) \in 𝑈. \text{ Then } 𝑥_1 + 2𝑥_2 + 3𝑥_3 = 0\\ &𝑦_1 + 2𝑦_2 + 3𝑦_3 = 0. \text{ Adding these equations, we have }\\ &(𝑥_1 +𝑦_1)+2(𝑥_2 +𝑦_2)+3(𝑥_3 +𝑦_3) = 0,\\ &\text{which means that } (𝑥_1 + 𝑦_1, 𝑥_2 + 𝑦_2, 𝑥_3 + 𝑦_3) \in 𝑈. \text{ Thus } 𝑈 \text{ is closed under addition.} \\ &\text{Next, suppose } (𝑥_1, 𝑥_2, 𝑥_3) \in 𝑈 \text{ and } 𝑎 \in 𝐅. \text{ Then }\\ &𝑥_1 + 2𝑥_2 + 3𝑥_3 = 0. \text{ Multiplying this equation by } 𝑎, \text{ we have }\\ &(𝑎𝑥_1) + 2(𝑎𝑥_2) + 3(𝑎𝑥_3) = 0,\\ &\text{which means that } (𝑎𝑥_1, 𝑎𝑥_2, 𝑎𝑥_3) \in 𝑈. \text{ Thus } 𝑈 \text{ is closed under scalar multiplication.}\\ &\text{Because } 𝑈 \text{ is a nonempty subset of } 𝐅^3 \text{ that is closed under addition and scalar multiplication, }\\ &𝑈 \text{ is a subspace of } 𝐅^3\\ \\ (b)\ &\text{Let } 𝑈=\{(𝑥_1,𝑥_2,𝑥_3)∈ 𝐅^3:𝑥 +2𝑥 +3𝑥 =0\}.\\ &\text{Then } (4,0,0) \in 𝑈 \text{ but } 0(4,0,0), \text{ which equals } (0,0,0),\\ &\text{ is not in } 𝑈.\\ &\text{ Thus, 𝑈 is not a subspace of } 𝐅^3 \\ (c)\ &\text{Let } 𝑈 = \{(𝑥_1,𝑥_2,𝑥_3) ∈ 𝐅^3 : 𝑥_1𝑥_2𝑥_3 = 0\}.\\ & \text{Then } (1,1,0) \in 𝑈 \text{ and } (0,0,1) \in 𝑈, \text{ but the sum of these two vectors, which equals } (1, 1, 1), \text{ is not in } 𝑈. \\ &\text{Thus } 𝑈 \text{ is not closed under addition.}\\ & \text{Thus } 𝑈 \text{ is not a subspace of } 𝐅^3\\ (d)\ &\text{Let}𝑈 = \{(𝑥_1,𝑥_2,𝑥_3) ∈ 𝐅 ∶ 𝑥_1 = 5𝑥_3\}.\\ &\text{To show that } 𝑈 \text{ is a subspace of } 𝐅^3 \text{ first note that } (0,0,0) \in 𝑈, \text{ so 𝑈 is nonempty}\\ &\text{Next, suppose } (𝑥_1, 𝑥_2, 𝑥_3) \in 𝑈 \text{ and } (𝑦_1, 𝑦_2, 𝑦_3) \in 𝑈. \text{ Then }\\ \\ &𝑥_1 = 5𝑥_3 \text{ and } 𝑦_1 = 5𝑦_3.\\ &\text{adding these equations, we have} \\ &𝑥_1 + 𝑦_1 = 5(𝑥_3 + 𝑦_3),\\ &\text{which means that } (𝑥_1 + 𝑦_1, 𝑥_2 + 𝑦_2, 𝑥_3 + 𝑦_3) \in 𝑈. \\ &\text{ Thus } 𝑈 \text{ is closed under addition.}\\ \\ &\text{Next, suppose } (𝑥_1, 𝑥_2, 𝑥_3) \in 𝑈 \text{ and } 𝑎 \in 𝐅. \text{ Then }\\ &𝑥_1 = 5𝑥_3. \text{ Multiplying this equation by } 𝑎, \text{ we have }\\ &𝑎𝑥_1 = 5(𝑎𝑥_3),\\ &\text{which means that } (𝑎𝑥_1, 𝑎𝑥_2, 𝑎𝑥_3) \in 𝑈. \\ &\text{Thus } 𝑈 \text{ is closed under scalar multiplication.} \\ &\text{Because U is a nonempty subset of} 𝐅^3 \\ &\text{ that is closed under addition and scalar multiplication, }\\ & 𝑈 \text{ is a subspace of } 𝐅^3. \end{align} $  --- Commentary: Here is the detailed LaTeX conversion for sol-2.pdf: $\textbf{Exercise 1.}$ For each of the following subsets of $\mathbb{F}^3$, determine whether it is a subspace of $\mathbb{F}^3$. (a) ${(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 + 2x_2 + 3x_3 = 0}$ (b) ${(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 + 2x_2 + 3x_3 = 4}$ (c) ${(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1x_2x_3 = 0}$ (d) ${(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 = 5x_3}$ $\textbf{Solution 1.}$ (a) Let $U = {(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 + 2x_2 + 3x_3 = 0}.$ To show that $U$ is a subspace of $\mathbb{F}^3$, first note that $(0, 0, 0) \in U$, so $U \neq \emptyset$. Next, suppose $(x_1, x_2, x_3) \in U$ and $(y_1, y_2, y_3) \in U$. Then $\begin{align*} x_1 + 2x_2 + 3x_3 &= 0 \\ y_1 + 2y_2 + 3y_3 &= 0. \end{align*}$ Adding these equations, we have $(x_1 + y_1) + 2(x_2 + y_2) + 3(x_3 + y_3) = 0,$ which means that $(x_1 + y_1, x_2 + y_2, x_3 + y_3) \in U$. Thus $U$ is closed under addition. Next, suppose $(x_1, x_2, x_3) \in U$ and $a \in \mathbb{F}$. Then $x_1 + 2x_2 + 3x_3 = 0.$ Multiplying this equation by $a$, we have $(ax_1) + 2(ax_2) + 3(ax_3) = 0,$ which means that $(ax_1, ax_2, ax_3) \in U$. Thus $U$ is closed under scalar multiplication. Because $U$ is a nonempty subset of $\mathbb{F}^3$ that is closed under addition and scalar multiplication, $U$ is a subspace of $\mathbb{F}^3$. $\textit{Commentary:}$ This part demonstrates that a subset of $\mathbb{F}^3$ defined by a homogeneous linear equation is a subspace. The key steps are verifying closure under addition and scalar multiplication, which follow from the properties of equality and the field operations. Geometrically, this subspace is a plane passing through the origin in $\mathbb{F}^3$. $\textit{Example:}$ In $\mathbb{R}^3$, the set ${(x, y, z) \in \mathbb{R}^3 \colon 2x - y + 3z = 0}$ is a subspace. It represents a plane through the origin. In $\mathbb{C}^3$, the set ${(z_1, z_2, z_3) \in \mathbb{C}^3 \colon iz_1 + (1+i)z_2 - z_3 = 0}$ is a subspace. It's a complex plane through the origin. In the space $\mathbb{F}_2^3$ (vectors with entries from the field with two elements), the set ${(a, b, c) \in \mathbb{F}_2^3 \colon a + b + c = 0}$ is a subspace. It contains the vectors $(0, 0, 0)$, $(1, 1, 0)$, $(1, 0, 1)$, and $(0, 1, 1)$. (b) Let $U = {(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 + 2x_2 + 3x_3 = 4}.$ Then $(4, 0, 0) \in U$ but $0(4, 0, 0)$, which equals $(0, 0, 0)$, is not in $U$. Thus $U$ is not closed under scalar multiplication. Hence $U$ is not a subspace of $\mathbb{F}^3$. $\textit{Commentary:}$ This part illustrates that a subset of $\mathbb{F}^3$ defined by a non-homogeneous linear equation is not a subspace. The key observation is that it's not closed under scalar multiplication, specifically by the scalar 0. Geometrically, this set is a plane not passing through the origin in $\mathbb{F}^3$. $\textit{Example:}$ In $\mathbb{R}^3$, the set ${(x, y, z) \in \mathbb{R}^3 \colon x + y - z = 1}$ is not a subspace. If $(1, 0, 0) \in U$, then $0(1, 0, 0) = (0, 0, 0)$ is not in $U$. In $\mathbb{C}^3$, the set ${(z_1, z_2, z_3) \in \mathbb{C}^3 \colon z_1 - iz_2 + z_3 = i}$ is not a subspace, for similar reasons. In $\mathbb{F}_2^3$, the set ${(a, b, c) \in \mathbb{F}_2^3 \colon a + b + c = 1}$ is not a subspace. It contains $(1, 0, 0)$, but $0(1, 0, 0) = (0, 0, 0)$ is not in the set. (c) Let $U = {(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1x_2x_3 = 0}.$ Then $(1, 1, 0) \in U$ and $(0, 0, 1) \in U$, but the sum of these two vectors, which equals $(1, 1, 1)$, is not in $U$. Thus $U$ is not closed under addition. Hence $U$ is not a subspace of $\mathbb{F}^3$. $\textit{Commentary:}$ This part shows that a subset of $\mathbb{F}^3$ defined by a nonlinear equation is typically not a subspace. Here, the equation involves the product of coordinates. The set is not closed under addition, as seen by adding two vectors in $U$ and getting a vector not in $U$. Geometrically, this set consists of the coordinate planes in $\mathbb{F}^3$. $\textit{Example:}$ In $\mathbb{R}^3$, the set ${(x, y, z) \in \mathbb{R}^3 \colon xyz = 0}$ is not a subspace. $(1, 0, 0)$ and $(0, 1, 0)$ are in the set, but their sum $(1, 1, 0)$ is not. In $\mathbb{C}^3$, the set ${(z_1, z_2, z_3) \in \mathbb{C}^3 \colon z_1z_2z_3 = 0}$ is not a subspace, for similar reasons. In $\mathbb{F}_2^3$, the set ${(a, b, c) \in \mathbb{F}_2^3 \colon abc = 0}$ is not a subspace. It contains $(1, 1, 0)$ and $(1, 0, 1)$, but their sum $(0, 1, 1)$ is not in the set. Note that in $\mathbb{F}_2$, $1 + 1 = 0$. (d) Let $U = {(x_1, x_2, x_3) \in \mathbb{F}^3 \colon x_1 = 5x_3}.$ To show that $U$ is a subspace of $\mathbb{F}^3$, first note that $(0, 0, 0) \in U$, so $U$ is nonempty. Next, suppose $(x_1, x_2, x_3) \in U$ and $(y_1, y_2, y_3) \in U$. Then $\begin{align*} x_1 &= 5x_3 \\ y_1 &= 5y_3. \end{align*}$ Adding these equations, we have $x_1 + y_1 = 5(x_3 + y_3),$ which means that $(x_1 + y_1, x_2 + y_2, x_3 + y_3) \in U$. Thus $U$ is closed under addition. Next, suppose $(x_1, x_2, x_3) \in U$ and $a \in \mathbb{F}$. Then $x_1 = 5x_3.$ Multiplying this equation by $a$, we have $ax_1 = 5(ax_3),$ which means that $(ax_1, ax_2, ax_3) \in U$. Thus $U$ is closed under scalar multiplication. Because $U$ is a nonempty subset of $\mathbb{F}^3$ that is closed under addition and scalar multiplication, $U$ is a subspace of $\mathbb{F}^3$. $\textit{Commentary:}$ This part demonstrates that a subset of $\mathbb{F}^3$ defined by a linear equation relating two coordinates is a subspace. The verification of closure under addition and scalar multiplication is similar to part (a). Geometrically, this subspace is a line in $\mathbb{F}^3$ passing through the origin. $\textit{Example:}$ In $\mathbb{R}^3$, the set ${(x, y, z) \in \mathbb{R}^3 \colon x = 2z}$ is a subspace. It's a line through the origin in the $xz$-plane. In $\mathbb{C}^3$, the set ${(z_1, z_2, z_3) \in \mathbb{C}^3 \colon z_2 = iz_1}$ is a subspace. It's a complex line through the origin. In $\mathbb{F}_2^3$, the set ${(a, b, c) \in \mathbb{F}_2^3 \colon b = a}$ is a subspace. It contains the vectors $(0, 0, 0)$, $(0, 0, 1)$, $(1, 1, 0)$, and $(1, 1, 1)$. These examples illustrate how the concept of a subspace unifies and generalizes various geometric notions (lines, planes, etc.) across different vector spaces and fields. The key properties are always closure under addition and scalar multiplication, which stem from the linearity of the equations defining the subset.