# differential inverse matrix

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 $f'(A)\cdot V=-A\cdot V\cdot A^{-1}$.

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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 $f'(A)\cdot H=-A^{-1}HA^{—1}$.

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I can't see how the $j=1$ term cancels, but I think this is because the result should be $f'(A).H=-A^{-1}HA^{-1}$: think about the case $n=1$! This form for $f'$ yields the required cancellation at $j=1$. – user12477 Sep 22 '12 at 20:50
@user12477 You are perfectly right. I've edited. – Davide Giraudo Sep 22 '12 at 20:52