--- Commentary: Here is the LaTeX conversion for sol-3.pdf, with detailed commentaries and examples: $\textbf{Exercise 3.}$ (a) Let $U$ be the subspace of $\mathbb{R}^5$ defined by $U = {(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5 : x_1 = 3x_2 \text{ and } x_3 = 7x_4}.$ Find a basis of $U$. (b) Extend the basis in (a) to a basis of $\mathbb{R}^5$. (c) Find a subspace $W$ of $\mathbb{R}^5$ such that $\mathbb{R}^5 = U \oplus W$. $\textbf{Solution 3.}$ (a) Note that $U = {(3x_2, x_2, 7x_4, x_4, x_5) : x_2, x_4, x_5 \in \mathbb{R}}.$ From this representation of $U$, we see easily that $(3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1)$ is a basis of $U$. Of course there are also other possible choices of bases of $U$. (b) The list $(3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1), (1, 0, 0, 0, 0), (0, 0, 1, 0, 0)$ is a basis of $\mathbb{R}^5$. (c) Let $W = {(x, 0, y, 0, 0) : x, y \in \mathbb{R}}.$ $\textit{Commentary:}$ This exercise explores a subspace $U$ of $\mathbb{R}^5$ defined by two linear equations, finds a basis for $U$, extends this basis to a basis of $\mathbb{R}^5$, and finds a complementary subspace $W$ such that $\mathbb{R}^5 = U \oplus W$. In part (a), the key observation is that the conditions $x_1 = 3x_2$ and $x_3 = 7x_4$ allow us to express $x_1$ and $x_3$ in terms of $x_2$ and $x_4$, respectively. This leads to the parametric representation of $U$ in terms of $x_2, x_4, x_5$. From this, we can read off a basis for $U$: $(3, 1, 0, 0, 0)$ corresponds to $x_2 = 1, x_4 = x_5 = 0$, $(0, 0, 7, 1, 0)$ corresponds to $x_2 = x_5 = 0, x_4 = 1$, and $(0, 0, 0, 0, 1)$ corresponds to $x_2 = x_4 = 0, x_5 = 1$. In part (b), we extend the basis of $U$ to a basis of $\mathbb{R}^5$ by adding vectors that introduce the missing dimensions. Here, $(1, 0, 0, 0, 0)$ introduces the first dimension and $(0, 0, 1, 0, 0)$ introduces the third dimension. In part (c), we choose $W$ to be the subspace spanned by the vectors we added in part (b). This ensures that $U \cap W = {0}$ and that every vector in $\mathbb{R}^5$ can be uniquely written as a sum of a vector in $U$ and a vector in $W$, which is the definition of a direct sum. $\textit{Examples:}$ 1. In $\mathbb{R}^4$, let $U = {(x, y, 2x, 3y) : x, y \in \mathbb{R}}$. A basis for $U$ is $(1, 0, 2, 0)$ and $(0, 1, 0, 3)$. This can be extended to a basis of $\mathbb{R}^4$ by adding $(0, 0, 1, 0)$ and $(0, 0, 0, 1)$. The subspace $W = {(0, 0, z, w) : z, w \in \mathbb{R}}$ satisfies $\mathbb{R}^4 = U \oplus W$. 2. In $\mathbb{C}^3$, let $U = {(z, iz, w) : z, w \in \mathbb{C}}$. A basis for $U$ is $(1, i, 0)$ and $(0, 0, 1)$. This can be extended to a basis of $\mathbb{C}^3$ by adding $(i, -1, 0)$. The subspace $W = {(ai, -a, 0) : a \in \mathbb{C}}$ satisfies $\mathbb{C}^3 = U \oplus W$. 3. In the space of polynomials $\mathbb{R}[x]$, let $U = {p(x) : p'(0) = p(1)}$. A basis for $U$ is $1, x^2, x^3, \ldots$. This can be extended to a basis of $\mathbb{R}[x]$ by adding $x$. The subspace $W = {ax : a \in \mathbb{R}}$ satisfies $\mathbb{R}[x] = U \oplus W$. These examples illustrate the same concepts in different vector spaces over different fields. The key steps are always to find a basis for the given subspace, extend it to a basis of the whole space, and then identify a complementary subspace that makes the whole space a direct sum of the two subspaces.