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
 A: 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. 
A: 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. 
A: 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.
