4
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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 $f'(A)\cdot H=-A^{-1}HA^{—1}$.

$\endgroup$
  • $\begingroup$ 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 $j=1$. $\endgroup$ – user12477 Sep 22 '12 at 20:50
  • $\begingroup$ @user12477 You are perfectly right. I've edited. $\endgroup$ – Davide Giraudo Sep 22 '12 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.