Finding the expression of $u^{-1}$ Suppose that $u=\lambda id-s$, as $s$ is a linear map as ($s\circ s = id)$ and $\lambda \in \mathbb{R} \setminus\{-1,1\}$.
How can prove that $u$ is invertible and find the expression of $u^{-1}$ (the inverse of u)?
 A: To approach this and related problems it can make sense to start with formal manipulations (the validity of which is established at the end, or possibly this can  even be circumvented). 
To simplify notation I set $c = \lambda^{-1}$ and $v= c u   = id - cs$. 
You like know that the inverse of $1 - q = \sum_{n=0}^{\infty} q^n$ (when this converges). So, let us be "optimistic" and start a formal calculation 
$$(id - cs)^{-1} = \sum_{n=0}^{\infty} (cs)^n.$$ 
It is unclear if this converges (it does not in general) but we are brave and continue
$$\sum_{n=0}^{\infty} (cs)^n = \sum_{n=0}^{\infty} (cs)^{2n}  + \sum_{n=0}^{\infty} (cs)^{2n+1} = \sum_{n=0}^{\infty} (c^2)^n id  + cs \sum_{n=0}^{\infty} (c^2)^n,$$
here we split into the odd and the even numbers and used that $s^2$ is the identity. 
Now, $\sum_{n=0}^{\infty} (c^2)^n = (1 - c^2)^{-1}$ (again formally) and so we get $(id - cs)^{-1} = (id + cs)/(1 - c^2)$. 
Changing back the notation we have (based on a formal calculation to be justified) 
$$u^{-1} = c v^{-1}= \lambda^{-1}  \frac{id + \lambda^{-1}s}{1 - \lambda^{-2}}=  - \frac{\lambda  id + s}{1 - \lambda^{2}}.$$
Now, we could either set up a framework where the above can be justified in a rigorous way, or we say the above was just our private heuristic and we now believe that 
$$u^{-1} =    - \frac{\lambda  id + s}{1 - \lambda^{2}}.$$
To justify this it suffices to note that $- \frac{\lambda  id + s}{1 - \lambda^{2}}$ makes sense (it is just some linear operator divide by a nonzero constant so that's fine) and to verify 
$$u (- \frac{\lambda  id + s}{1 - \lambda^{2}}) = -\frac{\lambda  id + s}{1 - \lambda^{2}} u = id,$$ 
which is just a direct calculation. 

Alternatively, you could have "guessed" that the inverse will be of the form $x \ id + y s$ and then determined $x,y$ by expanding and comparing coefficients. 
A: $u$ is a linear map, and as such is invertible if and only if it has a trivial kernel. Take a look at $x\in \ker(u)$, so $u(x) = 0$. Since $u(x) = \lambda x - s(x)$, this means that $$s(x) = \lambda x$$
But, from $s(x) = \lambda x$, it follows that $s(s(x))= s(\lambda x)$, meaning that (since $s$ is linear) $$s(s(x)) = s(\lambda x)\\ x = \lambda s(x)$$
Now, join $s(x) = \lambda x$ with $x=\lambda s(x)$ and you get that $s(x) = \lambda (\lambda s(x)),$ so either $x=0$ or $\lambda^2=1$. Since $\lambda ^2\neq 1$, it follows that $x=0$.
