# Show that the inner product exists and is unique

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.

The inner product on $V$ allows us to identify $V$ and the dual space $V^*$. Thus, we are searching for a linear form $\varphi$ such that $\varphi(\alpha_i)=c_i$. This form clearly exists and is unique, since any linear form is uniquely determined by its values on some basis. Now, use the identification of $V$ and $V^*$ to obtain a vector $\alpha$, which is related to $\varphi$ by $\langle \alpha, x \rangle = \varphi(x)$.
Your problem is equivalent to finding $\alpha=\sum_{j=1}^{n}\rho_j \alpha_j$ such that $$\sum_{j=1}^{n}\rho_j \langle \alpha_j,\alpha_i\rangle = c_i.$$ Being able to do this for all $\{ c_i \}$ is equivalent to the invertibility of the covariance matrix $[\langle \alpha_j,\alpha_i\rangle]$, which is equivalent to knowing that the covariance matrix has a trivial null space. The assertion of a trivial null space is easier to check because $\sum_{j=1}^{n}\rho_j\langle \alpha_j,\alpha_i\rangle =0$ for all $i$ implies $\langle \sum_{j=1}^{n}\rho_j \alpha_j,\alpha_i\rangle=0$ for all $i$, which gives $\langle \sum_{j=1}^{n}\rho_j\alpha_j,\sum_{i=1}^{n}\rho_i\alpha_i\rangle=0$ and, hence $\sum_{j=1}^{n}\rho_j\alpha_j=0$, which implies that $\rho_j=0$ for all $j$ because $\{ \alpha_j \}$ is a basis.