Application of the Frechet derivative [closed]

Question

$f\colon U\subset \mathbb{R}^{n}\longrightarrow\mathbb{R}^{m}$ 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

Answer 1

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 $$ DF(A)H=-A^{-1}HA^{-1}. $$