$\textbf{Exercise 7.}$ Suppose that $m$ is a nonnegative integer, $z_1, \ldots, z_{m+1}$ are distinct elements of $\mathbb{F}$, and $w_1, \ldots, w_{m+1} \in \mathbb{F}$. Prove that there exists a unique polynomial $p \in \mathcal{P}_m(\mathbb{F})$ such that $\begin{equation*} p(z_k) = w_k \text{ for each } k = 1, \ldots, m + 1. \end{equation*}$ This result can be proved without using linear algebra. However, try to find the clearer, shorter proof that uses some linear algebra. $\textbf{Solution 7.}$ Define $T: \mathcal{P}_m(\mathbb{F}) \to \mathbb{F}^{m+1}$ by $\begin{equation*} Tp = (p(z_1), \ldots, p(z_{m+1})). \end{equation*}$ We need to prove that $T$ is injective (which implies that at most one polynomial $p$ satisfies the condition required by the exercise) and surjective (which implies that at least one polynomial $p$ satisfies the condition required by the exercise). Clearly $T$ is a linear map. If $p \in \operatorname{null} T$, then $\begin{equation*} p(z_1) = \cdots = p(z_{m+1}) = 0, \end{equation*}$ which means that $p$ is a polynomial of degree $m$ with at least $m + 1$ distinct zeros, which means that $p = 0$ (by 4.8). Thus $T$ is injective, as desired. Now $\begin{align*} \dim \operatorname{range} T &= \dim \mathcal{P}_m(\mathbb{F}) - \dim \operatorname{null} T \\ &= (m + 1) - 0 \\ &= \dim \mathbb{F}^{m+1}, \end{align*}$ where the first equality comes from the fundamental theorem of linear maps (3.21) and the second equality holds because $\operatorname{null} T = \{0\}$. The last equality above implies that $\operatorname{range} T = \mathbb{F}^{m+1}$. Thus $T$ is surjective, as desired. $\textit{Comment:}$ Surjectivity of $T$ can also be proved by using an explicit construction. But linear algebra, specifically the fundamental theorem of linear maps, gives us surjectivity easily once we get injectivity. --- This exercise demonstrates the use of linear algebra to prove a result about the existence and uniqueness of a polynomial satisfying certain conditions. The linear map $T$ is defined to map a polynomial $p$ to the vector of its values at the points $z_1, \ldots, z_{m+1}$. Injectivity of $T$ shows that there can be at most one such polynomial, while surjectivity shows that there must be at least one. The proof of surjectivity uses the fundamental theorem of linear maps, which relates the dimensions of the domain, null space, and range of a linear map. ---  --- $\textbf{Exercise 7.}$ Suppose that $m$ is a nonnegative integer, $z_1, \ldots, z_{m+1}$ are distinct elements of $\mathbb{F}$, and $w_1, \ldots, w_{m+1} \in \mathbb{F}$. Prove that there exists a unique polynomial $p \in \mathcal{P}_m(\mathbb{F})$ such that $\begin{equation*} p(z_k) = w_k \text{ for each } k = 1, \ldots, m + 1. \end{equation*}$ This result can be proved without using linear algebra. However, try to find the clearer, shorter proof that uses some linear algebra. $\textbf{Solution 7.}$ Define $T: \mathcal{P}_m(\mathbb{F}) \to \mathbb{F}^{m+1}$ by $\begin{equation*} Tp = (p(z_1), \ldots, p(z_{m+1})). \end{equation*}$ We need to prove that $T$ is injective (which implies that at most one polynomial $p$ satisfies the condition required by the exercise) and surjective (which implies that at least one polynomial $p$ satisfies the condition required by the exercise). Clearly $T$ is a linear map. If $p \in \operatorname{null} T$, then $\begin{equation*} p(z_1) = \cdots = p(z_{m+1}) = 0, \end{equation*}$ which means that $p$ is a polynomial of degree $m$ with at least $m + 1$ distinct zeros, which means that $p = 0$ (by 4.8). Thus $T$ is injective, as desired. Now $\begin{align*} \dim \operatorname{range} T &= \dim \mathcal{P}_m(\mathbb{F}) - \dim \operatorname{null} T \\ &= (m + 1) - 0 \\ &= \dim \mathbb{F}^{m+1}, \end{align*}$ where the first equality comes from the fundamental theorem of linear maps (3.21) and the second equality holds because $\operatorname{null} T = \{0\}$. The last equality above implies that $\operatorname{range} T = \mathbb{F}^{m+1}$. Thus $T$ is surjective, as desired. $\textit{Comment:}$ Surjectivity of $T$ can also be proved by using an explicit construction. But linear algebra, specifically the fundamental theorem of linear maps, gives us surjectivity easily once we get injectivity. $\textit{Commentary:}$ This exercise demonstrates the power of linear algebra in solving a problem that could also be approached by direct construction. The problem is a classic one in interpolation theory: given $m+1$ distinct points and $m+1$ values, find a polynomial of degree at most $m$ that takes on those values at those points. The linear algebra approach is to define a linear map $T$ that evaluates a polynomial at the given points and to show that this map is bijective. Injectivity follows from the fact that a nonzero polynomial of degree $m$ can have at most $m$ zeros. Surjectivity follows from the fundamental theorem of linear maps, which relates the dimensions of the domain, kernel, and range of a linear map. This approach is much cleaner and more conceptual than attempting to construct the polynomial directly. $\textit{Example:}$ Let $m = 1$, $\mathbb{F} = \mathbb{R}$, $z_1 = 0$, $z_2 = 1$, $w_1 = 1$, and $w_2 = 2$. We want to find a unique polynomial $p(x) = ax + b$ such that $p(0) = 1$ and $p(1) = 2$. The linear map $T: \mathcal{P}_1(\mathbb{R}) \to \mathbb{R}^2$ is defined by $\begin{equation*} T(ax + b) = (a \cdot 0 + b, a \cdot 1 + b) = (b, a + b). \end{equation*}$ This map is injective because if $T(ax + b) = (0, 0)$, then $b = 0$ and $a + b = 0$, so $a = 0$. It's surjective because $\dim \mathcal{P}_1(\mathbb{R}) = 2 = \dim \mathbb{R}^2$. Therefore, there exists a unique polynomial $p(x) = ax + b$ such that $T(p) = (1, 2)$. Solving $b = 1$ and $a + b = 2$ gives $a = 1$ and $b = 1$, so $p(x) = x + 1$.