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Consider the inversion function $f:GL_n( \mathbb{R}) \rightarrow GL_n (\mathbb{R})$ , $f(X)=X^{-1}.$ Where $GL_n( \mathbb{R})$ denotes the set of invertible $ n \times n$ matrices over the reals.

The question wants me to show that it is a differentiable function and then to calculate its derivative. It says to think of the set as a subset of $\mathbb{R} ^{n^{2}}$.

I know that if the partials exist and are continuous then it is differentiable, I can't calculate the partials explicitly though since it seems too difficult, just thinking about it I know if I were to change 1 entry in the matrix keeping all others constant (this is how I interpret partial derivative of this function, is this correct?), I could find a neighbourhood around that entry such that the matrix is still invertible (since $det:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ is continuous? - this has been shown in my lecture notes) Is this the correct way to go about it? I have no solutions available to me so just seeking some clarification on here to make sure my understanding isn't completely all wrong, thanks!

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$Gl(n,\mathbb{R})$ is an open subspace of th vector space $M(n,\mathbb{R})$, the inverse $X\rightarrow X^{-1}$ is a rational function of its coordinates (expressed with the cofactor matrices) so it is differentiable.

You have $(X+h)^{-1}=X^{-1}(I+hX^{-1})^{-1}$ write $hX^{-1}=u$ with $\|u\|<1$, you obtain that $(I+u)^{-1}=\sum(-1)^nu^n$, this implies that t$(X+h)^{-1}=X^{-1}-X^{-1}hX^{-1}+O(h)$ and the differential is $h\rightarrow X^{-1}hX^{-1}$.

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  • $\begingroup$ Thanks for your answer! So I'm not sure what a rational function is, don't think we've actually defined that in class. I see how you've manipulated the increments to get what you got, and surely from what you end up with, the derivative is $h \rightarrow -X^{-1}hX^{-1}$ ? (minus sign) $\endgroup$ – Displayname Mar 7 at 10:11
  • $\begingroup$ Also, how did you know how to go about this question i.e use that power series representation of $(I+u)^{-1}$? Is it just experience in dealing with these types of problems? $\endgroup$ – Displayname Mar 7 at 10:13

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