Consider a finite-dimensional vector space $V$, which has some inner product $\langle \cdot \mid \cdot \rangle$. Let $\mathcal{B} = \{ \alpha_1, \dots, \alpha_n \}$ be a basis for $V$ (not necessarily, orthonormal). If $c_1, \dots, c_n$ are any scalars, show that there exists exactly one $\alpha$ such that $\langle \alpha \mid \alpha_i \rangle = c_i$.
I have seen the standard proof using Gram-Schmidt process. I am wondering if there is a more elegant way to do the proof.