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.
 A: 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 $\langle \alpha, x \rangle = \varphi(x)$.
A: 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.
