I need to find the well-know method of Schulz : $$X_{n+1} = X_{n}(2I - AX_{n})$$
The initial problem is to find the root of the function $f(X) = X^{-1} - A =0$ which is the inverse of A.
The Newton iteration is given by : $$X_{k+1} = X_{k} - [f'(X_{k})]^{-1}f(X_{k})$$
They tell me that the derivative f'(X) is equivalent to $g(X) = X^{-1}$ with g a linear operator defined as $$E \mapsto -X^{-1}EX^{-1}$$ In other words, the application of $g'(X)$ to a matrix E gives us the matrix $-X^{-1}EX^{-1}$.
So, from $X_{k+1} = X_{k} - [f'(X_{k})]^{-1}f(X_{k})$ we can rewrite it as $$X_{k} - [f'(X_{k})]^{-1}(X_{k}^{-1} -A)$$
but here is $f'(X_{k})$ applied to the matrix $(X_{k}^{-1} -A)$ ? What to do with the $f^{-1}$ ? Thanks in advance.
As you've correctly noted, for each $X$, $f'_X$ is a linear operator on matrices such that for all $Y$ $$f'_X[Y]=-X^{-1}YX^{-1}\text{.}$$ Therefore ${f'_X}^{-1}$ is also such a linear operator. It must satisfy $${f'_X}^{-1}[-X^{-1}YX^{-1}]=Y\text{.}$$ Therefore $${f'_X}^{-1}[Y]=-XYX\text{.}$$