To calculate the inverse of a quadratic matrix A, we could solve the equation $F(X):=X^{-1}-A=0$.
I need to show that if X is invertable, then $DF(X)(\Delta X)=-X^{-1}\Delta XX^{-1}$ where DF(X) is the Frechét derivative of F.
My idea would have been to use the standard recursive formula $x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}$, then transform to $f'$ and plug in the matrices but that didn't really work out. What could I do?
With matrix functions it is better to compute the Newton step $S$ in $X_{+1}=X+S$ as the solution of the linear system resulting from the tangent root equation $$ 0=F(X)+DF(X)[S]=X^{-1}-A-X^{-1}SX^{-1}. $$ Otherwise you would need to translate everything into $n^2$ long column vectors where then the derivative is a $n^2\times n^2$ matrix.