#math/linear-algebra If $v_{1}, \ldots{}, v_{n}$ is a basis of $V$, then the `dual basis` of $v_{1}, \ldots{}, v_{n}$ is the list $\varphi_{1}, \ldots{}, \varphi_{n}$ of elements of $V'$, (elements of $\mathcal{L}(V,\mathbf{F})$), where each $\varphi _{j}$ is the linear functional on $V$ such that $ \varphi _{j}(v_{k})=\begin{cases} 1 & \text{if }k=j \\ 0 & \text{if }k\neq{}j. \end{cases} $ And, with the [[3A3 - Linear Maps and Basis of Domain Theorem]] applied to each function, $\varphi _{j}$ is well-defined. Then for every $v\in V$, $v$ can be written as $\varphi_{1}(v)v_{1}+\ldots+\varphi _{n}(v)v_{n}$; the coefficients, (not all necessarily the value 1) come from the $\varphi _{i}(v)'s$. Now we can connect the dual basis as the basis of the dual space. Suppose $V$ is finite-dimensional. Then the dual basis of a basis of $V$ is a basis of $V'$. **Proof:** Suppose $a_{1}\varphi_{1}+\ldots+a_{n}\varphi _{n}=0$ (remember the right-hand side is the 0 function). We will now hand pick vectors, (in this case each vector from the basis $v_{1}, \ldots{}, v_{n}$) to show that $a_{1}=\ldots=a_{n}=0.$ Now, observe $(a_{1}\varphi_{1}+\ldots+a_{n}\varphi _{n})(v_{k})=a_{k}\varphi _{k}(v_{k})=a_{k}\cdot 1=a_{k}=0$ for $k=1,\ldots,n$.