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Given $V$ a finite dimensional vector space, let $R=\{f \in \operatorname{End}(V) : I+f \text{ is invertible}\}.$ We define the Cayley Transform $T$ such that $T: R \to \operatorname{End}(V)$ and $T(f) = (1-f)(1+f)^{-1}$. I am asked to show that the Cayley Transform is differentiable and to find its derivative. I've already showed that the Cayley Transform is an involution but I don't know how to use this to show that its differentiable.

I am trying to use the definition of differentiable but all I get is that $T(f+h) = (I-(f+h))(I+(f+h))^{-1}$ and I don't know what to do next

Bernard
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McNuggets666
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Don't validate this as an answer : it's just the copy of p. 128 of "Geometric Numerical Integration" E. Hairer C. Lubich G. Wanner, where the enlarged context is Cayley transform between a Lie group and its Lie algebra.

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Jean Marie
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    One must add that this derivative has much in common with the differential of ordinary real valued function $f$ defined by $f(x)=\dfrac{1+x}{1-x}$ which is $df=2\frac{dx}{(1-x)^2}=2(1-x)^{-1}dx(1-x)^{-1}$. – Jean Marie Jan 04 '23 at 18:02
  • An interesting paper here about the application of Cayley Transform to random orthogonal matrices. – Jean Marie May 05 '24 at 12:02