[proofplan] Fix an ordered basis $\mathcal{B}=(v_1,\ldots,v_n)$ of $V$ and construct the associated coordinate functionals $v_i^*:V\to k$. We prove that these functionals are linearly independent by evaluating a linear relation on each basis vector. We then prove that they span $V^*$ by showing that every linear functional is determined by its values on the basis vectors. Therefore $\mathcal{B}^*$ is a basis of $V^*$, and the dimension equality follows from counting the basis elements. [/proofplan] [step:Construct the coordinate functionals associated to the ordered basis] Let $\mathcal{B}=(v_1,\ldots,v_n)$ be an ordered basis of $V$. Since $\mathcal{B}$ is a basis, every vector $v\in V$ has a unique coordinate expansion \begin{align*} v=\sum_{m=1}^{n} c_m(v)v_m \end{align*} with scalars $c_m(v)\in k$. For each $i\in\{1,\ldots,n\}$, define $v_i^*:V\to k$ by \begin{align*} v_i^*(v)=c_i(v) \end{align*} for the unique coordinates of $v$ in the basis $\mathcal{B}$. The uniqueness of coordinates makes $v_i^*$ well-defined. If $v,w\in V$ and $a,b\in k$, then the coordinate expansion of $av+bw$ has $i$-th coordinate $a c_i(v)+b c_i(w)$, so \begin{align*} v_i^*(av+bw)=a v_i^*(v)+b v_i^*(w) \end{align*} and hence $v_i^*\in V^*$. For each $j\in\{1,\ldots,n\}$, the vector $v_j$ has coordinates $c_j(v_j)=1$ and $c_i(v_j)=0$ for $i\neq j$, so \begin{align*} v_i^*(v_j)=\delta_{ij}. \end{align*} Thus $(v_1^*,\ldots,v_n^*)$ is exactly the dual family $\mathcal{B}^*$. [/step] [step:Prove the dual family is linearly independent by evaluating on basis vectors] Suppose $a_1,\ldots,a_n\in k$ satisfy \begin{align*} \sum_{i=1}^{n} a_i v_i^*=0 \end{align*} as an element of $V^*$. For each fixed $j\in\{1,\ldots,n\}$, evaluate both sides at $v_j$. Since evaluation at $v_j$ is compatible with addition and scalar multiplication of linear functionals, \begin{align*} 0=\left(\sum_{i=1}^{n} a_i v_i^*\right)(v_j)=\sum_{i=1}^{n} a_i v_i^*(v_j)=\sum_{i=1}^{n} a_i\delta_{ij}=a_j. \end{align*} Since this holds for every $j\in\{1,\ldots,n\}$, all coefficients $a_j$ are zero. Hence $\mathcal{B}^*$ is linearly independent. [guided] We want to prove that the family $(v_1^*,\ldots,v_n^*)$ is linearly independent in the [vector space](/page/Vector%20Space) $V^*$. By definition, this means that whenever scalars $a_1,\ldots,a_n\in k$ satisfy \begin{align*} \sum_{i=1}^{n} a_i v_i^*=0, \end{align*} where $0:V\to k$ is the zero linear functional, then every coefficient $a_i$ must be zero. The useful feature of the [dual basis](/theorems/414) is that each $v_i^*$ detects exactly one basis vector. Fix $j\in\{1,\ldots,n\}$ and evaluate the assumed relation at the vector $v_j\in V$. Since the left side is the zero functional, its value at $v_j$ is $0$. Since addition and scalar multiplication in $V^*$ are defined pointwise, we get \begin{align*} 0=\left(\sum_{i=1}^{n} a_i v_i^*\right)(v_j)=\sum_{i=1}^{n} a_i v_i^*(v_j). \end{align*} By the defining property of the dual basis, $v_i^*(v_j)=\delta_{ij}$. Therefore the sum reduces to the single coefficient $a_j$: \begin{align*} \sum_{i=1}^{n} a_i v_i^*(v_j)=\sum_{i=1}^{n} a_i\delta_{ij}=a_j. \end{align*} Thus $a_j=0$. Since $j$ was arbitrary, every coefficient is zero, so the dual family is linearly independent. [/guided] [/step] [step:Show every linear functional is a linear combination of the dual basis] Let $f\in V^*$ be arbitrary. Define the linear functional $g:V\to k$ by \begin{align*} g=\sum_{i=1}^{n} f(v_i)v_i^*. \end{align*} For each $j\in\{1,\ldots,n\}$, we compute \begin{align*} g(v_j)=\sum_{i=1}^{n} f(v_i)v_i^*(v_j)=\sum_{i=1}^{n} f(v_i)\delta_{ij}=f(v_j). \end{align*} Thus $f$ and $g$ agree on every vector in the basis $\mathcal{B}$. If $v\in V$ has coordinate expansion \begin{align*} v=\sum_{j=1}^{n} c_j(v)v_j, \end{align*} then linearity of $f$ and $g$ gives \begin{align*} f(v)=\sum_{j=1}^{n} c_j(v)f(v_j)=\sum_{j=1}^{n} c_j(v)g(v_j)=g(v). \end{align*} Therefore $f=g$, so \begin{align*} f=\sum_{i=1}^{n} f(v_i)v_i^*. \end{align*} Since $f\in V^*$ was arbitrary, $\mathcal{B}^*$ spans $V^*$. [/step] [step:Count the dual basis elements to obtain the dimension equality] The family $\mathcal{B}^*=(v_1^*,\ldots,v_n^*)$ is linearly independent and spans $V^*$, so it is an ordered basis of $V^*$. Therefore $V^*$ is finite-dimensional and \begin{align*} \dim_k V^*=n. \end{align*} Since $\mathcal{B}=(v_1,\ldots,v_n)$ is a basis of $V$, we also have \begin{align*} \dim_k V=n. \end{align*} Hence \begin{align*} \dim_k V^*=\dim_k V. \end{align*} When $n=0$, the same argument says that the empty dual family is linearly independent and spans $V^*$, so the zero-dimensional case is included. This proves both the basis assertion and the dimension formula. [/step]