Matrix representation in a non-orthonormal basis Suppose I have a matrix representation of an operator in the standard basis. If I change the basis to one with basis vectors $\{u_1, u_2, u_3\}$, which are not orthonormal, how do I now get a matrix representation in this basis?
Specifically, I have
$$A = \sum_{ij} A_{ij}e_ie_j^*,$$
where $e_i^*$ is the transpose conjugate of $e_i$. If I now express the $e_i$ as linear combinations of $u_i$, I get
$$A = \sum_{ij}A'_{ij}u_{i}u^*_{j}$$
How are the matrices with entries $A_{ij}$ and $A'_{ij}$ related? Is there an invertible change-of-basis matrix (the one that goes from the standard basis to the $u_i$ basis) that relates the two?
EDIT: Just to clarify what I mean, an example of $u_i$ is
$$u_1 = \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}, u_2 = \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}, u_3 = \begin{pmatrix}0 \\ 1 \\ i\end{pmatrix}$$
and
$$u^*_1 = \begin{pmatrix}1 & 0 & 1\end{pmatrix}, u^*_2 = \begin{pmatrix}1 & 1 & 0\end{pmatrix}, u^*_3 = \begin{pmatrix}0 & 1 & -i\end{pmatrix}$$
Sorry if my notation was not correct in the first version of this question.
 A: Let $U$ denote the matrix whose columns are $u_1,u_2,u_3$. The matrix $A'$ with entries $A_{ij}'$ is given by
$$
A' = U^{-1} A(U^{-1})^*.
$$
$U$ can be regarded as a change of basis matrix for a bilinear form.

Edit: The indices in the question and in my answer are correct. With block-matrix multiplication, note that
$$
A = \sum_{ij}A_{ij}' u_iu_j^* = 
\pmatrix{u_1 & u_2 & u_3} 
\pmatrix{A_{11}' & A_{12}' & A_{13}'\\ A_{21}' & A_{22}' & A_{23}' \\ A_{31}' & A_{32}' & A_{33}'}\pmatrix{u_1^* \\ u_2^* \\ u_3^*}\\
= U A' U^*. 
$$
A: There is some matrix $C$ such that $u_i = Ce_i$ (namely, a matrix whose columns are the coordinate vectors $u_1, u_2, u_3$). I'll write $c_{ij}$ for the entries of this matrix. That gives
$$
u_i = \sum_j c_{ji}e_j.
$$
(this was previously swapped...sigh.) For what comes later, I'm going to shift an index and write
$$
u_i = \sum_k c_{ki}e_k.
$$
Note that
$$
u_j^{*} = e_j^{*} C^{*}.
$$
where $C^{*}$ denotes the conjugate-transpose of $C$, whose $ij$-entry is $\overline{c_{ji}}$; in coordinates, this means that
$$
u_j^{*} = \sum_\ell \overline{c}_{j\ell} e_{\ell}^{*}.
$$
Now let's substitute for the $u$s in the expression:
\begin{align}
A 
&= \sum_{ij}A'_{ij}u_{i}u^*_{j} \\
&= \sum_{ij}A'_{ij}\left(c_{ki}e_k\right)\left(\overline{c}_{j \ell} e_\ell^{*})\right) \\
&= \sum_{ijk\ell }A'_{ij}c_{ki}  \overline{c_{j\ell}} e_k e_\ell^{*} \\
&= \sum_{ijk\ell }c_{ki}A'_{ij} \overline{c_{j \ell }} e_i e_j^{*} 
\end{align}
But we have another expression for $A$ already, and equating coefficients of $e_i e_j^{*}$ gives
$$
A_{ij} = \sum_{k\ell} c_{ki} A'_{ij} \overline{c_{j \ell }}
$$
In matrix terms, this says that
$$
A = C A' \overline{C}
$$
if I've got the indexes right.
