--- Commentary: Mentioned in [[L7-text]] at minute 19:10 $\textbf{Exercise 4.}$ (a) Let $U = {p \in \mathcal{P}_4(\mathbb{R}) : p''(6) = 0}$. Find a basis of $U$. (b) Extend the basis in (a) to a basis of $\mathcal{P}_4(\mathbb{R})$. (c) Find a subspace $W$ of $\mathcal{P}_4(\mathbb{R})$ such that $\mathcal{P}_4(\mathbb{R}) = U \oplus W$. $\textbf{Solution 14.}$ (a) A basis of $U$ is $1, x - 6, (x - 6)^3, (x - 6)^4.$ 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, $\dim \mathcal{P}_4(\mathbb{R}) = 5$, which implies that $\dim U = 4$ or $\dim U = 5$. Because $U$ is a proper subspace of $\mathcal{P}_4(\mathbb{R})$, this implies that $\dim U = 4$. Hence the list above is a basis of $U$. (b) The polynomial $(x - 6)^2$ clearly is not in $U$. Thus $1, x - 6, (x - 6)^3, (x - 6)^4, (x - 6)^2$ is a linearly independent list in $\mathcal{P}_4(\mathbb{R})$ of length five. By 2.38, the list above is a basis of $\mathcal{P}_4(\mathbb{R})$. (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 $\mathcal{P}_4(\mathbb{R})$ consisting of the constant multiples of $(x - 6)^2$ gives a subspace $W$ such that $\mathcal{P}_4(\mathbb{R}) = U \oplus W$. $\textit{Commentary:}$ This exercise is similar to Exercise 13, but the subspace $U$ is now defined by a condition on the second derivative of the polynomials at $x = 6$. Specifically, $U$ consists of all polynomials in $\mathcal{P}_4(\mathbb{R})$ whose second derivative is zero at $x = 6$. In part (a), a basis for $U$ is found. The polynomials $1$, $x - 6$, $(x - 6)^3$, and $(x - 6)^4$ are chosen. These polynomials are in $U$ because their second derivatives at $x = 6$ are zero. They are linearly independent for the same reasons as in Exercise 13. The dimension of $U$ is argued to be 4 in the same way as before. In part (b), the basis of $U$ is extended to a basis of $\mathcal{P}_4(\mathbb{R})$ by adding the polynomial $(x - 6)^2$, which is not in $U$ because its second derivative at $x = 6$ is not zero. 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 $(x - 6)^2$. This ensures that $U$ and $W$ intersect only at the zero polynomial and that every polynomial in $\mathcal{P}_4(\mathbb{R})$ can be uniquely written as a sum of a polynomial in $U$ and a multiple of $(x - 6)^2$. This exercise further illustrates the process of finding a basis for a subspace defined by a condition, extending this basis, and identifying a complementary subspace. It also demonstrates that the conditions defining a subspace can involve derivatives of polynomials, not just their values. $\textit{Examples:}$ 1. In $\mathcal{P}_5(\mathbb{R})$, let $U = {p \in \mathcal{P}_5(\mathbb{R}) : p'(0) = p'''(0) = 0}$. A basis for $U$ is $1, x^2, x^4, x^5$. This can be extended to a basis of $\mathcal{P}_5(\mathbb{R})$ by adding $x$ and $x^3$. The subspace $W = \operatorname{span}(x, x^3)$ satisfies $\mathcal{P}_5(\mathbb{R}) = U \oplus W$. 2. In $\mathcal{P}_4(\mathbb{C})$, let $U = {p \in \mathcal{P}_4(\mathbb{C}) : p(i) = p''(i) = 0}$. A basis for $U$ is $1, x - i, (x - i)^3, (x - i)^4$. This can be extended to a basis of $\mathcal{P}_4(\mathbb{C})$ by adding $(x - i)^2$. The subspace $W = \operatorname{span}((x - i)^2)$ satisfies $\mathcal{P}_4(\mathbb{C}) = U \oplus W$. 3. In $\mathcal{P}_6(\mathbb{F}_3)$, let $U = {p \in \mathcal{P}_6(\mathbb{F}_3) : p(1) = p'(1) = p''(1) = 0}$. A basis for $U$ is $(x - 1)^3, (x - 1)^4, (x - 1)^5, (x - 1)^6$. This can be extended to a basis of $\mathcal{P}_6(\mathbb{F}_3)$ by adding $1$, $x - 1$, and $(x - 1)^2$. The subspace $W = \operatorname{span}(1, x - 1, (x - 1)^2)$ satisfies $\mathcal{P}_6(\mathbb{F}_3) = U \oplus W$. These examples demonstrate the principle in polynomial spaces over different fields, with subspaces defined by conditions involving values and derivatives of various orders at different points. The key steps remain the same, but the specific polynomials chosen for the bases depend on the conditions defining the subspace.