--- Commentary: $\textbf{Exercise 8.}$ Prove or give a counterexample: If $v_1, v_2, v_3, v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1, v_2 \in U$ and $v_3 \notin U$ and $v_4 \notin U$, then $v_1, v_2$ is a basis of $U$. $\textbf{Solution 8.}$ The statement above is false. For example, take $V = \mathbb{R}^4$, let $v_1, v_2, v_3, v_4$ be the standard basis of $\mathbb{R}^4$, and let $U = {(a, b, c, c) : a, b, c \in \mathbb{R}}.$ Then $v_1, v_2 \in U$ and $v_3 \notin U$ and $v_4 \notin U$. However, $v_1, v_2$ is not a basis of $U$, because $(0, 0, 1, 1) \in U$ but $(0, 0, 1, 1) \notin \operatorname{span}(v_1, v_2)$. $\textit{Commentary:}$ This exercise presents a claim about bases and subspaces and asks to either prove it or provide a counterexample. The claim is that if a basis of $V$ has two vectors in a subspace $U$ and two vectors not in $U$, then the two vectors in $U$ form a basis of $U$. The solution provides a counterexample to disprove the claim. In $\mathbb{R}^4$, we take $U$ to be the subspace of vectors whose third and fourth coordinates are equal. We take the basis of $V$ to be the standard basis. Then the first two basis vectors are in $U$ and the last two are not, but the first two vectors do not span $U$ (for example, they cannot span the vector $(0, 0, 1, 1)$, which is in $U$). This example illustrates that just because a subspace contains some basis vectors of the full space, those basis vectors do not necessarily form a basis of the subspace. In this case, the subspace $U$ has an additional "diagonal" direction $(0, 0, 1, 1)$ that is not captured by the first two standard basis vectors. The exercise highlights the importance of carefully checking the spanning and linear independence properties when determining whether a set of vectors forms a basis for a specific subspace. $\textit{Examples:}$ 1. In $\mathbb{R}^3$, let $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (0, 0, 1)$, and $U = {(a, a, b) : a, b \in \mathbb{R}}$. Then $v_1, v_3 \in U$ and $v_2 \notin U$, but $v_1, v_3$ is not a basis of $U$ because $(1, 1, 0) \in U$ is not in $\operatorname{span}(v_1, v_3)$. 2. In $\mathcal{P}_4(\mathbb{R})$, let $v_1 = 1$, $v_2 = x$, $v_3 = x^2$, $v_4 = x^3$, $v_5 = x^4$, and $U = {p \in \mathcal{P}_4(\mathbb{R}) : p(0) = p(1)}$. Then $v_1, v_3, v_5 \in U$ and $v_2, v_4 \notin U$, but $v_1, v_3, v_5$ is not a basis of $U$ because $x - x^2 \in U$ is not in $\operatorname{span}(v_1, v_3, v_5)$. These examples further demonstrate that having some basis vectors of the full space in a subspace does not guarantee that those vectors form a basis of the subspace. In each case, the subspace has additional directions that are not captured by the included basis vectors.