For a finite-dimensional vector space $V$. I am wondering how to prove that if $$\varphi_1, \varphi_2, \ldots, \varphi_n$$ is a basis of $V'$, then there exists a basis of $V$ $$v_1, v_2, \ldots, v_n$$ such that $\varphi_j v_k = \delta _{j,k}$.
I realize that the process should involve some normalization, maybe something like the Gram-Schmidt procedure. But I haven't read to the chapter about inner product space. So I am not very familiar with relevant concepts yet. Can someone provide a proof with relatively elementary methods (those that appear in the first 3 chapters of Linear Algebra Done Right).
As Mark indicated in the comments, since $V$ is finite-dimensional there is a natural isomorphism $V\to V''$ which sends each vector $v\in V$ to the linear functional on $V'$ given by $$\varphi\mapsto\varphi(v)$$
This allows us to naturally identify $V$ and $V''$, and shows that the duality relationship between $V$ and $V'$ is fundamentally symmetrical. The result you're trying to prove then follows from the fact that every basis has a dual basis.