Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $f'(A)\cdot H=-A^{-1}HA^{—1}$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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