I am trying to prove a matrix/operator identity but I can't and I was wondering if somebody could help me with it.
The identity is(for two generally non-commuting matrices/operators):
$\frac{1}{A-B}=\frac{1}{A}+\frac{1}{A}B\frac{1}{A}+\frac{1}{A}B\frac{1}{A}B\frac{1}{A}+...$.
In particular, my problem lies in trying to perform a Taylor expanion:
$\frac{1}{A-B}=\frac{1}{A}(\frac{1}{1-\frac{1}{A}B})$ which gives
$=\frac{1}{A}+\frac{1}{A^2}B+\frac{1}{A^2}B\frac{1}{A}B+\frac{1}{A^2}B\frac{1}{A}B\frac{1}{A}B+...$
I tried doing it like this:
$\frac{1}{A-B}=(\frac{1}{1-B\frac{1}{A}})\frac{1}{A}$ but it still given non-sense.
So, I am guessing that the right form to bring it in is $\frac{1}{A-B}=\frac{1}{A}(\frac{1}{1-B\frac{1}{A}})$ but I don't understand why this is.
Note: Please, don't be too rigorous as I am only an undergraduate physicist. Also, the expanion is found in scattering problems for weak potentials.
Your mistake was in the step where you factored out $A^{-1}$ before applying geometric series. Note that if $X$ and $Y$ are invertible operators, then $(XY)^{-1}=Y^{-1}X^{-1}$. Hence $(A-B)^{-1}=(A(1-A^{-1}B))^{-1}=(1-A^{-1}B)^{-1}A^{-1}$. Hence, provided $\sum\limits_{k=0}^{\infty}(A^{-1}B)^k$ converges, the identity holds.
Opinion/speculation: Using the notation $X^{-1}$ instead of $\frac1X$ when working with noncommuting variables makes it easier to keep things straight.