--- Commentary: $\textbf{Exercise 6.}$ (a) Let $U = {p \in \mathcal{P}_4(\mathbb{F}) : p(2) = p(5) = p(6)}$. 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 $\mathcal{P}_4(\mathbb{F}) = U \oplus W$. $\textbf{Solution 6.}$ (a) A basis of $U$ is $1, (x - 2)(x - 5)(x - 6), (x - 2)^2(x - 5)(x - 6).$ 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 three and thus $\dim U \geq 3$. However, $U$ is a proper subspace of ${p \in \mathcal{P}_4(\mathbb{F}) : p(2) = p(5)}$, which from the solution to Exercise 15 has dimension four. This implies that $\dim U = 3$. Hence the list above is a basis of $U$. (b) The list $1, (x - 2)(x - 5)(x - 6), (x - 2)^2(x - 5)(x - 6), x, x^2$ 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 $W = \operatorname{span}(x, x^2)$ gives a subspace $W$ such that $\mathcal{P}_4(\mathbb{F}) = U \oplus W$. $\textit{Commentary:}$ This exercise builds on Exercise 15, with the subspace $U$ now consisting of all polynomials in $\mathcal{P}_4(\mathbb{F})$ that take the same value at three points: $x = 2$, $x = 5$, and $x = 6$. In part (a), a basis for $U$ is found. The polynomials $1$, $(x - 2)(x - 5)(x - 6)$, and $(x - 2)^2(x - 5)(x - 6)$ are chosen. These polynomials are in $U$ because they are zero at $x = 2$, $x = 5$, and $x = 6$, so they certainly take the same value at these points. They are linearly independent as before. The dimension of $U$ is argued to be 3 by considering the subspace from Exercise 15 and noting that $U$ is a proper subspace of it. In part (b), the basis of $U$ is extended to a basis of $\mathcal{P}_4(\mathbb{F})$ by adding the polynomials $x$ and $x^2$, which are not in $U$ because they take different values at the three points. In part (c), the subspace $W$ is chosen to be the span of the polynomials added in part (b), which is the space of linear combinations of $x$ and $x^2$. 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 linear combination of $x$ and $x^2$. This exercise demonstrates that the conditions defining a subspace can involve the equality of a polynomial at more than two points. The basis for the subspace is constructed using polynomials that are zero at all the points in the condition. The complementary subspace is spanned by polynomials that take different values at these points. $\textit{Examples:}$ In $\mathcal{P}_5(\mathbb{R})$, let $U = {p \in \mathcal{P}_5(\mathbb{R}) : p(0) = p(1) = p(-1)}$. A basis for $U$ is $1, x(x - 1)(x + 1), x^2(x - 1)(x + 1)$. This can be extended to a basis of $\mathcal{P}_5(\mathbb{R})$ by adding $x, x^2, x^3$. The subspace $W = \operatorname{span}(x, x^2, x^3)$ satisfies $\mathcal{P}_5(\mathbb{R}) = U \oplus W$. In $\mathcal{P}_4(\mathbb{C})$, let $U = {p \in \mathcal{P}_4(\mathbb{C}) : p(i) = p(-i) = p(1)}$. A basis for $U$ is $1, (x - i)(x + i)(x - 1)$. This can be extended to a basis of $\mathcal{P}_4(\mathbb{C})$ by adding $x, x^2, x^3$. The subspace $W = \operatorname{span}(x, x^2, x^3)$ satisfies $\mathcal{P}_4(\mathbb{C}) = U \oplus W$. In $\mathcal{P}_6(\mathbb{F}_3)$, let $U = {p \in \mathcal{P}_6(\mathbb{F}_3) : p(0) = p(1) = p(2)}$. A basis for $U$ is $1, x(x - 1)(x - 2), x^2(x - 1)(x - 2), x^3(x - 1)(x - 2)$. This can be extended to a basis of $\mathcal{P}_6(\mathbb{F}_3)$ by adding $x, x^2, x^3$. The subspace $W = \operatorname{span}(x, x^2, x^3)$ satisfies $\mathcal{P}_6(\mathbb{F}_3) = U \oplus W$. These examples show the same principle in action, with subspaces defined by the condition that polynomials take the same value at three points. The basis for the subspace consists of polynomials that are zero at all three points, while the complementary subspace is spanned by lower-degree polynomials that take different values at these points. The specific polynomials in the bases depend on the field and the points in the condition. ---