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Consider the continuous image $f: GL_{n}(\Bbb R) \to GL_n \Bbb (R): A \mapsto A^{-1}$

I'm trying to proof with induction that $f$ is infinitely differentiable. I now understand how I can proof that $f$ is one time differentiable. I found that $Df(G) = H \mapsto -G^{-1}HG^{-1}$.

But I have a hard time figuring out how to find a good induction step. I tried figuring out what the derivative is of $Df$. But I don't really see what I need to do. I think I need to show that $Df(G+H)= X \mapsto -(G+H)^{-1}X(G+H)^{-1}$ is equal to $Df(G) + D^2(G)(H)+\epsilon(H)$.

Kasper
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    For a cheap way out you can use Cramer's rule, which tells you that each entry of $f(A)$ is a rational function of entries of $A$ with nonvanishing denominator. Therefore each entry of $f(A)$ is an infinitely differentiable function. – Giuseppe Negro Oct 10 '13 at 16:04
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    You know that $A^{-1}=\dfrac{1}{\det A}{\operatorname{adj} A}$. Both numerator and denominator are polynomial expressions in the coefficients of $A$. – Pedro Oct 10 '13 at 16:05
  • I'm trying to show this with induction. – Kasper Oct 12 '13 at 23:13

2 Answers2

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Hint

With a little work you can show $$D_{2}f(A) = H_{2} \to (H_{1} \to G^{-1}H_{1}G^{-1}H_{2}G^{-1} + G^{-1}H_{2}G^{-1}H_{1}G^{-1})$$

Using this you can find a general form of the $k$th derirative and prove the $k+1$th derirative fits this form.

Kasper
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Heiko Ubbink
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2

Hint

$(A+B)^{-1}=A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}BA^{-1}+\cdots$

Lennart
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