TITLE: differential inverse matrix
QUESTION [7 upvotes]: How to show that the application $$f:U=GL(\mathbb{R}^{2})\subset\mathbb{R}^{n^{2}}\longrightarrow \mathbb{R}^{n^{2}},$$ defined by $$f(A)=A^{-1}$$ is differentiable and and its derivative at point $A\in U$ is the linear transformation $$f'(A):\mathbb{R}^{n^{2}}\longrightarrow \mathbb{R}^{n^{2}},$$ defined by $f'(A)\cdot V=-A\cdot V\cdot A^{-1}$.

REPLY [7 votes]: Fix $N$ a sub-multiplicative norm on $\Bbb R^{n^2}$. We have 
$$f(A+H)-f(A)=(A+H)^{-1}-A^{-1}=(A(I+A^{—1}H))^{—1}-A^{—1}=\left((I+A^{—1}H)^{—1}-I\right)A^{—1}.$$
This gives, for $N(H)<\frac 1{2N(A^{—1})}$,
\begin{align}
f(A+H)-f(A)+A^{-1}HA^{—1}&=\left((I+A^{—1}H)^{—1}-I+A^{—1}H\right)A^{-1}\\
&=\left(\sum_{j=0}^{+\infty}(-1)^j(A^{—1}H)^j-I+A^{—1}H\right)A^{—1}\\
&=\sum_{j=2}^{+\infty}(-1)^j(A^{—1}H)^jA^{-1},
\end{align}
hence
\begin{align}
N(f(A+H)-f(A)+A^{—1}HA^{—1})&\leq \sum_{j=2}^{+\infty}N(A^{-1})^jN(H)^jN(A^{—1})\\
&=N(H)^2\sum_{k\geq 0}N(A^{—1})^{k+3}N(H)^k\\
&\leq N(H)^2N(A^{—1})^3\frac 1{1-1/2}\\
&=2N(H)^2N(A^{—1})^3.
\end{align}
This proves that $f'(A)\cdot H=-A^{-1}HA^{—1}$.