Let $V$ be an $n$-dimensional vector space and let $(v_1, \dots, v_n)$ denote any oriented basis for $V$. Also, let $g$ be an inner product on $V$ and let $G$ denote the Gram matrix of inner products $G = [g(v_i, v_j)]$. I am trying to show that if $v_j = A^k_je_k$, where $e_k$ denotes a basis that is orthonormal with respect to g, then $\det{(A^i_j)} = \sqrt{G}$.
I believe I have found a useful intermediate result, but I'm not really sure how to close the deal. For vectors $v_i$ and $v_j$ we have:
$ g(v_i, v_j) = g(A^k_i e_k, A^r_j e_r) = A^k_iA^r_j \delta_{kr} = \sum\limits_{m=1}^n A^m_iA^m_j = \langle A_i | A_j\rangle $
where $A_k$ denotes the $k^{th}$ column of $A$ and $\langle\cdot | \cdot\rangle$ denotes the standard Euclidean inner product. Therefore, the matrix $G$ is given by
$ G = [\langle A_i|A_j \rangle] $
At this point, I'm not sure what to do. I'm thinking there's some essential fact I need to know in order to continue.
So, my question is, am I on the right track and if so what should my next step be?
Edit: I updated this question to change the assumption that the $e_i$ are the standard basis vectors to the assumption that the $e_i$ are orthonormal with respect to $g$