![[sol-5.pdf#page=25]] --- $\textbf{Exercise 11.}$ Suppose $v_1, \ldots, v_n$ is a basis of $V$ and $\varphi_1, \ldots, \varphi_n$ is the corresponding dual basis of $V^*$. Suppose $\psi \in V^*$. Prove that $\begin{equation*} \psi = \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n. \end{equation*}$ $\textbf{Solution 11.}$ We have $\begin{equation*} (\psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n)(v_k) = \psi(v_k) \end{equation*}$ for each $k = 1, \ldots, n$. Because the linear functionals $\psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n$ and $\psi$ agree on a basis, they are equal by the uniqueness part of the linear map lemma (3.4). --- $\textbf{Exercise 11.}$ Suppose $v_1, \ldots, v_n$ is a basis of $V$ and $\varphi_1, \ldots, \varphi_n$ is the corresponding dual basis of $V^*$. Suppose $\psi \in V^*$. Prove that $\begin{equation*} \psi = \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n. \end{equation*}$ $\textbf{ Solution11.}$ We have $\begin{equation*} (\psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n)(v_k) = \psi(v_k) \end{equation*}$ for each $k = 1, \ldots, n$. Because the linear functionals $\psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n$ and $\psi$ agree on a basis, they are equal by the uniqueness part of the linear map lemma (3.4). $\textit{Commentary:}$ This exercise shows that every linear functional can be uniquely expressed as a linear combination of the dual basis functionals, with the coefficients being the values of the functional on the original basis vectors. This is a fundamental result in the theory of dual spaces and has many applications, such as in the representation of linear operators and the study of tensor products. The proof is a simple application of the linear map lemma, which states that a linear map is uniquely determined by its values on a basis. $\textit{Example:}$ Let $V = \mathbb{R}^2$ with the standard basis ${e_1, e_2}$, where $e_1 = (1, 0)$ and $e_2 = (0, 1)$. The dual basis of $V^* = (\mathbb{R}^2)^_$ is ${\varphi_1, \varphi_2}$, where $\varphi_1(x, y) = x$ and $\varphi_2(x, y) = y$. Let $\psi \in V^_$ be defined by $\psi(x, y) = 2x - 3y$. Then $\begin{align*} \psi &= \psi(e_1)\varphi_1 + \psi(e_2)\varphi_2 \ &= 2\varphi_1 - 3\varphi_2. \end{align*}$ You can check that this equality holds by evaluating both sides on any vector $(x, y) \in \mathbb{R}^2$.