--- Commentary: $\textbf{Exercise 10.}$ Suppose $V_1$ and $V_2$ are subspaces of $V$. Prove that the intersection $V_1 \cap V_2$ is a subspace of $V$. $\textbf{Solution 10.}$ The additive identity $0$ is in $V_1$ and in $V_2$. Thus $0 \in V_1 \cap V_2$. Suppose $u, v \in V_1 \cap V_2$. Then $u, v \in V_1$ and $u, v \in V_2$. Thus $u + v \in V_1$ and $u + v \in V_2$. Hence $u + v \in V_1 \cap V_2$. Therefore, $V_1 \cap V_2$ is closed under addition. Suppose $u \in V_1 \cap V_2$ and $a \in \mathbb{F}$. Then $u \in V_1$ and $u \in V_2$. Thus $au \in V_1$ and $au \in V_2$. Hence $au \in V_1 \cap V_2$. Therefore, $V_1 \cap V_2$ is closed under scalar multiplication. Thus $V_1 \cap V_2$ satisfies the three conditions of 1.34 and hence is a subspace of $V$. $\textit{Commentary:}$ This exercise demonstrates that the intersection of two subspaces is always a subspace. The proof verifies the three subspace conditions: the intersection contains the zero vector, and it's closed under addition and scalar multiplication. These follow directly from the fact that each of the original subspaces satisfies these conditions. This result is fundamental in linear algebra and has many applications. For instance, it's used in the study of eigenspaces (the intersection of the null spaces of $T - \lambda I$ for different eigenvalues $\lambda$), in the solution of systems of linear equations (the intersection of the hyperplanes defined by each equation), and in the decomposition of a vector space into a direct sum of subspaces. $\textit{Example:}$ In $\mathbb{Q}_p^3$, let $V_1 = {(x, y, z) : x + y + z = 0}$ and $V_2 = {(x, y, z) : x = y}$. Both $V_1$ and $V_2$ are subspaces of $\mathbb{Q}_p^3$, and their intersection $V_1 \cap V_2 = {(x, x, -2x) : x \in \mathbb{Q}_p}$ is also a subspace. In the space of polynomials $\mathbb{F}_q[x]$, let $V_1$ be the subspace of polynomials with even degree and $V_2$ be the subspace of polynomials with degree at most 3. Then $V_1 \cap V_2$ is the subspace of polynomials of the form $ax^2 + b$ for $a, b \in \mathbb{F}_q$. In the space of continuous functions $C([0,1], \mathbb{R})$, let $V_1$ be the subspace of functions satisfying $f(0) = f(1)$ and $V_2$ be the subspace of functions satisfying $\int_0^1 f(x) , dx = 0$. Then $V_1 \cap V_2$ is the subspace of continuous functions $f$ satisfying both $f(0) = f(1)$ and $\int_0^1 f(x) , dx = 0$. These examples illustrate the variety of situations where the intersection of subspaces arises naturally. The specific nature of the subspaces can vary greatly depending on the vector space and the application, but the basic principle that their intersection is a subspace remains universal.