--- Commentary: $\textbf{Exercise 5.}$ Suppose $V$ is finite-dimensional and $U, W$ are subspaces of $V$ such that $V = U + W$. Prove that there exists a basis of $V$ consisting of vectors in $U \cup W$. $\textbf{Solution 5.}$ Because $V$ is finite-dimensional, the subspaces $U$ and $W$ are also finite-dimensional (by 2.25). Thus there exists a list of vectors in $U$ that spans $U$ and there exists a list of vectors in $W$ that spans $W$. Put these two lists together to get a list of vectors in $U \cup W$ that spans $V$ (because $V = U + W$). By 2.30, this list can be reduced to a basis of $V$, thus producing a basis of $V$ consisting of vectors in $U \cup W$. $\textit{Commentary:}$ This exercise deals with the situation where a vector space $V$ is the sum of two subspaces $U$ and $W$. The goal is to prove that in this case, there exists a basis of $V$ that consists entirely of vectors from $U$ and $W$. The proof relies on several key facts: 1. If $V$ is finite-dimensional, then any subspace of $V$ is also finite-dimensional (2.25). 2. A finite-dimensional space has a spanning list (by definition). 3. If $V = U + W$, then a spanning list for $U$ together with a spanning list for $W$ forms a spanning list for $V$. 4. Any spanning list can be reduced to a basis (2.30). By combining these facts, we can construct a basis for $V$ that consists only of vectors from $U$ and $W$. This is a useful result because it allows us to build bases for a space by breaking it down into simpler subspaces. $\textit{Examples:}$ 1. In $\mathbb{R}^3$, let $U$ be the $xy$-plane and $W$ be the $z$-axis. Then $\mathbb{R}^3 = U + W$. A basis for $U$ is $(1, 0, 0)$ and $(0, 1, 0)$, and a basis for $W$ is $(0, 0, 1)$. Together, these form a basis for $\mathbb{R}^3$ consisting of vectors from $U \cup W$. 2. In the space of polynomials $\mathbb{R}[x]$, let $U$ be the subspace of polynomials with even powers of $x$ and $W$ be the subspace of polynomials with odd powers of $x$. Then $\mathbb{R}[x] = U + W$. A basis for $U$ is $1, x^2, x^4, \ldots$, and a basis for $W$ is $x, x^3, x^5, \ldots$. Together, these form a basis for $\mathbb{R}[x]$ consisting of vectors from $U \cup W$. These examples illustrate the theorem in concrete settings. In each case, we have a space that is the sum of two subspaces, and we can construct a basis for the whole space by taking basis elements from each subspace. $\textbf{Exercise 6.}$ Prove or give a counterexample: If $p_0, p_1, p_2, p_3$ is a list in $\mathcal{P}_3(\mathbb{F})$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2, then $p_0, p_1, p_2, p_3$ is not a basis of $\mathcal{P}_3(\mathbb{F})$. $\textbf{Solution 6.}$ To construct a counterexample, define $p_0, p_1, p_2, p_3 \in \mathcal{P}_3(\mathbb{F})$ by \begin{align*} p_0(z) &= 1, \ p_1(z) &= z, \ p_2(z) &= z^2 +