Advanced Linear An involution is a linear transformation $φ ∈ L(V, V )$ such that $φ^2 = Id_V$ . Show that the equation $φ = 2π−Id_V$ establishes a one-to-one correspondence between all projections $π$ and all involutions $φ$.
 A: If 
$\pi:V \to V \tag 1$
is a projection, then
$\pi^2 = \pi; \tag 2$
if we set
$\phi = 2\pi - I, \tag 3$
we have
$\phi^2 = (2\pi - I)(2\pi - I) = 4\pi^2 - 4\pi + I = 4\pi - 4\pi + I = I, \tag 4$
so $\phi$ is an involution on $V$.  
Likewise, if $\phi$ is an involution and (3) holds for some $\pi \in L(V, V)$, then
$4\pi^2 - 4\pi + I = (2\pi - I)(2\pi - I) = \phi^2 = I, \tag 5$
whence
$4\pi^2 - 4\pi = 0 \Longrightarrow 4\pi^2 = 4\pi \Longrightarrow \pi^2 = \pi, \tag 6$
so $\pi$ is a projection.
For every projection $\pi$, (3) defines and involution, and for every involution, (3) defines a projection, viz.
$\pi = \dfrac{1}{2}(\phi + I); \tag 7$
indeed this gives us
$\pi^2 = \dfrac{1}{4}(\phi + I)(\phi + I) = \dfrac{1}{4}(\phi^2 + 2\phi + I)$
$= \dfrac{1}{4}(I + 2\phi + I) = \dfrac{1}{4}(2\phi + 2I) = \dfrac{1}{2}(\phi + I) = \pi. \tag 8$
The mappings (3) and (7) are each clearly one-to-one, for
$\phi = 2\pi_1 - I = 2\pi_2 - I \Longrightarrow 2\pi_1 = 2\pi_2 \Longrightarrow \pi_1 = \pi_2, \tag 9$
and
$\pi = \dfrac{1}{2}(\phi_1 + I) = \dfrac{1}{2}(\phi_2 + I) \Longrightarrow \phi_1 + I = \phi_2 + I \Longrightarrow \phi_1 = \phi_2. \tag{10}$
