Prove that $(I-T)(I+T)^{-1}$ is an involution

I have to prove that if $V$ is a finite-dimensional vector space over a field of characteristic not 2, and $T$ is an endomorphism such that $\det(I+T) \neq 0$ then $T \mapsto (I-T)(I+T)^{-1}$ is an involution on the space of endomorphisms such that $\det(I+T) \neq 0$.

This is part of an exercise in Gadea and Masqué workbook. It seem trivial as they don't bother detailing the answer, but I can't find it.

Thanks,

JD

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You need to prove two things: (a) that $f(T)$ still has the property that $I+f(T)$ is invertible. (b) that $f(f(T))=T$. Can you make any progress on either of them? –  Henning Makholm Sep 24 '11 at 8:39
You also need an assumption that the scalar field does not have characteristic 2; otherwise $(I-T)(I+T)^{-1}=I$ which is not an involution. –  Henning Makholm Sep 24 '11 at 8:43
@HenningMakholm: But if the field has characteristic 2 then $I+T=I-T$ and $(I-T)(I+T)^{-1}=(I+T)(I+T)^{-1}=I$. And what is the function $f$ in this case? It is like this? : $X:=\{T : \det(I+T)\neq 0\}$ and $f : X \to X , f(T)=(I-T)(I+T)^{-1}$? –  Beni Bogosel Sep 24 '11 at 8:48
@Beni, yes, silly typo on my part. I managed to fix it within the edit window. –  Henning Makholm Sep 24 '11 at 8:49
@Henning: Within 13 seconds?! –  joriki Sep 24 '11 at 9:00

Let $K$ be the ground field and $X$ an indeterminate. It is straightforward to check that the formula $$\begin{pmatrix}a&b\\ c&d\end{pmatrix}X:=\frac{aX+b}{cX+d}$$ defines an action by $K$-automorphisms of the group $\text{GL}_2(K)$ of two by two invertible matrices with coefficients in $K$ on the field $K(X)$, and that the scalar matrices in $\text{GL}_2(K)$ act trivially.

The answer follows immediately from this observation.

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Hint: If char $\neq2$, then the formula $$\frac{1-\frac{1-T}{1+T}}{1+\frac{1-T}{1+T}}=\frac{(1+T)-(1-T)}{(1+T)+(1-T)}=\frac{2T}{2}=T$$ comes in handy. There are a number of details for you to check to make sure that all this algebra makes sense in the ring of endomorphisms.

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Instead of the full endomorphism ring, it may be best to interpret the above relation in the commutative ring $R=F[T,1/(1+T)]$. –  Jyrki Lahtonen Sep 24 '11 at 9:32

Compute $$[1+(1-T)(1+T)^{-1}]\frac{1+T}{2}$$ You should find $2I$. This means $f(T)$ is invertible (following the notation of Henning in the comment above) and its inverse is $\frac{1+T}{2}$. Next compute $$f(f(t))=[1-(1-T)(1+T)^{-1}][1+(1-T)(1+T)^{-1}]^{-1}=[1-(1-T)(1+T)^{-1}]\frac{1+T}{2}=T$$ and we are done.

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