Showing $A_{ij} = (Ae_i, e_j)$ for matrix $A$ of complex linear operator $\mathbb{C}^n \to \mathbb{C}^n$ and orthonormal basis $(e_i)_{i=1}^{n}$ I'm sure this is a simple question, but I get stuck in the algebra when I try to prove it from definition.
Suppose $A$ is an $n \times n$ matrix representing a complex linear operator $\mathbb{C}^n \to \mathbb{C}^n$, suppose $(e_i)_{i=1}^{n}$ is an orthonormal basis of $\mathbb{C}^n$ and suppose we equip $\mathbb{C}^n$ with the standard complex inner product, i.e. $(x, y) = \sum\limits_{i=1}^{n} x_i \overline{y_i}$.
I want to show the simple property (stated in my quantum mechanics lectures) $A_{ij} = (Ae_i, e_j)$ for all $i, j$, however after expanding the inner product and expanding $Ae_i$ I can't see how to finish the algebra. So far I have:
$(Ae_i, e_j) = \sum\limits_{k=1}^n (Ae_i)_k \overline{e_j}_k = \sum\limits_{k=1}^n (\sum\limits_{l=1}^n A_{kl} e_{i_{l}}) \overline{e_j}_k$
From here on any way I try to continue, particularly trying to separate entries of $A$ and elements in the basis $e_i$, seems to hit a dead end of messy algebra. Any hints/help greatly appreciated, thanks.
 A: First, let 
$$
a = Ae_i = 
\begin{bmatrix} A_{11} & \ldots & A_{1n} \\
A_{21}  &\ldots &A_{2n} \\
\vdots &\ddots &\vdots \\
A_{n1} & \ldots & A_{nn}
\end{bmatrix}
\begin{bmatrix} 0   \\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}
$$
$e_i$ has a $1$ at the $i$th row and $0$ everywhere else. Hence, on multiplying, it will "pick out" $A_{i1}, A_{i2}, \ldots A_{in}$
Hence,
$$
a = Ae_i = \begin{bmatrix}A_{i1} \\ A_{i2} \\ \vdots \\ A_{in} \end{bmatrix}
$$
Next,
$$
e_j = \begin{bmatrix} 0   \\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}
$$
The vector has a $1$ in the $jth$ row and $0$ everywhere else.
Hence, 
$$
\bar{e_j} = \begin{bmatrix} 0 & \ldots & 1 & \ldots & 0 & \end{bmatrix}
$$
So that
$$
(Ae_i, e_j) = Ae_i \bar{e_j} = \begin{bmatrix}A_{i1} \\ A_{i2} \\ \vdots \\ A_{in} \end{bmatrix} \times \begin{bmatrix} 0 & \ldots & 1 & \ldots & 0 & \end{bmatrix} = A_{ij}
$$
A: Note: It should be $A_{ij} = \langle A e_j, e_i \rangle$, not $\langle A e_i, e_j \rangle$.
Observation 1: The $i$-th component of the transform $Av$ of any column vector $v = [v_1 \quad v_2 \quad \cdots \quad v_n]'$ is $(Av)_i = {\large\sum\limits_k} A_{ik}v_k$.
Observation 2: The $i$-th component of the column vector representation of the $j$-th basis vector $e_j$, with respect to the same basis, is $0$ for all $i \ne j$ and $1$ for $i = j$.
From these two observations,
\begin{equation*}
(Ae_j)_i = A_{ij}.
\end{equation*}
Observation 3: For any vector $v = {\large\sum\limits_j} v_j e_j$, the $i$-th component can be written as an inner product (by using bilinearity of inner product and orthonormality of basis vectors)
\begin{equation*}
\langle v, e_i\rangle = \left\langle \sum_j v_j, e_j \right\rangle = \sum_j v_j \langle e_i, e_j \rangle = v_i.
\end{equation*}
Combining this with the earlier result,
\begin{equation*}
\large\boxed{
\langle Ae_j, e_i \rangle = A_{ij}
}.
\end{equation*}
