--- Commentary: $\textbf{Exercise 2.}$ Show that the subspaces of $\mathbb{R}^3$ are precisely ${0}$, all lines in $\mathbb{R}^3$ containing the origin, all planes in $\mathbb{R}^3$ containing the origin, and $\mathbb{R}^3$. $\textbf{Solution 2.}$ The set ${0}$, the set $\mathbb{R}^3$, all lines in $\mathbb{R}^3$ through the origin, and all planes in $\mathbb{R}^3$ through the origin are all subspaces of $\mathbb{R}^3$. To show that there are no other subspaces of $\mathbb{R}^3$, suppose $U$ is a subspace of $\mathbb{R}^3$. Then by 2.37, $\dim U$ equals 0, 1, 2, or 3. If $\dim U = 0$, then $U = {0}$. If $\dim U = 1$, then there is a basis of $U$ consisting of one vector $v$, and $U$ equals all scalar multiples of $v$; thus $U$ is a line through the origin. If $\dim U = 2$, then there is a basis $v_1, v_2$ of the subspace $U$, and hence $U$ equals $\operatorname{span}(v_1, v_2)$; thus $U$ is a plane through the origin. If $\dim U = 3$, then $U = \mathbb{R}^3$. $\textit{Commentary:}$ This exercise asks to classify all subspaces of $\mathbb{R}^3$. The key insight is that the subspaces can be categorized by their dimension, which can only be 0, 1, 2, or 3 in a 3-dimensional space. The proof first notes that ${0}$, $\mathbb{R}^3$, lines through the origin, and planes through the origin are indeed subspaces. This is geometrically intuitive and can be formally verified by checking the subspace axioms. Then, the proof shows that every subspace of $\mathbb{R}^3$ must fall into one of these categories. This is done by considering the four possible dimensions separately: - If the dimension is 0, the subspace must be ${0}$, the only 0-dimensional subspace. - If the dimension is 1, the subspace must be a line through the origin, as it's spanned by a single vector. - If the dimension is 2, the subspace must be a plane through the origin, as it's spanned by two linearly independent vectors. - If the dimension is 3, the subspace must be the entire space $\mathbb{R}^3$. This exhausts all possibilities. This exercise provides a complete picture of the subspace structure of $\mathbb{R}^3$. It illustrates the intimate connection between the algebraic concept of dimension and the geometric notions of lines, planes, and space. It also demonstrates the power of the concept of basis in classifying subspaces. $\textit{Examples:}$ 1. In $\mathbb{R}^2$, the subspaces are precisely ${0}$, all lines through the origin, and $\mathbb{R}^2$ itself. 2. In $\mathbb{C}^2$, viewed as a vector space over $\mathbb{C}$, the subspaces are precisely ${0}$, all complex lines through the origin, and $\mathbb{C}^2$ itself. 3. In the space of polynomials $\mathcal{P}_3(\mathbb{R})$, the subspaces are precisely ${0}$, all 1-dimensional subspaces (spanned by a single non-zero polynomial), all 2-dimensional subspaces (spanned by two linearly independent polynomials), all 3-dimensional subspaces (spanned by three linearly independent polynomials), and $\mathcal{P}_3(\mathbb{R})$ itself. These examples illustrate similar classifications in other spaces. The subspaces are always characterized by their dimension, but the geometric interpretation varies depending on the space. In $\mathbb{R}^2$, 1-dimensional subspaces are lines; in $\mathbb{C}^2$, 1-dimensional subspaces over $\mathbb{C}$ are complex lines; in $\mathcal{P}_3(\mathbb{R})$, 1-dimensional subspaces are sets of multiples of a single polynomial. Despite these differences, the underlying principle of categorizing subspaces by dimension remains the same.