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.
Best Answer
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 $$ A = \sum_{ij}A_{ij}' u_iu_j^* = \pmatrix{u_1 & u_2 & u_3} \pmatrix{A_{11}' & A_{12}' & A_{13}'\\ A_{21}' & A_{22}' & A_{23}' \\ A_{31}' & A_{32}' & A_{33}'}\pmatrix{u_1^* \\ u_2^* \\ u_3^*}\\ = U A' U^*. $$