[proofplan] Write the vector $v$ in its unique coordinate expansion with respect to the ordered basis $\mathcal B$. Then apply each [dual basis](/theorems/414) functional $e_i^*$ to this expansion. The defining Kronecker-delta relation for the dual basis shows that $e_i^*(v)$ is exactly the $i$th coordinate of $v$, and substituting these recovered coordinates back gives the claimed formula. [/proofplan] [step:Expand $v$ in the ordered basis $\mathcal B$] Fix $v\in V$. Since $\mathcal B=(e_1,\ldots,e_n)$ is a basis of $V$, there exist unique scalars $a_1,\ldots,a_n\in k$ such that \begin{align*} v=\sum_{j=1}^n a_j e_j. \end{align*} [/step] [step:Recover each coordinate by applying the corresponding dual basis functional] Fix $i\in\{1,\ldots,n\}$. Since $e_i^*:V\to k$ is a linear functional, applying $e_i^*$ to the basis expansion gives \begin{align*} e_i^*(v)=e_i^*\left(\sum_{j=1}^n a_j e_j\right). \end{align*} By linearity of $e_i^*$, \begin{align*} e_i^*(v)=\sum_{j=1}^n a_j e_i^*(e_j). \end{align*} Using the defining property $e_i^*(e_j)=\delta_{ij}$ of the dual basis, we obtain \begin{align*} e_i^*(v)=\sum_{j=1}^n a_j\delta_{ij}=a_i. \end{align*} [guided] Fix $i\in\{1,\ldots,n\}$. The goal is to identify the scalar multiplying $e_i$ in the basis expansion of $v$. The dual basis functional $e_i^*:V\to k$ is designed precisely to extract that coefficient: it takes value $1$ on $e_i$ and value $0$ on every other basis vector $e_j$. Starting from the expansion \begin{align*} v=\sum_{j=1}^n a_j e_j, \end{align*} we apply the [linear map](/page/Linear%20Map) $e_i^*:V\to k$ to both sides. This is valid because both sides are elements of $V$, and $e_i^*$ has domain $V$. Thus \begin{align*} e_i^*(v)=e_i^*\left(\sum_{j=1}^n a_j e_j\right). \end{align*} By linearity of $e_i^*$, scalar multiplication and finite sums pass through the functional: \begin{align*} e_i^*(v)=\sum_{j=1}^n a_j e_i^*(e_j). \end{align*} The defining property of the dual basis is \begin{align*} e_i^*(e_j)=\delta_{ij}, \end{align*} where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i\ne j$. Therefore every term in the sum vanishes except the term with $j=i$, and the remaining term is $a_i$. Hence \begin{align*} e_i^*(v)=\sum_{j=1}^n a_j\delta_{ij}=a_i. \end{align*} So $e_i^*(v)$ is exactly the $i$th coordinate of $v$ in the basis $\mathcal B$. [/guided] [/step] [step:Substitute the recovered coordinates into the basis expansion] The previous step proves that $a_i=e_i^*(v)$ for every $i\in\{1,\ldots,n\}$. Substituting these equalities into \begin{align*} v=\sum_{i=1}^n a_i e_i \end{align*} gives \begin{align*} v=\sum_{i=1}^n e_i^*(v)e_i. \end{align*} Since $v\in V$ was arbitrary, the formula holds for every $v\in V$. [/step]