Self inverse and self adjoint implies existence of an orthogonal projection.

Question

Let $R \in B(\textbf{H})$. Prove that the following are equivalent.
(a) The operator $R$ is self inverse and self adjoint: $R^{2} = I$ and $R^{*} = R$.
(b) There exists an orthogonal projection $P \in B(\textbf{H})$ such that $R + I = 2P$.

This is actually part of a bigger question involving more equivalent statements, but I'm having particular trouble with showing that (a) $\implies$ (b) which is what I need here. Part of the trouble is that the orthogonal projection defined in my lecture notes always comes accompanied with a subscript, indicating the closed subspace the projection is attributed to. In this question this doesn't seem to be the case. I'm also not entirely sure what $2P$ is supposed to mean, does this mean projecting twice or twice the projection? If it's projecting twice then isn't this just equal to $P$?

Answer 1

Denote $P=\frac12 (R+I)$, you have $$P^2=\left(\frac{R+I}2\right)^2=\frac14(R^2+I^2+2R)=\frac12(R+I)=P$$ and $P^*=P$ can also be calculated. This algebraic fact implies that $P$ is an orthogonal projection onto the orthogonal complement of its kernel. To see it:

Let $x$ lie in the orthogonal complement of $\ker(P)$. It is clear that $Px-x$ lies in the kernel of $P$ (from $P^2=P$), but since $x$ and $Px$ are both in the complement and $\ker(P)^\perp\cap\ker(P)=\{0\}$ you must have $Px=x$. It follows that $P$ acts as identity on the closed subspace $\ker(P)^\perp$ and as $0$ on $\ker(P)$. So $P$ is a projection onto $\ker(P)^\perp$.