--- Commentary: $\textbf{Exercise 5.}$ (a) Let $U = {p \in \mathcal{P}_4(\mathbb{F}) : p(2) = p(5)}$. Find a basis of $U$. (b) Extend the basis in (a) to a basis of $\mathcal{P}_4(\mathbb{F})$. (c) Find a subspace $W$ of $\mathcal{P}_4(\mathbb{F})$ such that $\mathcal{P}_4(\mathbb{F}) = U \oplus W$. $\textbf{Solution 5.}$ (a) A basis of $U$ is $1, (x - 2)(x - 5), (x - 2)^2(x - 5), (x - 2)^3(x - 5).$ Each polynomial in the list above is clearly in $U$. To verify that the list above is indeed a basis of $U$, first note that the list above is linearly independent (using the same reasoning as was used in Example 2.41 to show that the list in that example is linearly independent). Then note that the linearly independent list above has length four, and thus $\dim U \geq 4$. However, $\dim \mathcal{P}_4(\mathbb{F}) = 5$, which implies that $\dim U = 4$ or $\dim U = 5$. Because $U$ is a proper subspace of $\mathcal{P}_4(\mathbb{F})$, this implies that $\dim U = 4$. Hence the list above is a basis of $U$. (b) The polynomial $x$ clearly is not in $U$. Thus $1, (x - 2)(x - 5), (x - 2)^2(x - 5), (x - 2)^3(x - 5), x$ is a linearly independent list in $\mathcal{P}_4(\mathbb{F})$ of length five. By 2.38, the list above is a basis of $\mathcal{P}_4(\mathbb{F})$. (c) Using the idea of the proof of 2.33 and the answer above to (b), we see that taking $W$ to be the subspace of $\mathcal{P}_4(\mathbb{F})$ consisting of the constant multiples of $x$ gives a subspace $W$ such that $\mathcal{P}_4(\mathbb{F}) = U \oplus W$. $\textit{Commentary:}$ In this exercise, the subspace $U$ consists of all polynomials in $\mathcal{P}_4(\mathbb{F})$ that take the same value at $x = 2$ and $x = 5$. This condition is different from the ones in the previous exercises, as it involves the equality of the polynomial at two different points, not just a specific value or derivative at a single point. In part (a), a basis for $U$ is found. The polynomials $1$, $(x - 2)(x - 5)$, $(x - 2)^2(x - 5)$, and $(x - 2)^3(x - 5)$ are chosen. These polynomials are in $U$ because they are zero at both $x = 2$ and $x = 5$, so they certainly take the same value at these points. They are linearly independent for similar reasons as before. The dimension of $U$ is argued to be 4 as in the previous exercises. In part (b), the basis of $U$ is extended to a basis of $\mathcal{P}_4(\mathbb{F})$ by adding the polynomial $x$, which is not in $U$ because it takes different values at $x = 2$ and $x = 5$. In part (c), the subspace $W$ is chosen to be the span of the polynomial added in part (b), which is the space of multiples of $x$. This ensures that $U$ and $W$ intersect only at the zero polynomial and that every polynomial in $\mathcal{P}_4(\mathbb{F})$ can be uniquely written as a sum of a polynomial in $U$ and a multiple of $x$. This exercise demonstrates that the conditions defining a subspace can involve the equality of a polynomial at different points. It also shows that the polynomials chosen for the basis of the subspace should be designed to satisfy this equality condition. $\textit{Examples:}$ 1. In $\mathcal{P}_5(\mathbb{R})$, let $U = {p \in \mathcal{P}_5(\mathbb{R}) : p(1) = p(-1)}$. A basis for $U$ is $1, (x - 1)(x + 1), (x - 1)^2(x + 1), (x - 1)^3(x + 1), (x - 1)^4(x + 1)$. This can be extended to a basis of $\mathcal{P}_5(\mathbb{R})$ by adding $x$. The subspace $W = \operatorname{span}(x)$ satisfies $\mathcal{P}_5(\mathbb{R}) = U \oplus W$. 2. In $\mathcal{P}_3(\mathbb{C})$, let $U = {p \in \mathcal{P}_3(\mathbb{C}) : p(i) = p(-i)}$. A basis for $U$ is $1, (x - i)(x + i), (x - i)^2(x + i)$. This can be extended to a basis of $\mathcal{P}_3(\mathbb{C})$ by adding $x$. The subspace $W = \operatorname{span}(x)$ satisfies $\mathcal{P}_3(\mathbb{C}) = U \oplus W$. 3. In $\mathcal{P}_4(\mathbb{F}_5)$, let $U = {p \in \mathcal{P}_4(\mathbb{F}_5) : p(1) = p(4)}$ (note that in $\mathbb{F}_5$, 4 = -1). A basis for $U$ is $1, (x - 1)(x - 4), (x - 1)^2(x - 4), (x - 1)^3(x - 4)$. This can be extended to a basis of $\mathcal{P}_4(\mathbb{F}_5)$ by adding $x$. The subspace $W = \operatorname{span}(x)$ satisfies $\mathcal{P}_4(\mathbb{F}_5) = U \oplus W$. These examples illustrate the same principle in polynomial spaces over different fields, with subspaces defined by the condition that the polynomials take the same value at two different points. The specific polynomials chosen for the bases depend on the two points in the condition. The complementary subspace is always chosen to be the span of a polynomial that takes different values at the two points, ensuring a direct sum decomposition.