$f\colon U\subset \mathbb{R}^{n}\longrightarrow\mathbb{R}^{m}$ is differentiable at $x\_{0}$ if there exist a linear transformation $T\colon \mathbb{R}^{n}\longrightarrow\mathbb{R}^{m}$, such that:


\begin{equation}
 f(x\_{0}+h)=f(x\_{0})+T\cdot h +r(h) \quad\text{donde}\quad \lim\_{h\to 0}\frac{r(h)}{\Vert h\Vert}=0
\end{equation}


under this definition, I am trying to solve the following exercise.


Let $GL(\mathbb{R}^{n})=\{T\in \mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m}): T\quad \text{is invertible}\}$ and $Inv\colon GL(\mathbb{R}^{n})\longrightarrow GL(\mathbb{R}^{n})$ defined by $Inv(T)=T^{-1}$. Show that $Inv$ is differentiable and find $Inv'(T\_0)$.


Starting from the definition


\begin{align}
 Inv(T\_0+H)-Inv(T\_0)&=(T\_0+H)^{-1}-T\_0^{-1}\\
 &=(T\_0+H)^{-1}(I-(T\_0+H)T\_0^{-1})\\
 &=(T\_{0}+H)^{-1}(I-T\_0T\_{0}^{-1}+HT\_{0}^{-1})\\
&=-(T\_0+H)^{-1}HT\_{0}^{-1}
\end{align}


However, I cannot identify who the candidate for $Inv'(T\_0)\cdot H$ and $r(H)$.


I also know that another way to calculate $Inv'(T\_0)\cdot H$ is


\begin{equation}
 Inv'(T\_0)\cdot H=\lim\_{t\to 0}\frac{Inv(T\_{0}+tH)-Inv(T\_0)}{t}=\frac{d}{dt}(Inv(T\_{0}+tH))\big{|}\_{t=0}
\end{equation}
But I can't find it.


Any guidance is greatly appreciated



Application of the Frechet derivative
If $F:M^{n\times n}\to M^{n\times n}$ is differentiable, then for $A\in M^{n\times n}$, $DF(A)$ is a linear map from $M^{n\times n}$ to $M^{n\times n}$ and we denote by $DF(A)H$ value of this linear map at $H\in M^{n\times n}$.


If $F,G:M^{n\times n}\to M^{n\times n}$ are differentiable maps, then the product of the maps $F\cdot G:M^{n\times n}\to M^{n\times n}$ satisfies the product rule which is easy to prove:
$$
D(F\cdot G)(A)H=(DF(A)H)\cdot G(A) + F(A)\cdot (DG(A)H).
$$
Let $F:GL(\mathbb{R}^n)\to GL(\mathbb{R}^n)$, $F(A)=A^{-1}$. The formula for $A^{-1}$ shows that the entries of $F$ are rational functions and hence $F\in C^\infty$ (entries are $\pm$ minor determinant over matrix determinant and determinants are polynomials in coefficients).


Let $G(A)=A$. Since $G$ is the identity map, $DG(A)=Id$ i.e., $DG(A)H=H$. We have
$(F\cdot G)(A)=A^{-1}\cdot A=I$, so $F\cdot G$ is a constant map and hence $D(F\cdot G)=0$. Now. the profuct rule gives
$$
0=D(F\cdot G)(A)H=(DF(A)H)G(A)+F(A)(DG(A)H),
$$
$$
0=(DF(A)H)A+A^{-1}H
$$
and we get the formula for the derivative of the inverse map
$$
DF(A)H=-A^{-1}HA^{-1}.
$$


