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Given a continuous $f : GL_n(\mathbb{R}) \to GL_n(\mathbb{R}) : A \mapsto A^{-1}$, I want to show that for $g \in GL_n(\mathbb{R})$ the derivative $Df(g)$ equals $$ H \mapsto -g^{-1}Hg^{-1} $$ for $H \in Mat(n, \mathbb{R})$.

I tried proving $$ \lim_{X \to A} \frac{||f(X) - f(A) - Df(A)(X-A)||}{||X - A||} = \lim_{X \to A} \frac{||X^{-1} - A^{-1} + A^{-1}(X - A)A||}{||X - A||} = 0, $$ (with $g = A$) but did not succeed.

The syllabus suggests I use the chain-rule, but I don't see how.

Math Student 020
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1 Answers1

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In case of $n=1$, we deal with real numbers instead of matrices and it is about the $x\mapsto 1/x$ function, which has derivative $f'(x)=-1/x^2$. So, we would expect $g^{-2}$ in some arrangement for the matrix case, as ShuchangZhang commented, i.e. you probably meant $Df(g)=H\mapsto -g^{-1}Hg^{-1}$.

For that end, check that $\,X^{-1}-A^{-1} = X^{-1}(A-X)A^{-1} $, $\,$ so we have $$X^{-1}-A^{-1} \,+\,A^{-1}(X-A)A^{-1}\ =\ \dots \ =\ (X^{-1}-A^{-1})\,(A-X)\,A^{-1} \,.$$

Berci
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